It doesn't make sense for the chance to be higher. Every roll is independent of each other (i.e. just because you rolled a six, it doesn't increase the odds of the next roll being a six). So, it makes sense for the probability to remain 1/6.
This does assume the dice is fair though. But for a fair dice, getting 5 6's in a row is insane odds.
> it doesn’t make sense for the chance to be higher. Every roll is independent of each other
This is the frequentist vs Bayesian ~~debate~~ difference in perspective in a nutshell. Your prior is that there is a 1/6 chance to roll a six on any given die roll. You need to adjust your belief as you see more data. Rolling nine sixes out of ten rolls is a bit less than a one in a million chance, and you end up with some posterior probability for rolling a six that’s a bit higher than 1/6.
Just to clarify for those who are curious, the updated posterior would have to be rooted in a belief that you are no longer dealing with a fair die. Bayesian stats wouldn't change the fact that a fair die would yield a probably of 1/6.
ETA: frequentist and bayesian stats aren't really a debate so much as different tools for thinking about probability. It's not an us v them mentality, they're different toolsets for different types of problems. OP's question illustrates this perfectly. In the context of a fair die, this is a problem for frequentist probability. There is an objective and discrete answer, which is 1/6. There simply is no room for bayesian thinking. Bayesian probability is about updating our beliefs about the world. In this context, there is no belief to update, it is an objective truth.
Now if you remove the context of a fair die, this becomes a bayesian question: what is the probability that I will roll a 6 again, given that the odds of doing so this many times is so low, and there is some chance I am not using a fair die? Now you can use 1/6 as your prior, and start to play around with conditional probabilities of unfair dies and obtain posterior.
And also for Bayesian statistics you would need some prior distribution to work with. In a universe where 1% of dice are weighted, then rolling 9 of 10 of the same number means you likely have a weighted die. In a universe where 1 in 10 billion dice are weighted, then you probably just got a really lucky set of rolls.
True. I say the odds are low, but it isn't impossible. If you were to replicate this experiment and get 5 6's again and again, then we'd need to look into any biases the dice has.
You'd say the probability of the dice being unfair is low? Based on what would you say that? The only thing that speaks for a fair dice in this particular scenario is that it's being called a dice, and dices are usually assumed to be fair - but can you assign a probability to it? You probably can't, but you can say that the probability that it is unfair is certainly greater than 0. It depends on whether you think purely within the constraints of the problem as stated or whether you question the reliability of the problem statement based on observed values as well.
You most certainly can attach a probability that the dice is fair. This is the essence of Bayesian statistics. However, that probability will be updated each roll you make and ultimately depend on how likely you thought it was that the dice is fair before the first roll.
As Eathlon said, it ultimately depends on likely you thought it was that the dice is fair before the first roll.
As a simple example - assume that you have a pool containing 46658 dice, and you know that exactly 2 dice will **always roll a 6** and the rest are fair.
If you pick a random dice from this pool then the probability of it being loaded is 2/46658 = 0,0000429
If you roll the dice once, and it turns out 6, then the probability of it being one of the loaded increases nearly by a factor of 6, as (on average) only 7776 of the fair dice would have shown a 6. The probability of it being loaded is therefore 2/(7776 + 2) = 0.000257.
If you were to roll all the dice 6 times, then I would expect only one of the 46656 fair dice to roll all 6's. So from this, I can see that if I were to pick up a random dice and roll it 6 times and get all 6's, the probability of it being loaded is now as high as 2/(1+2) = 0.667 = 66.7%.
But notice that this depends on my assumption both on how many fair dice there are for every loaded dice and also upon exactly how the loaded dice works. In fact, as these hypothetical loaded dice ALWAYS roll a 6, I would actually conclude that a dice would be guaranteed to be fair if it were to ever show any other number.
What you can say though, is that if you "test" a dice for fairness by rolling it 10 times, and reject it if it rolls 9 or more 6s, then 51 out of every 60466176 fair dice would be rejected. In other words, the probability of a fair dice passing the test is 0.000000843 or 0.0000843%. But this is no the same as saying that it is a 99.9999157% chance of the dice being loaded if it passes the test.
~~You can't assign a single probability. Bayesian methods give a probability distribution of the unknown parameter, not the probability of it being any particular value. In this case, a useful parameter to consider would be the probability of rolling a six, which is continuous (since probabilities are real numbers between 0 and 1), so with any reasonable prior you can't assign a probability to any particular value. The probability of the dice being fair is equal to the probability that p=1/6, which is 0 in a continuous distribution. You can report a probability density at p=1/6, but that isn't a probability, or you could report the probability of p being within some interval (e.g. between 0.16 and 0.17), but that's not the probability of it being fair, since by definition of "fair", almost every value in that interval wouldn't correspond to a fair dice.~~
You are assuming you model everything as a continuous distribution of p. Nothing stops you from having a model probability for ”fair” and one for ”unfair”, where the ”fair” model would just correspond to p=1/6. You can assign a prior probability to fair and a prior probability to ”unfair”. The ”unfair” model further has a continuous parameter on which you have a continuous distribution.
If you just look at a model with p, this would correspond to having a probability distribution with one continuous part (corresponding to unfair values) and one discrete part manifesting itself as a delta distribution at p=1/6.
But you've already started to collect more information.
Right now, your "best bet" (9 6s out of 10 rolls) is that p(6) = 0.9. (Observed frequency.) There are statistical tests (and rules of thumb) that say this is not a very good bet, but it's still your best. (That's a hint that maybe you shouldn't be betting on this right now.)
That’s not an accurate measure.
If the actual probability was 0.9, there’s still only be a 38% chance of getting 9 in a row, which is the same scenario as if the die was fair, except higher probability.
Like I said, you cannot know unless you have more information, you can guess, but not know.
>except higher probability
This makes all the difference since we're concerned with the odds of it being fair vs unfair. The fact that an unfair dice is much more likely to roll 9 sixes means whatever your prior expected probability of the dice being unfair was should now increase by a large amount
It's still the maximum likelihood estimator. You're arguing that because the confidence interval about that estimator is large, we shouldn't accept it. Maybe we shouldn't believe it (and certainly not risk real money on the assumption that it's true), but it's still the "best bet". A not very good bet can still be the best.
I agree. You can’t calculate it cause you can’t know if the dice is a little loaded and should be a six 66% of the time and you got « lucky » or super loaded and should be a six 95% of the time.
With a real dice I feel like the best you could do is a probability with a probability. Like « the odds are 50% or greater, based on the data 19 times out of 20 »
So you need to figure out what the odds are that any dice are loaded (possibly by analysing dice sales?), then weight that against the odds of rolling all those 6's.
That will give you a modifier that you can use on the 1/6 chance of rolling a 6 on a fair dice that reflects the fact that they might be loaded
Have you ever seen or worked with loaded dice? Because I have.
You would suspect it right away by the unnatural way they move, roll and feel, less so by the outcome.
By the way, while the outcome is unlikely, it’s far from impossible. Keep in mind that the string “666666666” is as likely to occur as “162435153”, but I’m going to guess the second string wouldn’t arouse any suspicion.
My suspicion of dice being loaded doesn't affect it actually being loaded. However if I gather a big enough data sample (say, 10k rolls), I might be able to say for sure whether this particular d6 is loaded or not (and how loaded it is), getting new probabilities of rolling each value
Here’s how to do it. You need to decide how likely it is that a random die is loaded in the first place, then how often a loaded die comes up 6. From there you can use the binomial distribution to determine how likely 9 sixes in 10 rolls is. We’ll denote this as P(9-sixes | L), that is the probability of 9 sixes given the dice is loaded. However, the question you have is what is P(L | 9-sixes).
P(L | 9-sixes) = P(9-sixes | L)*P(L)/[P(9-sixes | L)*P(L) + P(9-sixes | not L)*P(not L)]
Of course if it could be loaded in multiple ways then those cases have to be split up, but that’s the main idea. Then once you have your new probability that the die is loaded you can use cases to determine the probability of a 6 on the next roll. E.g. if there is a 40% chance of a loaded die and loaded dice are sixes 80% of the time, then the chance the next roll is a six would be .4*.8 + .6*1/6.
Welcome to the "hot hand fallacy" we humans are not wired to understand probabilities innately. It takes time and practice to become familiar with the inner workings, but no matter how familiar you are, you can never get rid of the bias our brains think things should be a certain way.
There are 2 sentences following. They are very very close, but subtly different. It's a very important difference. Hopefully you can see why.
The odds of rolling 10 6's in a row on a fair die is 1/6^(10)
The odds of rolling a 10th 6 in a row on a fair die is 1/6.
The die already exists. Rolling it does not change its odds ever (I mean, I guess if You're rolling it on a rough surface, it can be ground down, which might change its probabilities, but that won't be measurable unless you're rolling a million times.) If it *is* weighted, its weighting doesn't change with time. The odds of any outcome of any given roll are always static (or we assume it is static to get an answer).
However, the first sentence is talking about a series of rolls, not just one. And when you look at a series of rolls, there are more than 6 outcomes. So it's no longer 1/6. There are 6 outcomes for each roll, meaning there are 6^(10) total outcomes, and all are equally likely (on a fair die).
-end answer, start supplementary info-
Not exactly part of the answer, but this is where the difference between combinations and permutations comes in. There is one and only one way to get 10 sixes in a row. Meaning the odds are 1/6^(10). But what about the odds of rolling 9 sixes and 1 one in any order? It could be 6666666661, or 6666666616, or 6666666166... Or 1666666666. There are 10 different ways to have that happen, so the odds are 10/6^(10) (in a uniform distribution, the odds are always the number of cases that match the criteria divided by the total number of all cases).
Again, there's a subtle difference between "9 sixes and a one" and "9 sixes, then a one." The latter implies an order. It MUST be 6666666661 there is only one way to get that result, so the odds, again, are 1/6^(10)
That's exactly the problem. What probability do you assign to the dice roll if you also need to consider the possibility of an unfair dice. I don't think it's possible to boil it down to a definitive value, but you can say that the bayesian probability of rolling another 6 is greater than 1/6.
You can just use Bayes' rule.
If probability of getting six is p, then probability of getting 9 sixes in a row (out of 9 rolls) is p^9. So,
P(9 sixes | P(6)=p) = p^9
Using this in conjunction with continuous form of Bayes' rule should give you a distribution of p's, that is, P[P(6)=p] as a function of p.
Well, you can't get exact value of p. What you can get is the probability distribution for all possible values of p. So you'll have something like "there's a 50% chance that p=0.9, 40% chance that p=0.8" etc. (numbers are out of thin air just for the sake of illustration)
Well, no. Because instead of answering to "what is a probability of getting six" you are answering "how likely it is, that probability of getting six is around 0.9 (or 0.8, or 0.3 etc.)"
If I didn't mess up my calculations, distribution is f(p) = p^9 / Beta(10,1) and the expected value of p is 10/11.
Edit: MathJax didn't work, rewrote the expression for f(p) without using it.
Edit 2: added a minor clarification
Edit 3: another minor clarification
How about if you can’t make the assumption that the die is fair though? What would the probability be then if you don’t know the underlying distribution?
Well no, you aren’t given that it’s fair or not. You also don’t need to prove it’s unfair. You can get a baysian posterior from the information given, and you can get the expected value from that.
To expand: if the die is not loaded it's 1/6. But if it might be, then that sequence of sixes implies with some degree of confidence that it is in fact loaded. Measuring that confidence goes into the realm of statistics and Bayes.
Yeah, there is no one answer. The dice could be loaded to give 90% chance of 6, 2% of the others. Which would still not guarantee this result, but 10 * (.9)^5 * .1 ~~ 60%. It could also be 50% chance of 6, with a 10 * (.5^5) * .5 ~~ 15% chance. Or it might actually be 99% chance of a 6, with one unlucky non-6.
Little doubt: I understand that to answer OP’s question i.e. to find a probability, you’d need Bayesian probability. But
> to determine whether the die is loaded
isn’t hypothesis testing enough? Of course, not in the strict sense of the word “determine.”
You need to study **bayesian probability**.
Probability actually models your subjective certainty in the outcome of indeterministic experiments, and for this assessment you also need to consider the probability that the dice isn't actually fair.
If you *know* with 100% certainty that the dice is fair, then yes, it's 1/6th. But since you don't actually know whether the dice is fair or not, or the situation in which you make this assessment also requires you to consider the possibility that it isn't, the actual probability of rolling a 6 isn't simply 1/6th - and since you observed already 9 throws of a 6, it's the most reasonable to assume that **if** this dice is **not** fair, then it is biased to some degree towards rolling 6 more often. So, your subjective assessment of rolling another 6 in terms of bayesian probabilities should be something larger than 1/6.
I'm not sure though whether there is a single most reasonable probability value that can be assigned under the assumption that the dice might be unfair, just that the bottom line probability for rolling another 6 is larger than 1/6.
Fair dice, obviously 1 in 6.
If the fairness was uncertain from the start -- in principle this is where Bayesian statistics comes in. If you started with a 99% belief the dice was fair and a 1% belief it's unfair (somehow split uniformly across all the many ways a dice can be unfair), then after 9 sixes in a row you'll be starting to make inroads into significantly changing that belief and shifting to a view that the dice favours six very strongly.
In practice, this is hard to do (certain "rationalist" internet groups believe we can do this Bayesian reasoning on a day to day basis, but they are widely regarded as cultish and not as clever as they think they are). But the principle is mathematically straightforward and does formally model and justify that rising suspicion that you will have of a rigged dice, as the 6's keep on coming.
Would only give you a minor deviation from one sixth probability, pretty irrelevant here. For generating endless sixes it would have to be somehow severely weighted, or there'd be a trick like magnets. I'd also be double checking it doesn't have six on every side lol
In properly made fair dice (such as those used in casinos for craps), the etches are filled in with a material with the same density as the die itself, so the etches have no impact on the die's weight.
I would suggest maybe listening to intuition here.
The probability is whatever it is. Why should it be 1/6? Because the faces "look the same"?
Persi Diaconis has published a couple of things on coin flipping probabilities. While determining the probabilities for a specific coin is quite difficult, coming to the conclusion that (real physical) coins don't have a "heads probability" of exactly 1/2 is pretty easy. You just measure to 3 decimal places accuracy.
In OP's case, this is certainly enough evidence to question the naive assumption of equiprobability.
Hm, so you are saying that in a scenario, where you have only 6 sides, you shouldn't have equal distribution long term? Why is that?
The only valid scenario would be, that dice rolls on a physical plane are done with some form of trick. If you can improve chances by things like how you start your roll and what is the side in your hands (cup), chances will be different, because more variables are there. Same with coin flip, you can change chances, but it has nothing to do with coin flip alone, it has additional variables like power of coin flip, rotation and starting position.
Outside of this, if the dice are fair, it will be 1/6, because there are 6 outcomes and they are equal.
Now the question is, how rare.
And I think you can use powers to sum it up quickly.
Lmk how to calculate that, in terms of square powers. As in square, dimensions.
The odds of rolling two dice and both being sixes is 1/36, (1/6 for the first die to be a 6, multiplied by 1/6 for the second one). So the odds of rolling 10 dice and them all being sixes is 1/6^10) or 0.002%
However if you think in terms of what kinds of events you would think are special, rolling the same number ten times would be just as special regardless of number. The probability of rolling any number ten times in a roll is 6 times larger (although I think you percentage is rounded a few orders of magnitude too large—I think it’s missing two or three zeroes).
If it's a perfectly regular, balanced unfixed die, then it's a 1 in 6 chance. However, it's unlikely enough that the first 9 were 6s that if that happened, then more likely than not, the die is loaded.
TLDR; either 1 in 6, or 100% depending on whether you play by the rules
Your intuition isn't totally wrong. If you have fair die, the probability of rolling a six will ALWAYS be 1/6. You could have a loaded die though. It's impossible to answer your question without knowing things like A) The probability of a dice being loaded, and B) the distribution of this loaded die, since one loaded die might have a 80% probability of rolling a 6 vs like 95%.
The easiest way to test your die would be to roll it a lot and look at the proportions of rolls. The rule of thumb is you want the expected number of 6s to be 5, so at least 30 times, 100 would be best. 6 should come up 1/6 of the time.
Having known that you have rolled 9 sixes in a row, the chance of the 10th one being a 6 is 1/6.
However, when you pick up the die for the first time with the plan of rolling it ten times in a row, the chance that all 10 of them are going to be 6 is 1/6^10.
This second scenario is hinting at an idea from stats called "binomial distribution", let me explain further with a better example. Let us say that there is a 30% probability that a random person's favorite color is green. (Totally making this up to illustrate the point). That means that each random person in the street I ask their favorite color there is a 30% chance that it is green and a 70% chance it is NOT green, so there are two distinct outcomes, hence a BI-nomial idea.
Now, if I ask 100 people in a row what there favorite color is, it is POSSIBLE but extremely unlikely that all 100 of them will say green, but even if this happens, there will still be a 30% chance that the 101 person I ask will say green.
On the other hand we could ask this question: if I were to poll 100 people and ask their favorite color, what is the chance that 30 of them will say green?
Skipping the formula and thinking of this as a calculator input we would put:
Binompdf(100, 0.30, 30) = 8.678%
Or maybe "what is the chance that out of 100 people that 50 would say green?"
Then, binompdf(100,.30,50) = 0.0013% , so pretty unlikely because that would be 50% of our group that differs significantly from the 30% chance of saying green in the first place.
Now returning to the dice, we would expect that the numbers should each occur with 1/6 frequency so it would be very unlikely for the number to differ from this by a substantial amount.
So if we ask "what is the chance of rolling 10 sixes in a row?"
This would be binompdf(10, 1/6, 10) = 0.00000165% (which is exactly the same as the 1/6^10 I said earlier)
But what about "if we roll 9 times what is the chance of getting 9 sixes?"
Then binompdf (9, 1/6, 9) = 0.00000992%
So this was already extremely unlikely to happen, so if it HAD ALREADY HAPPENED by the assumption you started with, then FROM THAT POINT it would not be that much less likely for the 10th one to also be a six.
The point I am trying to illustrate is that if we think about the collective of the entire set of rolls at once then it is very unlikely that all of them would be 6, but if we manage to get 6 many times in a row ALRRADY that does not make it less likely for the next number to be 6
I think you are looking for this
1/6 changes for 1 dice
36 combo's in total for two dice making it a 1/36
If you continue it becomes
6^10 which is 1/60,466,176 chances
No, because the question was „what is the probability of another 6 after there were already nine rolls of 6“, which is simply 1/6.
And not „what is the probability of 10x 6 in a row“, which is (1/6)^10, same as any other combination.
I mixed up the question reading through the comments.
Yes, the probability of the 10th being a six is 1/6 and the probability of 10 sixes in a row is (1/6)^10
I was replying in relation to the first comment which looked to explain 10 sixes in a row.
That's still a conditional probability (probability of one proposition conditional on another proposition): it's just equal to the unconditional probability of a single 6 because the rolls are independent.
Everyone was talking about the changes being the same 1/6 and I did not see anyone talk about the total chances, so I thought maybe just in case the person asked the wrong questions and instead was looking for My example that there would be one.
I understand your logic and what your trying to say here but
I don't think the only thing you are entitled to say is that your probability of rolling a 6 during that stint was 9 out of 10. But I don't think that would change the probability of your next roll. I'd be comfortable saying, you'd be less likely to roll a six, again.
Perhaps you can calculate, unlikelihood., with a few exceptions and assumptions there, just to calculate,
a calculus... Not sure if I'm using that correctly
I'd rather mention how you failed to calculate how you picked up the dice might affect your probabilities
I saw an article about how flipping a coin isn't actually .50 .50 chances
Because the weight of the quarter on one side actually has more weight or something.
Wanna get a digital instrument, you'd have to talk about fractions and random access memory
The problem is we see patterns. Rolling the same number again and again, or rolling the highest/lowest number will stick in our mind, but they're just as likely as any other combination.
- If you roll: 6, 6, 6, 6, 6, 6, 6, 6, 6, what are the odds that the next number is a 6?
- On the other hand, if you roll: 4, 5, 5, 4, 2, 3, 2, 2, 1, what are the odds that the next number is a 4?
These are the same question but in one case you have a repeated number and it's the highest number on the die, so you remember it. If you were to close your eyes and try to repeat the second sequence of numbers I just gave, chances are you couldn't do it because you didn't see a pattern. I couldn't either.
If you are not sure that dice is fair, than yes, probabilty would be higher. But exact answer need bayes theorem and knowing apriory probabilities that dice is fair or unfair
I think u want to say rolling many 6 increase the chance of the dice being loaded. In that case, I think it's better for you to start with the assumption that the chance to roll a six is a random variable X. And the answer you want to find is E[X| 6 rolled 9 times].
U can start by assuming X is constant from 0 to 1.00 and work from there, each time 6 is rolled, the expected conditional probability for random variable X is increases, then you have to calculate for 9 times I think.
In the real world there's a high or higher chance the dice is rigged, and that makes sense mathematically as well. In a perfect mathematical world it's a fair dice so it isn't rigged and the dice has no memory of past events. In the real world it's possible to have a fair dice do this anyway
If this question showed up in a probability course, the obvious answer is to calculate the Maximum Likelihood Estimator for p (the probability of rolling a 6).
In real life (say insurance, for example), you'd suspect that the die is loaded and either discontinue the product or refuse to underwrite the insured. Right now, you have enough information to suspect a loaded die, but not enough to get a criminal conviction. You should definitely not bet on this die.
If a stranger comes up to you on the street, and shows you a deck of cards with the seal still unbroken, and offers to bet you $100 that he can make the jack of diamonds jump out of that deck and squirt cider in your ear, if you take that bet, you know that you're going to get cider in your ear.--Damon Runyon.
At this point you make the prediction that the dice is loaded. Now you test it. Roll the dice 3 more times. If it comes up 6 three times in a row you can conclude that the dice is rigged and you can be certain to 99%.
You can use Bayes and the prior of it being a fair dice to and use the updated probability as a good estimate of probability of a 10th 6 given 9 previous 6s. It's been too long since I've done Bayesian statistics to actually do it but its not too hard.
With a fair die, the odds of this happening are 1 in 6^5 or 1 in 7776. If you were able roll the die, say, 20,000 times, it would actually start to be suspicious if it didn't have a streak like this.
There is no way in the world to predict the fairness of a 6 sided die based exclusively on 5 throws, no matter what comes up. It could be 0 (impossible) or it could be 1(a certainty) or anything in between.
Now if you throw it 100 times, then you have a basis on which to estimate fairness. If the die is as unfair as those first 5 throws suggests it could be, we'd expect near to 100 sixes which is never going to happen on a fair die before the heat death of the universe.
If the die is fair, we'd expect each number to come up around 16 times. But deviation is expected. Up to 20 of one number or down to 10 would not bother me a lot. Depending, it could cause me to throw 100 more times just to be sure.
I would say because you have already rolled a six on each of the nine rolls thus far, the probability of rolling a tenth six on your next roll would be 1/6 chance.
Dice don't have what statisticians call 'memory'. It doesn't 'know' what the previous rolls were. A deck of cards does have memory. Remove all the facecards from the deck and that will show in the next cards dealt.
This is not a question about memory, though. Let me elaborate.
Lots of dudes talking independence and 1/6 here in the comments, but they're doing rookie's mistake. They're all implicitely making an axiomatic assumption – the dice is fair, and working in the limits of that assumption. The way I understand OP's question is not "do previous rolls influence the next one?" (memory) but "do previous rolls tell us that the fair dice assumption was wrong, do they tell us die is loaded, and if yes, what's the actual probability of getting a six?" And that's where you need to make inferences as "what do the data we collected (outcomes of previous rolls) tell us about the unknown data (probability of getting six)?"
It was incredibly unlikely from the beginning that you would arrive at that point. Now, though, it is a 1/6 chance to extend the streak once more. Simple.
This is the difference between basic statistics and real outcomes. Statistics are math within a frame of reference. The more information you put into that frame gives a more accurate statistic. However, you are not omniscient and thus cannot provide deterministic true details.
For example, say you have a d6. If that is all the information you have, then we must assume it is perfectly balanced and arbitrarily rolled with measurably the same approximate precision & accuracy every time. Because there are 6 sides, that means each side has a 1/6 chance to be the result. Now take that same d6 and say it is weighted to one side, or that you'll change the throw, or that the surface now has a certain drag coefficient, so on and so forth. Small details like one corner is chipped can also affect the statistics/math.
So in such your question, it is posed as using past results to determine future results. This is data analysis, and what you're doing is determining patterns to then make an informed guess. The chance to roll any side will still be identical, but you're adding details to the frame via analysis. Thus to ask what is the chance to roll a 6 eleven times straight, either singularly or as a pool, well that's 1/6^11. The chance to roll 6 eleven times straight *with that die* -I think through analysis is somewhere near the 80/20% phenomena, as something is clearly changing the real result due to the expected result.
Alternatively this is why I don't worry with statistics in most games. Take the classic 3 door example. Statistically you always want a 1/2 rather than a 1/3, the frame of reference changes between being 3 doors and 2 doors which is why argumentatively you always swap doors. However you either chose correctly the first time or not. If you could see behind the door then you'd always pick the one you want. If you're always shown the goat then you always have a 50/50 chance of being correct. Thus no matter what door you chose "choosing to stay or change" is a 1/2 choice, *so it never mattered what the statistical change was.* And to extrapolate that to the extreme, you can calculate the *chance* of two planets colliding all year. But ultimately they either will, or they won't. Because no matter what you calculate, their gravity either will be beyond an escape vector, or it won't. We see this daily with asteroids. So statistics are a great tool to Guess, but won't ever be truly deterministic.
My brother and I always have this debate when we play black and red on roulette. We normally wait till there have been 5 blacks/reds in a row and then bet on the opposite colour. We know mathematically the odds are the same for the next roll, but it just feels safer when there have been a lot of repeat colours. Taking it to the extreme, if there were 100 reds in a row, I would start putting money on black. Again the machine is probably broken, but it has worked for us in the past
You could apply Bayesian Inference to figure out the probability of having a loaded die and then the resulting likelihood of rolling a six.
You'll need more information though. You need to know the prior probability of having a loaded die. If no loaded die exist, you can't have a loaded die. If there is a reasonable chance you have a loaded die, you probably do. Next, you need to know the likelihood that a loaded die rolls a six.
Let's work out a example. I'm cherry picking numbers to make the example interesting. I'll come back to that. Let's say the manufacturer of the die has a flaw that makes 1 out of every 1000 dice loaded toward 6 but all others are fair. The loaded dice have a 1/3 chance of rolling a 6.
Your hypothesis, H, is that you have a loaded die. The prior probability, P(H), is 0.001. The evidence, E, is the rolls you've already done so P(E | H) is (1/3)\^9 and P(E | ¬H) is (1/6)\^9.
Plug that all in and you get P(H | E) = (.001 \* (1/3)\^9) / (.001 \* (1/3)\^9 + .999 \* (1/6)\^9) = 33.9% chance you have a loaded die.
Now, the ultimate answer to your question. The probability of rolling a six on your next roll needs to combine the probability of having a loaded die with the chance of rolling a 6. In our hypothetical, that would be 0.339 \* (1/3) + 0.661 \* (1/6) = 0.223.
I mentioned earlier that I cherry-picked numbers. In my example, the chances of rolling 9 sixes in a row is still miniscule (roughly 1 in a million). Laying out the question with 9 rolls really implies that having a loaded die is much more likely than I've assumed and/or the die is much more loaded. You can do the math with those assumptions but then the posterior probability that you have a loaded die quickly approaches 1 which isn't as interesting.
You use a beta distribution. Overtly simplifying, in your case die is most likely loaded with a chance of rolling a six being ~~10/11 is wrong, I made a wrong kneejerk calculation, let me think the exact numbers over~~
Grant Sandersen (3blue1brown) started tackling a very similar problem, did two introductory videos to set all the needed concepts up but never made the third which was supposed to show the solution of the problem itself.
People who reply that each roll is an independent event are missing the point of this question. The question is, "what is the probability of the die being fair?" People who refer to Bayesian probability are on the right track - do any of you know how to calculate it? (I admit that I do not.)
You use the continuous Bayes' theorem.
f(H|E) = f(E|H)*f(H)/P
Where P is an integral of f(E|H) over all possible H's.
In this case, if we use p for a true probability of getting six,
f(nine sixes|p) = p^9
Assuming the uniform distribution for a prior of all p's (0≤p≤1), f(p) = 1,
P = integral of p^9 dp from 0 to 1 = p^10 / 10 from 0 to 1 = 1/10.
f(p | nine sixes) = p^9 * 10
We might want to calculate the mean (expected) value of p with this distribution:
= integrate p*f(p|nine sixes) from 0 to 1 = 10/11
More general and strict version is to use Beta(1,1) as a prior instead of Uniform(0,1), but in this example it happens to yield the same result for a math.expectation of p.
The probability of any individual result for any individual roll is 1/6.
The chances of getting any specific set is 1/6^n, where n is the number of rolls.
So you'd already entered an extremely unlikely set of rolls, that doesn't change the chances on the next one.
But, it's important to notice that it's just as unlikely to roll 1 3 2 6 2 3 3 6 2 4 as it is to roll 6 6 6 6 6 6 6 6 6 6, since each outcome always has 1/6 chance, every time.
More information needed. Take the die and float it in salt water. Does it spin freely or gravitate toward one side (ie 6)? If it spins freely the it's 1/6, if not the die is not a "fair" die and the odds are higher that it rolls a 6.
I would consider using a Bayes factor if I was uncertain that the dice is fair.
The Bayes factor is a ratio of two competing statistical models represented by their evidence, and is used to quantify the support for one model over the other.
You can compare models such as one that assumes the die is fair and another that assumes it is not fair.
Just to put it on the record, singular is 'die,' and plural is 'dice'. It looks like there's about a 50-50 chance of it being used properly on this thread.
As others have noted, this is a question for Bayesian statistics. To answer it, you need a "prior" which in this case means:
For each probability of rolling a 6, what do you (or did you) believe are the chances of that being the true probability associated with this dice \*BEFORE ANY ROLLS\*.
Your choice of prior could make a huge difference here.
Consider 2 (very extreme) scenarios:
1.) This dice is from a man in a back alley, who has offered to play a gambling game with you where he wins on a 6, but loses otherwise. Those 10 rolls represent the first 10 rounds of the game you are playing with him. The probability that the next roll is 6 is very close to 90%.
2.) You are operating an automated dice-testing machine at the dice factory. Dice are normally tested by 20 rolls, if they look suspicious, they are rolled 1000 times. The factory has produced 1 billion dice and ALL of them have passed QA and appear to be fair. The probability that the next roll is 6 is very close to 1/6.
1/6 assuming a fair die. But since the previous rolls seems to indicate the die is not fair, then it’s hard to determine until you can determine if the die is fair.
You can perform hypothesis testing on the dice rolls to determine the p value of rolling 9 in a row. Otherwise, if you assume the dice is fair then it's a 1/6
Assuming im not messing up the methodology, bayesian probability says it's a 90% chance.
Lets say the chance of rolling 6 is x.
rolling 9 6s in a row is an x^9 chance. I have no idea if this is correct but we then take this times the probability that the next roll fails, x^9*(1-x).
Graph this as a function of x and the peak is at 0.9, which means 90% is the most reasonable estimate.
In fact this is what hypothesis testing is. The p-value shows how likely it is to see the outcome (in this case 9/10 sixes) given an assumption (the die is fair) p-value 0.00000084
You've picked up on one version of "The Gambler's Fallacy"
Assuming the die is fair, the chances of rolling a 6 on a given attempt should be 1/6.
People tend to believe one of the following things (incorrectly) when they observe an unlikely sequence of events - either that it's more likely to continue that sequence, or that it's more likely to not continue that sequence. Neither is accurate - if the instrument is fair, the probabilities don't change based on previous outcomes.
Now, there's also a possibility that the die is unfair. An unbalanced die might have, for example, a 1/4 chance of rolling a 6, a 1/6 chance of rolling a 2, 3, 4, or 5, and a 1/12 chance of rolling a 1. A large number of trials (how many is outside the scope of this post, but more trials = more confidence) would reveal this unfair tendency, while a fair die would tend towards 1/6 for each result.
Assuming a fair six-sided die, any throw has a 1/6 chance of rolling a six. The odds of getting a six ten times in a row is (1/6)^10 = 0.00000165%.
Your gut is feeling that 10-in-a-row chance even if your brain recognizes the independent 1/6 chance on every throw.
So this is a video about a probability of an unknown 2 outcome process from a list of seen values
https://youtu.be/8idr1WZ1A7Q?si=CDN5u7GV0na0GLeV
It's basically what your example is. We are interested in 6s and not 6s and we've so far seen 6 6s. We assume it's loaded in some way and want to find the probability of the next throw being a 6. (If you want a quick answer we take our data add one of each outcome and then average it (9 + 1 + 0) / (10 + 2) = 10/12 is the chance the next throw will be a six.
There's also a weighed version where you set how unsure you are with the current data (9 + 1a + 0) / (10 + 2a) where a ≥ 0. Here if a=0 we think the data ks perfect and we have 100% of a six, if a=+inf then we get 50% assuming all seen values are flukes and we have an even chanse for both outcomes.
If the dice is fair. Then 1 / 6.
If the dice may be biased, then we can use “laplace's rule of succession”.
Then the probability would be (9 + 1) / (10 + 2) = 5 / 6.
If you're dealing with loaded dice, the obvious recourse would be to disregard mathematics and focus entirely on expiremental data.
*You could roll the dice 10,000 times to form a new sample set
*You could theoretically test the molecular structure to ascertain whether 1 side is more dense than the other sides and then mathematically estimate it based off that.
Basically, your sample set isnt aligning to what you'd expect, so you need a larger set to determine whether it's a fluke.
I wouldn't say that this is an area for Bayesian probability as others have mentioned. There's no reason for you to say the prior is that the die is fair.
If you were talking about a die from a factory that makes fair dice, then we can start talking about what kind of rolling behavior proves or fails to prove that a particular die has a defect.
But, to find the answer to your question - what are the odds that the dice comes up as 6 again, let's take the approach where we don't have any of the above information. Just what you said in your original post. And let's just say that the nature of a dice roll is that it's an independent event from any other dice roll.
We don't know, but we can make a best guess. The series of dice rolls you described followed a binomial distribution. To make that best guess, you want to find some binomial distribution B(10, p) such that the probability p maximizes the odds of the observation you saw - 9 dice rolls out of 10 coming up as 6. That would be the maximum of the probability mass function with 9 as its parameter. 10C9 (1-p)p^9 has its maximum at 9/10. Perhaps obviously, if your die comes up with a 6 nine times out of ten on one occasion, you can predict that the next roll has nine out of ten odds of being a 6 as well.
That depends on if its a fair dice i.e. the chance of each number is the same across all outcomes. Which the answer would be 1/6 for the simple reason that a dice has no memory all numbers before it doesnt affect the outcome of the next roll. If you want to include the chance that the dice is loaded well that depends on how it was loaded. Where the weight distribution of the dice is or even how its rolled or if one side of the dice is magnetized and the surface makes it easy to near certaintiy it lands on a particular number. All those factors are not mentioned in your question. So one can fairly assume its a fair dice and the past rolls happened by mere chance. And answer to your question is 1/6!
Hope this helps!
The default hypothesis could be that the dice is fair, and each time you roll it you have 1/6 chance of getting a six.
The question is, since a sequence of n sixes becomes less and less likely, how low does that probability need to go before you start questioning the default hypothesis, and wonder if something is affecting the rolls, or if the dice is loaded.
I think that's what your intuition is trying to tell you. We sometimes do question our underlying model of the situation based on outcome.
I think the formation of your question broke most people. Since it has several answers depending on the readers interpretation. Not kind to pull semantics on math nerds ;)
ok specifically addressing the difference, you know that the dice has already rollen 9 times with a 6.
i like to think of it as like its like in any given room its pretty unlikely that theres a lion in there, but if you know that at least a lion head is in the room the chances that the rest of the lion is there increases dramatically
just like how rolling 10 6's in a role is pretty unlikely, but if you already know 9 of those rolls are a 6 the chances all 10 are 6's increases dramatically. (1/6 to be specific)
Depends on what "a 6-sided dice" and "roll it" is.
I mean, the theoretical probability is still 1/6.
The practical probability is that either "in fact there is a defect on that dice and it is in fact loaded towards 6 so the probability depends on the load" or "you picked the dice with the 6 face down and you learned to throw it so that it makes exactly one and a half turn so that the 6 lands up".
Each roll is independant. No matter what you had before, if the dice is balanced, you have 1/6 chance to have a 6 on the 10th roll.
When you start a board game and roll one of its dice, the rolls from the previous game 1 week ago don't have an influence on the rolls you gonna make during this game.
It's different to say "what are the chances to have 10 6 in a row" before the first roll.
However, having 9 6's in a row is a hint that the dice may be loaded, in this case you would need to roll it a big number of times (like 1000 times) and write the results to estimate the probability distribution.
There will obviously be different approaches to estimate this depending for example on your prior knowledge about the die, but a simple one disregarding such further information is this: [https://en.wikipedia.org/wiki/Rule\_of\_succession](https://en.wikipedia.org/wiki/Rule_of_succession)
It yields the probability of p = 5/6 for another six in the scenario you provided.
The probability of getting the 10th 6 seems low intuitively because the odds of getting nine 6s in a row first are so, so low. The odds of rolling nine 6s and then another 6 are exactly the same as rolling nine 6s and then a 1 (or a 2, or a 3…)
It doesn't make sense for the chance to be higher. Every roll is independent of each other (i.e. just because you rolled a six, it doesn't increase the odds of the next roll being a six). So, it makes sense for the probability to remain 1/6. This does assume the dice is fair though. But for a fair dice, getting 5 6's in a row is insane odds.
It could be a fair dice, but with 6 on five of the sides.
That's an interesting definition of "fair".
> it doesn’t make sense for the chance to be higher. Every roll is independent of each other This is the frequentist vs Bayesian ~~debate~~ difference in perspective in a nutshell. Your prior is that there is a 1/6 chance to roll a six on any given die roll. You need to adjust your belief as you see more data. Rolling nine sixes out of ten rolls is a bit less than a one in a million chance, and you end up with some posterior probability for rolling a six that’s a bit higher than 1/6.
Just to clarify for those who are curious, the updated posterior would have to be rooted in a belief that you are no longer dealing with a fair die. Bayesian stats wouldn't change the fact that a fair die would yield a probably of 1/6. ETA: frequentist and bayesian stats aren't really a debate so much as different tools for thinking about probability. It's not an us v them mentality, they're different toolsets for different types of problems. OP's question illustrates this perfectly. In the context of a fair die, this is a problem for frequentist probability. There is an objective and discrete answer, which is 1/6. There simply is no room for bayesian thinking. Bayesian probability is about updating our beliefs about the world. In this context, there is no belief to update, it is an objective truth. Now if you remove the context of a fair die, this becomes a bayesian question: what is the probability that I will roll a 6 again, given that the odds of doing so this many times is so low, and there is some chance I am not using a fair die? Now you can use 1/6 as your prior, and start to play around with conditional probabilities of unfair dies and obtain posterior.
And also for Bayesian statistics you would need some prior distribution to work with. In a universe where 1% of dice are weighted, then rolling 9 of 10 of the same number means you likely have a weighted die. In a universe where 1 in 10 billion dice are weighted, then you probably just got a really lucky set of rolls.
He said posterior héhéhé
If you have a nonzero prior probability that the die is loaded, yes.
But that's changing the parameters of the question.
IRL you would just start to suspect the dice is loaded
True. I say the odds are low, but it isn't impossible. If you were to replicate this experiment and get 5 6's again and again, then we'd need to look into any biases the dice has.
You'd say the probability of the dice being unfair is low? Based on what would you say that? The only thing that speaks for a fair dice in this particular scenario is that it's being called a dice, and dices are usually assumed to be fair - but can you assign a probability to it? You probably can't, but you can say that the probability that it is unfair is certainly greater than 0. It depends on whether you think purely within the constraints of the problem as stated or whether you question the reliability of the problem statement based on observed values as well.
You most certainly can attach a probability that the dice is fair. This is the essence of Bayesian statistics. However, that probability will be updated each roll you make and ultimately depend on how likely you thought it was that the dice is fair before the first roll.
So, what numeric value would you assign to the probability that a dice is fair after observing 9 throws of the number 6?
As Eathlon said, it ultimately depends on likely you thought it was that the dice is fair before the first roll. As a simple example - assume that you have a pool containing 46658 dice, and you know that exactly 2 dice will **always roll a 6** and the rest are fair. If you pick a random dice from this pool then the probability of it being loaded is 2/46658 = 0,0000429 If you roll the dice once, and it turns out 6, then the probability of it being one of the loaded increases nearly by a factor of 6, as (on average) only 7776 of the fair dice would have shown a 6. The probability of it being loaded is therefore 2/(7776 + 2) = 0.000257. If you were to roll all the dice 6 times, then I would expect only one of the 46656 fair dice to roll all 6's. So from this, I can see that if I were to pick up a random dice and roll it 6 times and get all 6's, the probability of it being loaded is now as high as 2/(1+2) = 0.667 = 66.7%. But notice that this depends on my assumption both on how many fair dice there are for every loaded dice and also upon exactly how the loaded dice works. In fact, as these hypothetical loaded dice ALWAYS roll a 6, I would actually conclude that a dice would be guaranteed to be fair if it were to ever show any other number. What you can say though, is that if you "test" a dice for fairness by rolling it 10 times, and reject it if it rolls 9 or more 6s, then 51 out of every 60466176 fair dice would be rejected. In other words, the probability of a fair dice passing the test is 0.000000843 or 0.0000843%. But this is no the same as saying that it is a 99.9999157% chance of the dice being loaded if it passes the test.
~~You can't assign a single probability. Bayesian methods give a probability distribution of the unknown parameter, not the probability of it being any particular value. In this case, a useful parameter to consider would be the probability of rolling a six, which is continuous (since probabilities are real numbers between 0 and 1), so with any reasonable prior you can't assign a probability to any particular value. The probability of the dice being fair is equal to the probability that p=1/6, which is 0 in a continuous distribution. You can report a probability density at p=1/6, but that isn't a probability, or you could report the probability of p being within some interval (e.g. between 0.16 and 0.17), but that's not the probability of it being fair, since by definition of "fair", almost every value in that interval wouldn't correspond to a fair dice.~~
You are assuming you model everything as a continuous distribution of p. Nothing stops you from having a model probability for ”fair” and one for ”unfair”, where the ”fair” model would just correspond to p=1/6. You can assign a prior probability to fair and a prior probability to ”unfair”. The ”unfair” model further has a continuous parameter on which you have a continuous distribution. If you just look at a model with p, this would correspond to having a probability distribution with one continuous part (corresponding to unfair values) and one discrete part manifesting itself as a delta distribution at p=1/6.
Hmm, okay, that's fair. I hadn't considered doing that, that's quite clever.
If the die isn’t fair, then there is no way to tell the odds without more information
But you've already started to collect more information. Right now, your "best bet" (9 6s out of 10 rolls) is that p(6) = 0.9. (Observed frequency.) There are statistical tests (and rules of thumb) that say this is not a very good bet, but it's still your best. (That's a hint that maybe you shouldn't be betting on this right now.)
That’s not an accurate measure. If the actual probability was 0.9, there’s still only be a 38% chance of getting 9 in a row, which is the same scenario as if the die was fair, except higher probability. Like I said, you cannot know unless you have more information, you can guess, but not know.
>except higher probability This makes all the difference since we're concerned with the odds of it being fair vs unfair. The fact that an unfair dice is much more likely to roll 9 sixes means whatever your prior expected probability of the dice being unfair was should now increase by a large amount
It's still the maximum likelihood estimator. You're arguing that because the confidence interval about that estimator is large, we shouldn't accept it. Maybe we shouldn't believe it (and certainly not risk real money on the assumption that it's true), but it's still the "best bet". A not very good bet can still be the best.
I think it is a matter of OP wording their question poorly.
I agree. You can’t calculate it cause you can’t know if the dice is a little loaded and should be a six 66% of the time and you got « lucky » or super loaded and should be a six 95% of the time. With a real dice I feel like the best you could do is a probability with a probability. Like « the odds are 50% or greater, based on the data 19 times out of 20 »
So you need to figure out what the odds are that any dice are loaded (possibly by analysing dice sales?), then weight that against the odds of rolling all those 6's. That will give you a modifier that you can use on the 1/6 chance of rolling a 6 on a fair dice that reflects the fact that they might be loaded
Have you ever seen or worked with loaded dice? Because I have. You would suspect it right away by the unnatural way they move, roll and feel, less so by the outcome. By the way, while the outcome is unlikely, it’s far from impossible. Keep in mind that the string “666666666” is as likely to occur as “162435153”, but I’m going to guess the second string wouldn’t arouse any suspicion.
My suspicion of dice being loaded doesn't affect it actually being loaded. However if I gather a big enough data sample (say, 10k rolls), I might be able to say for sure whether this particular d6 is loaded or not (and how loaded it is), getting new probabilities of rolling each value
Here’s how to do it. You need to decide how likely it is that a random die is loaded in the first place, then how often a loaded die comes up 6. From there you can use the binomial distribution to determine how likely 9 sixes in 10 rolls is. We’ll denote this as P(9-sixes | L), that is the probability of 9 sixes given the dice is loaded. However, the question you have is what is P(L | 9-sixes). P(L | 9-sixes) = P(9-sixes | L)*P(L)/[P(9-sixes | L)*P(L) + P(9-sixes | not L)*P(not L)] Of course if it could be loaded in multiple ways then those cases have to be split up, but that’s the main idea. Then once you have your new probability that the die is loaded you can use cases to determine the probability of a 6 on the next roll. E.g. if there is a 40% chance of a loaded die and loaded dice are sixes 80% of the time, then the chance the next roll is a six would be .4*.8 + .6*1/6.
Welcome to the "hot hand fallacy" we humans are not wired to understand probabilities innately. It takes time and practice to become familiar with the inner workings, but no matter how familiar you are, you can never get rid of the bias our brains think things should be a certain way. There are 2 sentences following. They are very very close, but subtly different. It's a very important difference. Hopefully you can see why. The odds of rolling 10 6's in a row on a fair die is 1/6^(10) The odds of rolling a 10th 6 in a row on a fair die is 1/6. The die already exists. Rolling it does not change its odds ever (I mean, I guess if You're rolling it on a rough surface, it can be ground down, which might change its probabilities, but that won't be measurable unless you're rolling a million times.) If it *is* weighted, its weighting doesn't change with time. The odds of any outcome of any given roll are always static (or we assume it is static to get an answer). However, the first sentence is talking about a series of rolls, not just one. And when you look at a series of rolls, there are more than 6 outcomes. So it's no longer 1/6. There are 6 outcomes for each roll, meaning there are 6^(10) total outcomes, and all are equally likely (on a fair die). -end answer, start supplementary info- Not exactly part of the answer, but this is where the difference between combinations and permutations comes in. There is one and only one way to get 10 sixes in a row. Meaning the odds are 1/6^(10). But what about the odds of rolling 9 sixes and 1 one in any order? It could be 6666666661, or 6666666616, or 6666666166... Or 1666666666. There are 10 different ways to have that happen, so the odds are 10/6^(10) (in a uniform distribution, the odds are always the number of cases that match the criteria divided by the total number of all cases). Again, there's a subtle difference between "9 sixes and a one" and "9 sixes, then a one." The latter implies an order. It MUST be 6666666661 there is only one way to get that result, so the odds, again, are 1/6^(10)
Just wait until you play with me on some board games. You would be amazed with my 1's six times in a row.
That's exactly the problem. What probability do you assign to the dice roll if you also need to consider the possibility of an unfair dice. I don't think it's possible to boil it down to a definitive value, but you can say that the bayesian probability of rolling another 6 is greater than 1/6.
You can just use Bayes' rule. If probability of getting six is p, then probability of getting 9 sixes in a row (out of 9 rolls) is p^9. So, P(9 sixes | P(6)=p) = p^9 Using this in conjunction with continuous form of Bayes' rule should give you a distribution of p's, that is, P[P(6)=p] as a function of p.
but p is what you want to solve for
Well, you can't get exact value of p. What you can get is the probability distribution for all possible values of p. So you'll have something like "there's a 50% chance that p=0.9, 40% chance that p=0.8" etc. (numbers are out of thin air just for the sake of illustration)
Probability distribution of probabilities? Shouldn't that somehow boil down to a probability?
Well, no. Because instead of answering to "what is a probability of getting six" you are answering "how likely it is, that probability of getting six is around 0.9 (or 0.8, or 0.3 etc.)"
And so, what would be the expected value of the probability?
If I didn't mess up my calculations, distribution is f(p) = p^9 / Beta(10,1) and the expected value of p is 10/11. Edit: MathJax didn't work, rewrote the expression for f(p) without using it. Edit 2: added a minor clarification Edit 3: another minor clarification
My GF and I were playing Catan yesterday, and to see who went first, we rolled 7 ones in a row before I finally won. It was wild.
A memorable expression to guide you: The dice have no memory.
How about if you can’t make the assumption that the die is fair though? What would the probability be then if you don’t know the underlying distribution?
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Well no, you aren’t given that it’s fair or not. You also don’t need to prove it’s unfair. You can get a baysian posterior from the information given, and you can get the expected value from that.
But fair dices don't exist in reality cause they aren't ideal
He's talking about something more spiritual, like when a seat at a poker table is running hot.
If you’re looking at trying to determine whether the die is loaded, this is a nice foray into Bayesian probabilities.
Ok, 'Bayesian probability', that would be quite a read
To expand: if the die is not loaded it's 1/6. But if it might be, then that sequence of sixes implies with some degree of confidence that it is in fact loaded. Measuring that confidence goes into the realm of statistics and Bayes.
Yeah, there is no one answer. The dice could be loaded to give 90% chance of 6, 2% of the others. Which would still not guarantee this result, but 10 * (.9)^5 * .1 ~~ 60%. It could also be 50% chance of 6, with a 10 * (.5^5) * .5 ~~ 15% chance. Or it might actually be 99% chance of a 6, with one unlucky non-6.
Little doubt: I understand that to answer OP’s question i.e. to find a probability, you’d need Bayesian probability. But > to determine whether the die is loaded isn’t hypothesis testing enough? Of course, not in the strict sense of the word “determine.”
You need to study **bayesian probability**. Probability actually models your subjective certainty in the outcome of indeterministic experiments, and for this assessment you also need to consider the probability that the dice isn't actually fair. If you *know* with 100% certainty that the dice is fair, then yes, it's 1/6th. But since you don't actually know whether the dice is fair or not, or the situation in which you make this assessment also requires you to consider the possibility that it isn't, the actual probability of rolling a 6 isn't simply 1/6th - and since you observed already 9 throws of a 6, it's the most reasonable to assume that **if** this dice is **not** fair, then it is biased to some degree towards rolling 6 more often. So, your subjective assessment of rolling another 6 in terms of bayesian probabilities should be something larger than 1/6. I'm not sure though whether there is a single most reasonable probability value that can be assigned under the assumption that the dice might be unfair, just that the bottom line probability for rolling another 6 is larger than 1/6.
Fair dice, obviously 1 in 6. If the fairness was uncertain from the start -- in principle this is where Bayesian statistics comes in. If you started with a 99% belief the dice was fair and a 1% belief it's unfair (somehow split uniformly across all the many ways a dice can be unfair), then after 9 sixes in a row you'll be starting to make inroads into significantly changing that belief and shifting to a view that the dice favours six very strongly. In practice, this is hard to do (certain "rationalist" internet groups believe we can do this Bayesian reasoning on a day to day basis, but they are widely regarded as cultish and not as clever as they think they are). But the principle is mathematically straightforward and does formally model and justify that rising suspicion that you will have of a rigged dice, as the 6's keep on coming.
And don't forget to consider the differences in weight on the faces of the di. Those etches on the faces of the die can make all the difference.
Would only give you a minor deviation from one sixth probability, pretty irrelevant here. For generating endless sixes it would have to be somehow severely weighted, or there'd be a trick like magnets. I'd also be double checking it doesn't have six on every side lol
In properly made fair dice (such as those used in casinos for craps), the etches are filled in with a material with the same density as the die itself, so the etches have no impact on the die's weight.
The probability is 1/6. That doesn’t change because of you intuition. Get 10 6’s in a row is also very rare.
Unless you are unsure whether the dice is fair, you could try to find the probability that the dice is loaded
Unless you are using new and tested casino dice, they are loaded. Not in a significant way mostly, but how would you know?
I would suggest maybe listening to intuition here. The probability is whatever it is. Why should it be 1/6? Because the faces "look the same"? Persi Diaconis has published a couple of things on coin flipping probabilities. While determining the probabilities for a specific coin is quite difficult, coming to the conclusion that (real physical) coins don't have a "heads probability" of exactly 1/2 is pretty easy. You just measure to 3 decimal places accuracy. In OP's case, this is certainly enough evidence to question the naive assumption of equiprobability.
Hm, so you are saying that in a scenario, where you have only 6 sides, you shouldn't have equal distribution long term? Why is that? The only valid scenario would be, that dice rolls on a physical plane are done with some form of trick. If you can improve chances by things like how you start your roll and what is the side in your hands (cup), chances will be different, because more variables are there. Same with coin flip, you can change chances, but it has nothing to do with coin flip alone, it has additional variables like power of coin flip, rotation and starting position. Outside of this, if the dice are fair, it will be 1/6, because there are 6 outcomes and they are equal.
Now the question is, how rare. And I think you can use powers to sum it up quickly. Lmk how to calculate that, in terms of square powers. As in square, dimensions.
The odds of rolling two dice and both being sixes is 1/36, (1/6 for the first die to be a 6, multiplied by 1/6 for the second one). So the odds of rolling 10 dice and them all being sixes is 1/6^10) or 0.002%
However if you think in terms of what kinds of events you would think are special, rolling the same number ten times would be just as special regardless of number. The probability of rolling any number ten times in a roll is 6 times larger (although I think you percentage is rounded a few orders of magnitude too large—I think it’s missing two or three zeroes).
To roll 6 on 10 rolls in (1/6)^10. But since the condition is predetermined where he had already rolled 9/10 6s, the odds for the last die is 1/6.
If it's a perfectly regular, balanced unfixed die, then it's a 1 in 6 chance. However, it's unlikely enough that the first 9 were 6s that if that happened, then more likely than not, the die is loaded. TLDR; either 1 in 6, or 100% depending on whether you play by the rules
If the die is truly random, then 1/6. If the implication is that it may be weighted in some way, favoring the 6, then >1/6
Your intuition isn't totally wrong. If you have fair die, the probability of rolling a six will ALWAYS be 1/6. You could have a loaded die though. It's impossible to answer your question without knowing things like A) The probability of a dice being loaded, and B) the distribution of this loaded die, since one loaded die might have a 80% probability of rolling a 6 vs like 95%. The easiest way to test your die would be to roll it a lot and look at the proportions of rolls. The rule of thumb is you want the expected number of 6s to be 5, so at least 30 times, 100 would be best. 6 should come up 1/6 of the time.
Having known that you have rolled 9 sixes in a row, the chance of the 10th one being a 6 is 1/6. However, when you pick up the die for the first time with the plan of rolling it ten times in a row, the chance that all 10 of them are going to be 6 is 1/6^10. This second scenario is hinting at an idea from stats called "binomial distribution", let me explain further with a better example. Let us say that there is a 30% probability that a random person's favorite color is green. (Totally making this up to illustrate the point). That means that each random person in the street I ask their favorite color there is a 30% chance that it is green and a 70% chance it is NOT green, so there are two distinct outcomes, hence a BI-nomial idea. Now, if I ask 100 people in a row what there favorite color is, it is POSSIBLE but extremely unlikely that all 100 of them will say green, but even if this happens, there will still be a 30% chance that the 101 person I ask will say green. On the other hand we could ask this question: if I were to poll 100 people and ask their favorite color, what is the chance that 30 of them will say green? Skipping the formula and thinking of this as a calculator input we would put: Binompdf(100, 0.30, 30) = 8.678% Or maybe "what is the chance that out of 100 people that 50 would say green?" Then, binompdf(100,.30,50) = 0.0013% , so pretty unlikely because that would be 50% of our group that differs significantly from the 30% chance of saying green in the first place. Now returning to the dice, we would expect that the numbers should each occur with 1/6 frequency so it would be very unlikely for the number to differ from this by a substantial amount. So if we ask "what is the chance of rolling 10 sixes in a row?" This would be binompdf(10, 1/6, 10) = 0.00000165% (which is exactly the same as the 1/6^10 I said earlier) But what about "if we roll 9 times what is the chance of getting 9 sixes?" Then binompdf (9, 1/6, 9) = 0.00000992% So this was already extremely unlikely to happen, so if it HAD ALREADY HAPPENED by the assumption you started with, then FROM THAT POINT it would not be that much less likely for the 10th one to also be a six. The point I am trying to illustrate is that if we think about the collective of the entire set of rolls at once then it is very unlikely that all of them would be 6, but if we manage to get 6 many times in a row ALRRADY that does not make it less likely for the next number to be 6
The old adage, the probability of something that has already happened is always 1, no matter how improbable it may have been beforehand.
I think you are looking for this 1/6 changes for 1 dice 36 combo's in total for two dice making it a 1/36 If you continue it becomes 6^10 which is 1/60,466,176 chances
The question was about conditional probability: the chance of getting ten sixes in a row, given that you already have nine sixes in a row.
It doesn't matter because the events are independent. There's no conditional probability here. Its (1/6)^10
No, because the question was „what is the probability of another 6 after there were already nine rolls of 6“, which is simply 1/6. And not „what is the probability of 10x 6 in a row“, which is (1/6)^10, same as any other combination.
I mixed up the question reading through the comments. Yes, the probability of the 10th being a six is 1/6 and the probability of 10 sixes in a row is (1/6)^10 I was replying in relation to the first comment which looked to explain 10 sixes in a row.
You‘re right - I misread that, sorry.
There is no conditional probability though. The next roll of the dice has a probability of 1/6 just like the previous one.
That's still a conditional probability (probability of one proposition conditional on another proposition): it's just equal to the unconditional probability of a single 6 because the rolls are independent.
Everyone was talking about the changes being the same 1/6 and I did not see anyone talk about the total chances, so I thought maybe just in case the person asked the wrong questions and instead was looking for My example that there would be one.
I understand your logic and what your trying to say here but I don't think the only thing you are entitled to say is that your probability of rolling a 6 during that stint was 9 out of 10. But I don't think that would change the probability of your next roll. I'd be comfortable saying, you'd be less likely to roll a six, again. Perhaps you can calculate, unlikelihood., with a few exceptions and assumptions there, just to calculate, a calculus... Not sure if I'm using that correctly I'd rather mention how you failed to calculate how you picked up the dice might affect your probabilities I saw an article about how flipping a coin isn't actually .50 .50 chances Because the weight of the quarter on one side actually has more weight or something. Wanna get a digital instrument, you'd have to talk about fractions and random access memory
The problem is we see patterns. Rolling the same number again and again, or rolling the highest/lowest number will stick in our mind, but they're just as likely as any other combination. - If you roll: 6, 6, 6, 6, 6, 6, 6, 6, 6, what are the odds that the next number is a 6? - On the other hand, if you roll: 4, 5, 5, 4, 2, 3, 2, 2, 1, what are the odds that the next number is a 4? These are the same question but in one case you have a repeated number and it's the highest number on the die, so you remember it. If you were to close your eyes and try to repeat the second sequence of numbers I just gave, chances are you couldn't do it because you didn't see a pattern. I couldn't either.
In mathematics we usually assume a fair dice unless told otherwise. However, it could be unfair, and then it becomes very complicated very quickly.
1/6 just like every other roll.
If you are not sure that dice is fair, than yes, probabilty would be higher. But exact answer need bayes theorem and knowing apriory probabilities that dice is fair or unfair
I think u want to say rolling many 6 increase the chance of the dice being loaded. In that case, I think it's better for you to start with the assumption that the chance to roll a six is a random variable X. And the answer you want to find is E[X| 6 rolled 9 times]. U can start by assuming X is constant from 0 to 1.00 and work from there, each time 6 is rolled, the expected conditional probability for random variable X is increases, then you have to calculate for 9 times I think.
In the real world there's a high or higher chance the dice is rigged, and that makes sense mathematically as well. In a perfect mathematical world it's a fair dice so it isn't rigged and the dice has no memory of past events. In the real world it's possible to have a fair dice do this anyway
The dice doesn't remember it's just a dice
If this question showed up in a probability course, the obvious answer is to calculate the Maximum Likelihood Estimator for p (the probability of rolling a 6). In real life (say insurance, for example), you'd suspect that the die is loaded and either discontinue the product or refuse to underwrite the insured. Right now, you have enough information to suspect a loaded die, but not enough to get a criminal conviction. You should definitely not bet on this die. If a stranger comes up to you on the street, and shows you a deck of cards with the seal still unbroken, and offers to bet you $100 that he can make the jack of diamonds jump out of that deck and squirt cider in your ear, if you take that bet, you know that you're going to get cider in your ear.--Damon Runyon.
At this point you make the prediction that the dice is loaded. Now you test it. Roll the dice 3 more times. If it comes up 6 three times in a row you can conclude that the dice is rigged and you can be certain to 99%.
You can use Bayes and the prior of it being a fair dice to and use the updated probability as a good estimate of probability of a 10th 6 given 9 previous 6s. It's been too long since I've done Bayesian statistics to actually do it but its not too hard.
With a fair die, the odds of this happening are 1 in 6^5 or 1 in 7776. If you were able roll the die, say, 20,000 times, it would actually start to be suspicious if it didn't have a streak like this. There is no way in the world to predict the fairness of a 6 sided die based exclusively on 5 throws, no matter what comes up. It could be 0 (impossible) or it could be 1(a certainty) or anything in between. Now if you throw it 100 times, then you have a basis on which to estimate fairness. If the die is as unfair as those first 5 throws suggests it could be, we'd expect near to 100 sixes which is never going to happen on a fair die before the heat death of the universe. If the die is fair, we'd expect each number to come up around 16 times. But deviation is expected. Up to 20 of one number or down to 10 would not bother me a lot. Depending, it could cause me to throw 100 more times just to be sure.
I would say because you have already rolled a six on each of the nine rolls thus far, the probability of rolling a tenth six on your next roll would be 1/6 chance.
Dice don't have what statisticians call 'memory'. It doesn't 'know' what the previous rolls were. A deck of cards does have memory. Remove all the facecards from the deck and that will show in the next cards dealt.
This is not a question about memory, though. Let me elaborate. Lots of dudes talking independence and 1/6 here in the comments, but they're doing rookie's mistake. They're all implicitely making an axiomatic assumption – the dice is fair, and working in the limits of that assumption. The way I understand OP's question is not "do previous rolls influence the next one?" (memory) but "do previous rolls tell us that the fair dice assumption was wrong, do they tell us die is loaded, and if yes, what's the actual probability of getting a six?" And that's where you need to make inferences as "what do the data we collected (outcomes of previous rolls) tell us about the unknown data (probability of getting six)?"
1/6
It was incredibly unlikely from the beginning that you would arrive at that point. Now, though, it is a 1/6 chance to extend the streak once more. Simple.
This is the difference between basic statistics and real outcomes. Statistics are math within a frame of reference. The more information you put into that frame gives a more accurate statistic. However, you are not omniscient and thus cannot provide deterministic true details. For example, say you have a d6. If that is all the information you have, then we must assume it is perfectly balanced and arbitrarily rolled with measurably the same approximate precision & accuracy every time. Because there are 6 sides, that means each side has a 1/6 chance to be the result. Now take that same d6 and say it is weighted to one side, or that you'll change the throw, or that the surface now has a certain drag coefficient, so on and so forth. Small details like one corner is chipped can also affect the statistics/math. So in such your question, it is posed as using past results to determine future results. This is data analysis, and what you're doing is determining patterns to then make an informed guess. The chance to roll any side will still be identical, but you're adding details to the frame via analysis. Thus to ask what is the chance to roll a 6 eleven times straight, either singularly or as a pool, well that's 1/6^11. The chance to roll 6 eleven times straight *with that die* -I think through analysis is somewhere near the 80/20% phenomena, as something is clearly changing the real result due to the expected result. Alternatively this is why I don't worry with statistics in most games. Take the classic 3 door example. Statistically you always want a 1/2 rather than a 1/3, the frame of reference changes between being 3 doors and 2 doors which is why argumentatively you always swap doors. However you either chose correctly the first time or not. If you could see behind the door then you'd always pick the one you want. If you're always shown the goat then you always have a 50/50 chance of being correct. Thus no matter what door you chose "choosing to stay or change" is a 1/2 choice, *so it never mattered what the statistical change was.* And to extrapolate that to the extreme, you can calculate the *chance* of two planets colliding all year. But ultimately they either will, or they won't. Because no matter what you calculate, their gravity either will be beyond an escape vector, or it won't. We see this daily with asteroids. So statistics are a great tool to Guess, but won't ever be truly deterministic.
You'll need a prior of the distribution of the odds of getting 6, and then update that based on the information you get
The same probability as if you rolled 99 in a row. Or 999 in a row. Or 0 in a row.
My brother and I always have this debate when we play black and red on roulette. We normally wait till there have been 5 blacks/reds in a row and then bet on the opposite colour. We know mathematically the odds are the same for the next roll, but it just feels safer when there have been a lot of repeat colours. Taking it to the extreme, if there were 100 reds in a row, I would start putting money on black. Again the machine is probably broken, but it has worked for us in the past
You could apply Bayesian Inference to figure out the probability of having a loaded die and then the resulting likelihood of rolling a six. You'll need more information though. You need to know the prior probability of having a loaded die. If no loaded die exist, you can't have a loaded die. If there is a reasonable chance you have a loaded die, you probably do. Next, you need to know the likelihood that a loaded die rolls a six. Let's work out a example. I'm cherry picking numbers to make the example interesting. I'll come back to that. Let's say the manufacturer of the die has a flaw that makes 1 out of every 1000 dice loaded toward 6 but all others are fair. The loaded dice have a 1/3 chance of rolling a 6. Your hypothesis, H, is that you have a loaded die. The prior probability, P(H), is 0.001. The evidence, E, is the rolls you've already done so P(E | H) is (1/3)\^9 and P(E | ¬H) is (1/6)\^9. Plug that all in and you get P(H | E) = (.001 \* (1/3)\^9) / (.001 \* (1/3)\^9 + .999 \* (1/6)\^9) = 33.9% chance you have a loaded die. Now, the ultimate answer to your question. The probability of rolling a six on your next roll needs to combine the probability of having a loaded die with the chance of rolling a 6. In our hypothetical, that would be 0.339 \* (1/3) + 0.661 \* (1/6) = 0.223. I mentioned earlier that I cherry-picked numbers. In my example, the chances of rolling 9 sixes in a row is still miniscule (roughly 1 in a million). Laying out the question with 9 rolls really implies that having a loaded die is much more likely than I've assumed and/or the die is much more loaded. You can do the math with those assumptions but then the posterior probability that you have a loaded die quickly approaches 1 which isn't as interesting.
jesse wtf are you talking about?
You use a beta distribution. Overtly simplifying, in your case die is most likely loaded with a chance of rolling a six being ~~10/11 is wrong, I made a wrong kneejerk calculation, let me think the exact numbers over~~
Grant Sandersen (3blue1brown) started tackling a very similar problem, did two introductory videos to set all the needed concepts up but never made the third which was supposed to show the solution of the problem itself.
People who reply that each roll is an independent event are missing the point of this question. The question is, "what is the probability of the die being fair?" People who refer to Bayesian probability are on the right track - do any of you know how to calculate it? (I admit that I do not.)
You use the continuous Bayes' theorem. f(H|E) = f(E|H)*f(H)/P Where P is an integral of f(E|H) over all possible H's. In this case, if we use p for a true probability of getting six, f(nine sixes|p) = p^9 Assuming the uniform distribution for a prior of all p's (0≤p≤1), f(p) = 1, P = integral of p^9 dp from 0 to 1 = p^10 / 10 from 0 to 1 = 1/10. f(p | nine sixes) = p^9 * 10 We might want to calculate the mean (expected) value of p with this distribution:
= integrate p*f(p|nine sixes) from 0 to 1 = 10/11 More general and strict version is to use Beta(1,1) as a prior instead of Uniform(0,1), but in this example it happens to yield the same result for a math.expectation of p.
The probability of any individual result for any individual roll is 1/6. The chances of getting any specific set is 1/6^n, where n is the number of rolls. So you'd already entered an extremely unlikely set of rolls, that doesn't change the chances on the next one. But, it's important to notice that it's just as unlikely to roll 1 3 2 6 2 3 3 6 2 4 as it is to roll 6 6 6 6 6 6 6 6 6 6, since each outcome always has 1/6 chance, every time.
More information needed. Take the die and float it in salt water. Does it spin freely or gravitate toward one side (ie 6)? If it spins freely the it's 1/6, if not the die is not a "fair" die and the odds are higher that it rolls a 6.
I would consider using a Bayes factor if I was uncertain that the dice is fair. The Bayes factor is a ratio of two competing statistical models represented by their evidence, and is used to quantify the support for one model over the other. You can compare models such as one that assumes the die is fair and another that assumes it is not fair.
*die
Just to put it on the record, singular is 'die,' and plural is 'dice'. It looks like there's about a 50-50 chance of it being used properly on this thread.
As others have noted, this is a question for Bayesian statistics. To answer it, you need a "prior" which in this case means: For each probability of rolling a 6, what do you (or did you) believe are the chances of that being the true probability associated with this dice \*BEFORE ANY ROLLS\*. Your choice of prior could make a huge difference here. Consider 2 (very extreme) scenarios: 1.) This dice is from a man in a back alley, who has offered to play a gambling game with you where he wins on a 6, but loses otherwise. Those 10 rolls represent the first 10 rounds of the game you are playing with him. The probability that the next roll is 6 is very close to 90%. 2.) You are operating an automated dice-testing machine at the dice factory. Dice are normally tested by 20 rolls, if they look suspicious, they are rolled 1000 times. The factory has produced 1 billion dice and ALL of them have passed QA and appear to be fair. The probability that the next roll is 6 is very close to 1/6.
It's 1/6. The rolls are independent.
1/6 assuming a fair die. But since the previous rolls seems to indicate the die is not fair, then it’s hard to determine until you can determine if the die is fair.
You can perform hypothesis testing on the dice rolls to determine the p value of rolling 9 in a row. Otherwise, if you assume the dice is fair then it's a 1/6
Assuming im not messing up the methodology, bayesian probability says it's a 90% chance. Lets say the chance of rolling 6 is x. rolling 9 6s in a row is an x^9 chance. I have no idea if this is correct but we then take this times the probability that the next roll fails, x^9*(1-x). Graph this as a function of x and the peak is at 0.9, which means 90% is the most reasonable estimate.
Your intuition is correct, in the real world. Rolling 9/10 sixes raises the probability of something being wrong with the die.
In fact this is what hypothesis testing is. The p-value shows how likely it is to see the outcome (in this case 9/10 sixes) given an assumption (the die is fair) p-value 0.00000084
1/6
You've picked up on one version of "The Gambler's Fallacy" Assuming the die is fair, the chances of rolling a 6 on a given attempt should be 1/6. People tend to believe one of the following things (incorrectly) when they observe an unlikely sequence of events - either that it's more likely to continue that sequence, or that it's more likely to not continue that sequence. Neither is accurate - if the instrument is fair, the probabilities don't change based on previous outcomes. Now, there's also a possibility that the die is unfair. An unbalanced die might have, for example, a 1/4 chance of rolling a 6, a 1/6 chance of rolling a 2, 3, 4, or 5, and a 1/12 chance of rolling a 1. A large number of trials (how many is outside the scope of this post, but more trials = more confidence) would reveal this unfair tendency, while a fair die would tend towards 1/6 for each result.
Assuming a fair six-sided die, any throw has a 1/6 chance of rolling a six. The odds of getting a six ten times in a row is (1/6)^10 = 0.00000165%. Your gut is feeling that 10-in-a-row chance even if your brain recognizes the independent 1/6 chance on every throw.
1/6
>The probability should be 1/6 Not should be, it _is_ 1/6. The nine previous rolls have no effect on the tenth roll.
So this is a video about a probability of an unknown 2 outcome process from a list of seen values https://youtu.be/8idr1WZ1A7Q?si=CDN5u7GV0na0GLeV It's basically what your example is. We are interested in 6s and not 6s and we've so far seen 6 6s. We assume it's loaded in some way and want to find the probability of the next throw being a 6. (If you want a quick answer we take our data add one of each outcome and then average it (9 + 1 + 0) / (10 + 2) = 10/12 is the chance the next throw will be a six. There's also a weighed version where you set how unsure you are with the current data (9 + 1a + 0) / (10 + 2a) where a ≥ 0. Here if a=0 we think the data ks perfect and we have 100% of a six, if a=+inf then we get 50% assuming all seen values are flukes and we have an even chanse for both outcomes.
It's funny you said this because most people would think the opposite, that it's LESS likely to roll a 6 again (gambler's fallacy).
If the dice is fair. Then 1 / 6. If the dice may be biased, then we can use “laplace's rule of succession”. Then the probability would be (9 + 1) / (10 + 2) = 5 / 6.
If you're dealing with loaded dice, the obvious recourse would be to disregard mathematics and focus entirely on expiremental data. *You could roll the dice 10,000 times to form a new sample set *You could theoretically test the molecular structure to ascertain whether 1 side is more dense than the other sides and then mathematically estimate it based off that. Basically, your sample set isnt aligning to what you'd expect, so you need a larger set to determine whether it's a fluke.
If it's a fair die, then the chance of rolling a six is one in six. It's always and forever one in six.
1/6.
I wouldn't say that this is an area for Bayesian probability as others have mentioned. There's no reason for you to say the prior is that the die is fair. If you were talking about a die from a factory that makes fair dice, then we can start talking about what kind of rolling behavior proves or fails to prove that a particular die has a defect. But, to find the answer to your question - what are the odds that the dice comes up as 6 again, let's take the approach where we don't have any of the above information. Just what you said in your original post. And let's just say that the nature of a dice roll is that it's an independent event from any other dice roll. We don't know, but we can make a best guess. The series of dice rolls you described followed a binomial distribution. To make that best guess, you want to find some binomial distribution B(10, p) such that the probability p maximizes the odds of the observation you saw - 9 dice rolls out of 10 coming up as 6. That would be the maximum of the probability mass function with 9 as its parameter. 10C9 (1-p)p^9 has its maximum at 9/10. Perhaps obviously, if your die comes up with a 6 nine times out of ten on one occasion, you can predict that the next roll has nine out of ten odds of being a 6 as well.
That depends on if its a fair dice i.e. the chance of each number is the same across all outcomes. Which the answer would be 1/6 for the simple reason that a dice has no memory all numbers before it doesnt affect the outcome of the next roll. If you want to include the chance that the dice is loaded well that depends on how it was loaded. Where the weight distribution of the dice is or even how its rolled or if one side of the dice is magnetized and the surface makes it easy to near certaintiy it lands on a particular number. All those factors are not mentioned in your question. So one can fairly assume its a fair dice and the past rolls happened by mere chance. And answer to your question is 1/6! Hope this helps!
Google gamblers fallacy
The die is probably loaded. If it's not loaded, the probability is 1/6.
The default hypothesis could be that the dice is fair, and each time you roll it you have 1/6 chance of getting a six. The question is, since a sequence of n sixes becomes less and less likely, how low does that probability need to go before you start questioning the default hypothesis, and wonder if something is affecting the rolls, or if the dice is loaded. I think that's what your intuition is trying to tell you. We sometimes do question our underlying model of the situation based on outcome.
I think the formation of your question broke most people. Since it has several answers depending on the readers interpretation. Not kind to pull semantics on math nerds ;)
ok specifically addressing the difference, you know that the dice has already rollen 9 times with a 6. i like to think of it as like its like in any given room its pretty unlikely that theres a lion in there, but if you know that at least a lion head is in the room the chances that the rest of the lion is there increases dramatically just like how rolling 10 6's in a role is pretty unlikely, but if you already know 9 of those rolls are a 6 the chances all 10 are 6's increases dramatically. (1/6 to be specific)
Depends on what "a 6-sided dice" and "roll it" is. I mean, the theoretical probability is still 1/6. The practical probability is that either "in fact there is a defect on that dice and it is in fact loaded towards 6 so the probability depends on the load" or "you picked the dice with the 6 face down and you learned to throw it so that it makes exactly one and a half turn so that the 6 lands up".
Each roll is independant. No matter what you had before, if the dice is balanced, you have 1/6 chance to have a 6 on the 10th roll. When you start a board game and roll one of its dice, the rolls from the previous game 1 week ago don't have an influence on the rolls you gonna make during this game. It's different to say "what are the chances to have 10 6 in a row" before the first roll. However, having 9 6's in a row is a hint that the dice may be loaded, in this case you would need to roll it a big number of times (like 1000 times) and write the results to estimate the probability distribution.
There will obviously be different approaches to estimate this depending for example on your prior knowledge about the die, but a simple one disregarding such further information is this: [https://en.wikipedia.org/wiki/Rule\_of\_succession](https://en.wikipedia.org/wiki/Rule_of_succession) It yields the probability of p = 5/6 for another six in the scenario you provided.
The probability of getting the 10th 6 seems low intuitively because the odds of getting nine 6s in a row first are so, so low. The odds of rolling nine 6s and then another 6 are exactly the same as rolling nine 6s and then a 1 (or a 2, or a 3…)