Let's say I have $n$ fair 6-sided dice. What is the probability that the same set of outcomes is obtained when $n$ fair dice are rolled twice?


$n=8$: Suppose $8$ dice are rolled. If the first roll is $[1, 3, 6, 2, 3, 4, 3, 1]$, there are two 1's, one 2, three 3's, one 4, and one 6.

Rolling the same dice again, what is the probability that the next roll will also have two 1's, one 2, three 3's, one 4, and one 6?

The above example uses 8 dice, but I'm curious about the probability for any positive $n$.


My question is different than Two dice throw probability since I'm looking for the probability of $n$ dice having the same result in two consecutive throws

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    $\begingroup$ @Jneven Those don't look the same to me. $\endgroup$ – Arthur Aug 16 '19 at 9:37
  • $\begingroup$ @Jneven Die is singular; dice is plural; dices is the third person singular form of the verb to dice, meaning to cut into small cubes. $\endgroup$ – N. F. Taussig Aug 16 '19 at 9:46
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    $\begingroup$ Possible duplicate of Two dice throw probability $\endgroup$ – Paul Frost Aug 16 '19 at 13:51
  • $\begingroup$ @PaulFrost I already explained how my question is different than Two dice throw probability $\endgroup$ – alexdriedger Aug 16 '19 at 14:04

You have multinomial distribution with $n$ trials, $k = 6$ outcomes and all $p_i = \frac{1}{6}$. So probability of getting two equal rolls is $\sum\limits_{x_1 + \ldots + x_6 = n} \left(\frac{n!}{x_1! \ldots x_6!}\cdot \frac{1}{6^n}\right)^2$. It's unlikely there is good closed form for it. For asymptotic see, for example, this answer on mathoverflow.

For example, if we have $n = 2$, we need to sum over all variants to partitioning $2$ into $6$ non-negative terms. There are $15$ of them with $2$ ones and $4$ zeroes: $1+1+0+0+0+0$, $1+0+1+0+0+0$, ..., $0+0+0+0+1+1$ and $6$ with $1$ two and $5$ zeroes: $2+0+0+0+0+0$, $0+2+0+0+0+0$, ...

First variant will give in our sum $15$ terms equal to $\left(\frac{2!}{1!1!0!0!0!0!} \cdot\frac{1}{36}\right)^2 = \frac{1}{18^2}$. Second will give $6$ terms equal to $\left(\frac{2!}{2!0!0!0!0!0!} \cdot \frac{1}{36^2}\right)^2 = \frac{1}{36^2}$. So the answer is $\frac{1}{18^2}\cdot 15 + \frac{1}{36^2}\cdot 6 \approx 0.05$.

  • $\begingroup$ I'm coming from an engineering background, but I'm not sure how to put numbers into that equation to get a percent chance. Could you show an example (ex: for 8 dice, these numbers go here => calculate => 4%)? $\endgroup$ – alexdriedger Aug 16 '19 at 9:53
  • $\begingroup$ What is the first part confusing you? $\sum$ symbol? Limit of the sum? Factorial ($!$) symbol? $\endgroup$ – mihaild Aug 16 '19 at 10:15
  • $\begingroup$ The partitioning part was confusing, but it makes a lot more sense now! $\endgroup$ – alexdriedger Aug 16 '19 at 10:28
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    $\begingroup$ The example you gave gives an answer of 1.83. Is this number a percentage? And if so, shouldn't it be a number between 0 and 1? $\endgroup$ – alexdriedger Aug 16 '19 at 13:22
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    $\begingroup$ Sorry, the $\frac{1}{6^n}$ part should be inside brackets (we need to sum square of probability of result over all resutls). Fixed now. Good catch! $\endgroup$ – mihaild Aug 16 '19 at 13:46

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