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I am having trouble with how to do this question.

What is the probability that if 5 people put their names into a hat, that they will all pick out their own name (Assume they do not put their names back)? What about if they do put their names back?

I know that when they do not put their names back, that the size of the sample space is 5!. Would I do something like 5!/1!? Does that make sense?

Your help is appreciated.

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2 Answers

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• If they do not put their name back: There is one configuration (event) where everyone gets his own name; there are, in total, $5!$ configurations (events) possible (i.e., $5!$ permutations of the 5 names to the 5 people). Since all these events have the same probability of happening, the probability you are looking for is $\frac{1}{5!}=\frac{1}{120}$.

• If they do: The first one has a probability $1/5$ of picking his name. After that, the second one has also a probability $1/5$ of picking his name, since there are still 5 names in the hat. Same for the other ones; as they all draw a name independently (the 5 events are independent), the probability that all 5 events hold is $\left(\frac{1}{5}\right)^5=\frac{1}{3125}$.

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Does a person put his name back in the hat after selecting it? This is important, because in this case sampling is with replacement, and the probability is $\frac{1}{5^5}$. If this is not the case, sampling is without replacement, and the probability is $\frac{1}{5} \cdot \frac{1}{4} \cdots 1 = \frac{1}{5!}$.

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