Let's first determine the probability that the first person picks his own name and the remaining 4 people dopick their own names.NOT

Knowing this, we cast the first person aside for right now, and consider the the following question: Four people remove their name tags and put them in a hat. Then they each randomly choose one of the tags. What is the probability that exactlynonepicks their own name?

Let E denote the event thatNOmatches occur. Since the probability depends on n, the number of people choosing, we write . We are going to go about this solution recursively, so we begin by conditioning on whether or not the first (of 4) people draws his own name tag. Let's denote these events . So,

Note that , and thus

.

Now, is the probability that none of n-1 people pick their own name tag when selecting from a hat containing n-1 tags with exactly one of the tags not belonging to any of the n-1 people. (So, there 1 "extra person" whose name tag is not in the hat, and there is 1 extra tag in the hat which does not belong to any of the men choosing.) This can happen in two mutually exclusive ways:

1) No matches and extra man does NOT select the extra hat (hat of man who chose first). The probability of this event is (which we get if we assume the extra hat belongs the extra person).

2) No matches and extra man DOES INDEED select the extra hat. The probability of this event is .

Since the two ways described above are mutually exclusive, is simply the sum of the two probabilities; i.e.,

.

Now, we can plug this information into the equation for :

Aha! Recursion! It can do the rest of our dirty work.

So we have is the probability of n people all selecting a name tag that is NOT their own (no matches). Thus,

Now we take advantage of that recursion.

Therefore, the probability of 4 people each selecting name tags (each person's name included) that are NOT their own is .

Now, let's bring back the real first person to select from the hat. The probability that he, selecting first, selected his name own tag is 1/5. So the probability of (HIT, MISS, MISS, MISS, MISS) is (3/8)/(1/5). However, notonlymust we consider this person being "the one" to select his own hat (while the others don't), we must also consider the separate events whereby each of the other four people is "the only one" to select his name tag. So, we multiply (3/8)/(1/5) by (1/5), which cancels out the (1/5) in (3/8)/(1/5). Thereforeforeforefore, the probability of EXACTLY ONE person selecting his own name tag in our problem is 3/8 = 0.375.

-Andy