Probability problem: throwing 6n dice at random

Question

You throw 6n dice at random. Show that the probability that each number appears exactly n times is

Answer 1

We assume that the dice are fair (each number on the dice has the same probability to be on top, and the numbers on one dice are independent of the numbers on the other dice).

Then the probability of a desired outcome equals the ratio of

number of ways to achieve the desired outcome / number of ways to achieve all possible outcomes

The number of ways to achieve the desired outcomes equals the number of ways 6n dice can be divided into 6 equal groups (corresponding to each of the 6 numbers) of n dice in each group.

The number of ways to choose n dice from 6n is

_6nC_n = (6n)! / (n! x (5n)!)

The number of ways to choose the next group of n dice from the remaining 5n dice is

_5nC_n = (5n)! / (n! x (4n)!)

Repeating this calculation for the remaining four groups of n dice and multiplying all the results will give

_6nC_n x _5nC_n x _4nC_n x _3nC_n x _2nC_n x _nC_n =

(6n)! / (n! x (5n)!) x (5n)! / (n! x (4n)!) x (4n)! / (n! x (3n)!) x (3n)! / (n! x (2n)!) x (2n)! / (n! x (n)!) x n! / (n! x 0!) =

= (6n)! / ((n!)⁶)

This number needs to be divided by the number of all outcomes 6⁶ⁿ (six possible outcomes for each of 6n dice). Thus the final probability is

P = (6n)! / ((n!)⁶ x 6⁶ⁿ)