[Aren Siekmeier]

Originally Posted by Ether

If a fair die is thrown 420 times, what's the probability of getting exactly 70 of each number (1 thru 6)?

Here's a crack at it before reading responses

Spoiler for Answer:

tl;dr final answer:

420! / (70!^6 * 6^420)
or about
6.04e-7

A little rusty on my abstract algebra but here goes.

The odds of 70 ones followed by 70 twos followed by .... followed by 70 sixes is:
(1/6)^420

There are 420! permutations of this, i.e. the order of the symmetric group of order 420. However, some of these permutations don't matter. The symmetric group is generated by individual swaps, one pair at a time. Swapping two ones isn't actually another permutation (ones are indistinguishable), and so on with twos, threes, ... These irrelevant swaps generate six distinct symmetric groups of order 70 within the larger S_420. Their direct sum is isomorphic to all elements in S_420 generated by the irrelevant swaps. Call this subgroup T<S_420. Then the right cosets of T in S_420 are the equivalence classes of indistinguishable permutations, and we need to count these. This is the index of T in S_420, which is the quotient of their orders: 420! / (70!)^6

Then multiply this by the probability for the ordered case, final result:

420! / (70!^6 * 6^420)

Stirling's approximation on the factorials cancels lots of stuff and we get:

sqrt( 6 / (140*pi) ^ (5) ) ~ 0.000000604 (6.04e-7)

For an n sided die rolled n*k times, this becomes:

(nk)! / (k!^n * n^(nk)) ~ sqrt( n / (2*pi*k)^(n-1) )

Edit: euhlmann method for me