Probability of winning Pirate Roulette game 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?

 A: 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:
$$\frac{6\cdot 23!}{24!}=\frac14$$
A: The number of ways to select $18$ safe slots equals ${23 \choose 18} = 33649$, while the number of ways to select 18 slots equals ${24 \choose 18} = 134596$. As such, the probability of winning the game equals:
$$\frac{33649}{134596} = \frac{1}{4}$$
You can also look at it as follows: if you select $18$ slots out of $24$, then the probability of not having selected the losing slot equals:
$$1 - \frac{18}{24} = \frac{6}{24} = \frac{1}{4}$$
A: Think reversely. What if you first pick your spots, and only afterwards does the mechanism decide which one is the special spot? (As a justification for why this is OK to do, from the outside this is indistinguishable from if the spot was decided beforehand, but the mechanism includes a delay long enough for you to stick all your swords, so the probabilities involved are the same.) Of the $24$ spots it could choose, there are $6$ that don't have a sword in them. What's the probability that one of them is chosen?
