Let’s play a game with a single, ordinary die. The goal is to get each of the numbers 1 through 6, in any order, in exactly six rolls. If you lose – you roll a number you previously rolled – then you give me \$1. If you win, by rolling all six numbers, then I give you \$10.

I’m trying to determine what the prize for winning should be. Would you play for \$10? For \$100? I think I have determined the probability of winning to be about 1.54%, which would suggest a prize of about \$65 if the game is to be a fair game.

I am expecting this to be wrong, but this is how I determined the determined the probability of winning. On the first roll, you cannot lose; any result will allow you to roll again. On the second roll, there is a $\frac{1}{6}$ probability of losing by rolling what was previously rolled. This suggests the second roll has a $\frac{5}{6}$ probability of success (allowing you to roll again). Likewise, the third roll has a $\frac{2}{6}$ probability of rolling a duplicate, so the third roll has a $\frac{2}{3}$ probability of winning. Following this logic through, the probability of winning is $\frac{5}{6} \cdot \frac{2}{3} \cdot \frac{1}{2} \cdot \frac{1}{3} \cdot \frac{1}{6} = \frac{10}{648} = \frac{5}{324} \approx 0.0154 = 1.54\%$. Does that seem right to you?


Yes, that is the right answer and your argument is valid. Here is another way of thinking about the problem.

How many 'winning' outcomes are there? To win, we need the numbers $1, 2, \ldots, 6$ in some order, so there are $6!$ winning outcomes.

How many outcomes are there? Each roll has $6$ outcomes, and all the rolls are independent, so there are $6^6$ possible outcomes.

Hence the answer is $\frac{6!}{6^6} = \frac{5}{324}$.

It's worth noting that if you don't reduce your fractions, your calculation becomes $\frac{5}{6} \times \frac{4}{6} \times \cdots \times \frac{1}{6}$ and then it is clear that the answers are the same.


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