-1

Let's say there are $12$ people who each roll a fair $6$-sided die. What is the probability that every face has been seen?

I first attempted to find this as a string counting problem: let's say there $12\choose6$ ways of arranging the $6$ unique faces among the $12$ rollers and $6^6$ outcomes for the remaining ones. The total arrangements of the rolls would be $6^{12}$ leaving us with a total of $\frac{{12\choose{6}}\cdot6^6}{6^{12}}$

However, let's say you roll first. From your perspective, you have rolled something unique and are waiting for the remaining $11$ people to roll which would result in an expression such as $\frac{{11\choose{5}}\cdot6^6}{6^{11}}$

Why are these two approaches not equal (assuming one of them is correct)

mathematics2x2life
  • 13,322
shrizzy
  • 754
  • See https://math.stackexchange.com/questions/379525. (Not closing as a duplicate because you specifically asked about the error in your own approach.) – joriki Apr 01 '23 at 00:53
  • As the answer of Robert Shore indicates, Inclusion-Exclusion is the way to go here. See this article for an introduction to Inclusion-Exclusion. Then, see this answer for an explanation of and justification for the Inclusion-Exclusion formula. – user2661923 Apr 01 '23 at 01:13

1 Answers1

6

There are $6^{12}$ possible combinations all told. Of those, there are $6 \cdot 5^{12}$ possible combinations that omit any particular number. But this is an overcount of "bad" combinations because, for example, you are double-counting the combinations that omit both $1$ and $2$.

Use the principle of inclusion and exclusion to see that the total number of "good" combinations is:

$$6^{12}-\binom 61 5^{12}+ \binom 62 4^{12}-\binom 63 3^{12}+\binom 64 2^{12}-\binom 65.$$

Divide this number by $6^{12}$ and you'll have the probability you want.

The problem with your proposed solution is that you are overcounting "good" rolls. For example, it's possible in your example that the other six rolls also give you all six faces, in which case there's a clear overcount.

Robert Shore
  • 23,332