Suppose I roll a six-sided die $10$ times and each time it shows a $6$. What is the probability of the next roll coming up $6$?

You might say $1/6$. But it was never declared to be a fair die. In fact it seems from the data available that there is a high probability of it being an unfair dice and the probability of another $6$ is quite high. Nevertheless it could still be coincidence.

Is there any way to model this situation and describe mathematically the probability distribution of the next throw?

  • 1
    $\begingroup$ There is a Bayesian approach to such problems that I do not find persuasive. $\endgroup$ – André Nicolas Mar 15 '14 at 21:26
  • 3
    $\begingroup$ Think about a different question: if you keep flipping a coin, how many times does it have to show heads and no tails before you check both sides to see whether it is double-headed? $\endgroup$ – Henry Mar 15 '14 at 21:48
  • $\begingroup$ @ST3 (about the edit): According to Wikipedia, both "die" and "dice" are acceptable as singular. $\endgroup$ – celtschk Mar 24 '14 at 10:26
  • $\begingroup$ @celtschk thanks for information, didn't know that. $\endgroup$ – ST3 Mar 24 '14 at 10:44

You can only model that mathematically if you specify what else it could be. For example, say you know (or believe) that one of 1000 dice is manipulated to give a 6 every time, while the others are fair. Then this would give you a prior probability of 0.001 for an all-6 die, and 0.999 for a true die. Now every time you throw a 6, you can use Bayes' rule to update the probability of the die being an all-6 one:

$$p_{n+1}(\text{all-6}) = \frac{p_n(\text{all-6}) \cdot 1}{p_n(\text{all-6}) \cdot 1 + p_n(\text{fair}) \cdot 1/6}$$

So after the first 6, you'd have $p_1(\text{all-6}) = \frac{2}{335} \approx 0.006$. After 10 times throwing a 6, iterating this function gives a probability for an all-6 die of $2239488/2239525 \approx 0.999983$, that is, under the above assumptions you can be fairly sure the die is actually an all-6 one.

Of course with that assumption, as soon as you throw a non-6, you know the die is fair.

Another possibility is to assume that all probability distributions are initially equally likely (that is, you assume you literally know nothing about the probabilities), and then you can calculate that the probability of throwing another 6 after throwing a 6 $k$ times out of $n$ is $p(k,n)=(k+1)/(n+6)$, that is, after throwing ten times a 6, it's $7/12 \approx 0.583$.

Of course you can use any other prior and get other results.


There is an unrigorous approach to such questions, which has an ultimate base in hypothesis testing, and which is prone to being misused.

Suppose the probability of throwing $6$ is $p$, then the probability of throwing ten sixes is $p^{10}$. If my tolerance of error is $\frac 1{20}=5\%$, then I say $p^{10}\ge 0.05$ so that $p\gt 0.74$.


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