Probability problem

**chil2e** · January 13th 2010, 08:56 AM

I would appreciate if you solve this too. Thanks.

Five probabilists at a meeting remove their name tags and put them in a bag. Then they each randomly choose one of the tags. What is the probability that exactly one of them gets the correct name tag?

**abender** · January 15th 2010, 08:33 AM

Originally Posted by chil2e

Five probabilists at a meeting remove their name tags and put them in a bag. Then they each randomly choose one of the tags. What is the probability that exactly one of them gets the correct name tag?

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

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 exactly none picks their own name?

Let E denote the event that NO matches occur. Since the probability depends on n, the number of people choosing, we write $P_n=P(E)$ . 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 $I \text{ and } I^c$ . So,

$ P_n = P(E) = P(E|I) P(I) + P(E|I^c) P(I^c) $

Note that $P(E|I) = 0$ , and thus

$P_n = P(E|I^c) P(I^c) = P(E|I^c) (\tfrac{n-1}{n})$ .

Now, $P(E|I^c)$ 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 $P_{n-1}$ (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 $\tfrac{1}{n-1}P_{n-2}$ .

Since the two ways described above are mutually exclusive, $P(E|I^c)$ is simply the sum of the two probabilities; i.e.,

$P(E|I^c) = P_{n-1} + \tfrac{1}{n-1} P_{n-2}$

.

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

$ P_n = P(E|I^c) \tfrac{n-1}{n} = \tfrac{n-1}{n} P_{n-1} + \tfrac{1}{n} P_{n-2} \implies P_n - P{n-1} = -\tfrac{1}{n} (P_{n-1} - P_{n-2}) $

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

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

$ P_1 = 0 $

$ P_2 = \tfrac{1}{2} $

Now we take advantage of that recursion.

$ P_3 - P_2 = -\tfrac{p_2 - P_1}{3} = -\tfrac{1}{3!} \implies P_3 = \tfrac{1}{2!} - \tfrac{1}{3!} $

$ P_4 - P_3 = -\tfrac{p_3 - P_2}{4} = -\tfrac{1}{4!} \implies P_4 = \tfrac{1}{2!} - \tfrac{1}{3!} + \tfrac{1}{4!} $

Therefore, the probability of 4 people each selecting name tags (each person's name included) that are NOT their own is $P_4 = \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} = 0.375$ .

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, not only must 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

**Plato** · January 15th 2010, 08:52 AM

That reply is too long to even quote.
There arer five ways to select the one getting his/her own tag.
There are $D(4)=9$ derangements the other four.
So the probability of exactly one is $\frac{5\cdot 9}{5!}=0.375$ .

**Archie Meade** · January 15th 2010, 11:20 AM

Originally Posted by Plato

There are $D(4)=9$ derangements the other four.

That was neat!

Here is another logical way...

Label them A,B,C,D,E.
If A strikes, then the other 4 must miss.
The ways this can happen are as follows....

A picks A in all following cases
B picks C, C picks B so D must pick E to leave E picking D
B picks C, C picks D so D must pick E to leave E picking B
B picks C, C picks E so D must pick B to leave E picking D.

In the second of these cases, if D picked B, E would be left picking E.

There are 3 ways, for which the probabilities are...

$\frac{1}{5}\ \frac{1}{4}\ \frac{1}{3}\ \frac{1}{2}$

in all cases.

Repeat the process for B picks D and B picks E,
therefore, we treble the probabilty for the B picks C case.
Finally multiply by 5 as any of the 5 can strike.

$P=5(3)(3){\frac{1}{5}\ \frac{1}{4}\ \frac{1}{3}\ \frac{1}{2}}=\frac{3}{8}$

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