In the pirate roulette game, there are 24 slots. Only 1 slot leads to the pirate popping up when a sword is stuck in it.
Suppose I have 18 swords, what is the probability that I win the game by using all 18 swords without hitting the one slot that results in me losing?
There are 23 safe slots and I have to pick 18 of them, so there are $\frac{23!}{(23-18)!} ways of winning.
The total number of combinations is $\frac{24!}{6!}$, so the probability of winning is $\frac{\frac{23!}{6!}}{\frac{24!}{6!}} = \frac{1}{24}$, which is about 4%. This is a smaller number than I expected, and does not seem right. What I am doing wrong?
Best Answer
We give each slot a number and write them down in the order that they're used. If you haven't lost after sticking $18$ swords in, the 'bad' slot must be one of the last $6$. There are $24!$ ways to order the slots, but when the 'bad' slot is among the last $6$, there are $6\cdot 23!$ possiblities, so the chance of this happening is: $$\frac{6\cdot 23!}{24!}=\frac14$$