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Question

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?

Example

$n=8$: 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 $n$.

Edit

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, 2019 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. $\endgroup$ Aug 16, 2019 at 9:46
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    $\begingroup$ Possible duplicate of Two dice throw probability $\endgroup$
    – Paul Frost
    Aug 16, 2019 at 13:51
  • $\begingroup$ @PaulFrost I already explained how my question is different than Two dice throw probability $\endgroup$ Aug 16, 2019 at 14:04

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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 $\frac{1}{18^2}\cdot 15 + \frac{1}{36^2}\cdot 6 \approx 0.05$.

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  • $\begingroup$ 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%)? $\endgroup$ Aug 16, 2019 at 9:53
  • $\begingroup$ What is the first part confusing you? $\sum$ symbol? Limit of the sum? Factorial ($!$) symbol? $\endgroup$
    – mihaild
    Aug 16, 2019 at 10:15
  • $\begingroup$ The partitioning part was confusing, but it makes a lot more sense now! $\endgroup$ Aug 16, 2019 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? $\endgroup$ Aug 16, 2019 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, 2019 at 13:46

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