Six dice are thrown. The six dice are thrown a second time. What is the probability of getting the same numbers as in the first throw? If the order of the six numbers matters, the problem is easy, but if the order does not matter, I find myself in troubles, because I should consider too many cases depending on the number of repeated numbers and don't know how to proceed.
Probability of throwing the same multiset twice in a row with six dice
-
0Yes, it helps in the sense that now it is immediate to conclude, but is there some slick way to do this? I mean, how can one produce your table during an examination, when time is limited? (this is part of an exercise taken from an exam paper in probability) – 2012-03-28
2 Answers
The number $a_n$ of favourable events for $n$ $n$-sided dice is OEIS sequence A033935. There's a formula given there,
$a_n=[x^n]n!^2\left(\sum_{k=0}^n\frac{x^k}{k!^2}\right)^n$
(where $[x^n]$ denotes extraction of the coefficient of $x^n$), which is a succinct statement of the perhaps more obvious formula
$a_n=\sum_{1n_1+2n_2+\dotso+kn_k=n}\frac{n!}{(n-(n_1+\dotso+n_k))!}\frac1{n_1!\cdots n_k!}\left(\frac{n!}{1!^{n_1}\cdots k!^{n_k}}\right)^2\;,$
where if we have $n_j$ groups of $j$ identical dice each, the first factor gives the number of ways of assigning values to the groups, the second factor accounts for the fact that it doesn't matter which group a value is assigned to as long as it has a certain number of dice, and the third factor gives the number of favourable events for each assignment, which is the square of the number of distinct permutations of the dice given the assignment.
Based initially on Mark Dominus's table, you get
Pattern Ways Different Ways/Different (Ways/Different)^2 Ways^2/Different AAAAAA 6 6 1 1 6 AAAAAB 180 30 6 36 1080 AAAABB 450 30 15 225 6750 AAAABC 1800 60 30 900 54000 AAABBB 300 15 20 400 6000 AAABBC 7200 120 60 3600 432000 AAABCD 7200 60 120 14400 864000 AABBCC 1800 20 90 8100 162000 AABBCD 16200 90 180 32400 2916000 AABCDE 10800 30 360 129600 3888000 ABCDEF 720 1 720 518400 518400 Sum 46656 462 1602 708062 8848236
And the answer is $\dfrac{8848236}{46656^2} \approx 0.0040648\ldots$
Note, as joriki suggests, that $8848236$ is the sixth term of OEIS A033935, though if you are using a number of six-sided dice then OEIS A169715 is more relevant.