I have a solved question from Ross as stated below.

Q : Suppose that each of three men at a party throws his hat into the center of the room. The hats are first mixed up and then each man randomly selects a hat. What is the probability that none of the three men selects his own hat?

Sol: We shall solve this by first calculating the complementary probability that at least one man selects his own hat......

I want to start with basics and identify the sample space first.

$$\mathrm{Space} = \{h_1p_1, h_1p_2, h_1p_3, h_2p_1, h_2p_2, h_2p_3, h_3p_1, h_3p_2, h_3p_3\}$$

where $h_ip_j$ is the event of picking up a hat of person $i$ by person $j$.

To satisfy the condition that nobody picks his own hat, $i$ should not be equal to $j$.

Why the complement is "at least one man selects his own hat" and not that "all guys select their own hat".

  • 1
    $\begingroup$ You can easily generalizes the answer of JonMark Perry for $n$. Note that $P_0$ is the number of derangements divided by $n!$, that $P_{n - 1}$ is zero, and that $P_n$ is $1/{n!}$. Other probabilities can be expressed with partial derangements, which can be expressed with full derangements. $\endgroup$
    – BoLe
    Commented Aug 8, 2017 at 9:18

2 Answers 2

  • The sample space that you defined is not relevant to the question.

Each event in the space must be one possible outcome of the experiment "throwing hats and picking them at random". One of these event, for instance is "man 1 gets hat C, man 2 gets hat A, man 3 gets hat B", which we can note "A2B3C1".

There are 6 such events in your universe (which is in bijection with the set of all permutations of $\{1,2,3\}$): $U=\{A1B2C3,A1B3C2,A2B1C3,A2B3C1,A3B1C2,A3B2C1\}$

The event "nobody picks his own hat" is the subset $E=\{A2B3C1,A3B1C2\}$

  • The complement of "nobody picks his own hat" is (in english) the sentence that say the opposite of "nobody picks his own hat" i.e. "at least someone picks his own hat". One is true if and only if the other is false.

In maths, it is $U-E=\{A1B2C3,A1B3C2,A2B1C3,A3B2C1\}$

When you run the experiment, one and only one of the two events "nobody picks his own hat" and "at least someone picks his own hat" will happen: that is the very definition of "complementing events".


Let the number of men who get their own hat be represented by the random variable $M$.

Then, using $==$ for 'is exactly': $$P(M==0)+P(M==1)+P(M==3)=1$$

$P(M==2)=0$ because if two men get their own hat, so does the third.

The complement of $P(M==0)$ is $1-P(M==0)$, and

$$1-P(M==0)=P(M==1)+P(M==3)=P(M\ge 1)$$


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .