Each succeeding person takes his own hat if it is not taken. If it is taken, he picks a hat at random from the remaining hats. What is the probability that the last person gets their own hat?
I'm just a college student trying to make my way in life. This is a homework question, so I'd appreciate some hints! Thank you all very much.
What I've attempted so far: I tried to make a conditional probability tree for all the possibilities that Mark could have. I.e., he could pick the first guest's hat, the second guest's hat, etc., or he could pick his own hat, with each 'branch' having probability $\frac{1}{100}$.
Then the first guest would have 99 hats to choose from. If Mark picked out the first guest's hat, then the first guest has $\frac{1}{99}$ probability of choosing any hat from the remaining. (or, $\frac{1}{99}$ probability of choosing the last hat).
Continuing on this logic, I thought that the probability that the last guest doesn't get their hat would be the sequence $\frac{1}{100} + \frac{1}{100 * 99} + \frac{1}{100 * 99 * 98}$ and so on and so forth.
i.e P(the last person does not get their hat) = ${1} - \sum_{k=0}^{100} \frac{k!}{100!}$
I know this is wrong, because just because an nth guest gets their own hat does not mean that the last guest gets theirs, if I'm making any sense.