# Probability of two individuals getting their own hats, after 6 people put their hats on the table.

Two friends, Joe and Enzo, are members of a group of 6 persons who have placed their hats on a table. What is the probability that both Joe and Enzo get their own hats. (5 marks).

I am unsure about this problem, as we do not know the order in which they take the hats, or whether they are the first ones who get the hats. Any hints would be appreciated.

• The order people take hats does not matter, so you might as well have Joe pull first and Enzo pull second. Can you do that? – Ross Millikan Jan 19 at 15:43
• "as we do not know the order in which they take the hats, or whether they are the first ones who get the hats." Neither of those matter in any way. – fleablood Jan 19 at 16:41
• The outcome is Enzo has a hat. Joe has a different hat. The 4 other people have different hats. How they came to that result doesn't matter. All that matters is there are only so many outcomes and that will be the same no matter what order you do them in. – fleablood Jan 19 at 16:47
• "Two friends, Joe and Enzo, are members of a group of 6 persons who have placed their hats on a table.... missing action..... What is the probability that both Joe and Enzo get their own hats" I would quibble and say the probability is $0$. The put their hats on the table. It never says they ever picked them back up. Or I could quibble it is $1$. When it comes time to retrieve their hats every one would pick up their own. What kind of weirdo would try to take another persons hat? And why? And why would he think he'd get away with it. – fleablood Jan 19 at 17:22

We have six people: Joe, Enzo, $$W$$, $$X$$, $$Y$$, and $$Z$$.

If Joe and Enzo take their own hat, then there are $$4!$$ ways that $$W$$, $$X$$, $$Y$$, and $$Z$$ can take the other four hats, so that each of them took one hat.

And altogether there are $$6!$$ ways that these six people can take the hats, so that each person took one hat. If each of these possibilities is equally likely, then the probability that Joe and Enzo get their own hats is

$$\frac{4!}{6!}=\frac{1}{30}.$$

The order in which the people take the hats does not matter, given that we are effectively creating six random people-hat pairs, where every pairing is equally likely.

It may help to look at this from the perspectives of the hats, instead of the people: every time a person picks a hat, each of the leftover hats is equally likely to be taken by that person. Hence, every hat is as likely to be chosen at any point as any other hat, and by this symmetry between the hats, that also means that any specific hat is equally likely to be the first hat that is taken, as it is the second hat that is taken, etc. Thus, the probability of Enzo's hat being taken by the first person that takes a hat is the same as the probability as it being taken by the second person, and is thus $$\frac{1}{6}$$ for all people involved, no matter at what time they pick their hat. Therefore, it doesn't matter whether Enzo picks a hat first, second, fourth, or whatever: he always has a $$\frac{1}{6}$$ chance of ending up with his own hat.

Now, by the same logic, Joe also has a $$\frac{1}{6}$$ chance of ending up with his own hat, so you might think that the probability that both Joe and Enzo get their own hat is $$\frac{1}{6} \times \frac{1}{6}=\frac{1}{36}$$, but that would be incorrect! You can only multiply probabilities like that if the two events are independent, and in this case they are not: Enzo ending up with his own hat means that Joe can no longer end up with Enzo's hat, and has thus increased his chances of getting his own hat to $$1$$ out of $$5$$.

Indeed, the formula we should be using is:

$$P(J\land E) = P(J|E) \times P(E) = \frac{1}{5} \times \frac{1}{6}=\frac{1}{30}$$

Enzo has a hat. It doesn't matter if he picked first randomly out of the six hats on the table, or if he picked last and it was the only hat left. The end outcome is the same. Enzo has a hat.

There are $$6$$ possible hats it could be. And again it doesn't matter how it came about there are six possible hats it could be. And $$1$$ of the possibility is that it is his own hat.

Joe has a hat. It is a different hat than the hat Enzo has. Of the $$6$$ possible options for the hat that Enzo has, there are $$5$$ possible options for Joes hat. So there are $$6*5 =30$$ possible combinations for the two hats Enzo and Joe have.

Of the $$1$$ possibility that Enzo has has own hat that is $$1$$ possibility that Joe also has his own hat.

Now at this point, I could say, If we assume what hats the other four people have is equally likely so we only need to concern ourselves with Joe and Enzo, we could say. Of the $$30$$ options of their two hats, $$1$$ is that they both have their own so the probability is $$\frac 1{30}$$ that they both have their own hats.

BUT that is asking you to take a leap of faith that the hats the other four people have are equally likely.

Let's continue and show just that.

There is another person. Let's call her Consuella. She has a hat. Again it doesn't matter whether she got that hat by picking first or last. One way or another, she has a hat. And it is a different hat than the ones Enzo and Joe have.

For each off the $$6*5$$ options of hats that Enzo and Joe have there are $$4$$ options for the hat that Consuella has. So there are $$6*5*4=120$$ options of what hats the three people may have.

For each of the $$1$$ option that Enzo and Joe have their own hats, there are $$4$$ other options for the hat Consuella has. So there are $$1*4=4$$ options of ways the have hats where Enzo and Joe have their own hats.

Let's say the remaining people are named Carla, Michael, and Flippitybingbong.

For each of the $$6*5*4$$ options of hats for Enzo, Joe, COnsuella there are $$3$$ options of hats for Carla so there are $$6*5*4*3$$ ways the four can have hats.

For each of the $$1*4$$ ways that Enzo, Joe, Consuella can have hats and Enzo and Joe have there one hats there are $$3$$ possible hats that Carla may have, so there are $$1*4*3$$ ways the four have hats and Enzo and Joe have their own.

For each of the $$6*5*4*3$$ ways those four can have hats, there are $$2$$ options of hats for Michael, or $$6*5*4*3*2$$ ways those five have hats. For each of the $$1*4*3$$ ways that Enzo and Joe have there own hats and Consuella and Carlo have hats there are $$2$$ options of hats for Michael, or $$1*4*3*2$$ ways for the five to have hats and Enzo and Joe have their own.

And for each of $$6*5*4*3*2$$ ways there is one option of hat for Flippitybingbong. Even if Flippitybingbong chose first he will have chosen the one hat that the rest of the people in the future will not choose. By choosing first he assured that it would be the one the others will not have. Order doesn't matter. So there are $$6*5*4*3*2*1$$ ways total for the six people to have hats.

And for each of the $$1*4*3*2$$ ways where Joe and Enzo have there own hats and Consuella, Carla, and Michael have other hats, there is one option for the hat Flippitybingbong has. So there are $$1*4*3*2*1$$ ways that Joe and Enzo have their own hat.

So the probability that Joe and Enzo have their own hats is $$\frac {1*4*3*2*1}{6*5*4*3*2*1} =\require{cancel} \frac {1*\cancel{4*3*2*1}}{6*5*\cancel{4*3*2*1}} = \frac 1{6*5} = \frac 1{30}$$.