Imagine that each of N=6 men has a hat. Also imagine that these hats are identical except that each man’s name is written inside his hat. Finally, imagine that the 6 hats are taken up and then later, because they look alike, randomly returned to the men.

As the 6 hats are returned to the 6 men, there’s a chance that no man will receive his own hat. The chance of this happening is a tad greater than 1 in 3. To be more precise, the probability (to 3 decimal places) of all 6 hats going to the *wrong* individuals is .368.

Now, let’s add a new wrinkle to this imaginary situation. Suppose the number of men (each with a hat) is greater than 6. What if there are 7 men? Or 8? Or more? As N increases, what happens to the probability that no hat will be returned to its proper owner? Some people guess that this probability goes up as N increases. Others guess that this probability goes down.

Both thoughts are wrong.

That’s because the likelihood of no correct “match” is virtually the same for any N > 5, whether N = 6 or N = 600 or N = 600,000!

The actual probability (p) of having no hat returned to its proper owner is given by this formula:

p = 1/(2!) – 1/(3!) + 1/(4!) – 1/(5!) + . . .

where there are N-1 terms on the right side of the equation. With the symbol “!” standing for “factorial,” we could rewrite the above formula as

p = 1/2 – 1/6 + 1/24 – 1/120 + . . .

As either of the above formulas shows, additional terms on the right side of the equation have a smaller and smaller impact on the value of p. Moreover, the drop-off of this impact is sharp, not gradual. This fact is made clear by the following chart showing the value of p, to 6 decimal places, for the case where N = 2, 3, 4, … , 10.

N = 2 p = .500000

N = 3 p = .333333

N = 4 p = .375000

N = 5 p = .366666

N = 6 p = .368054

N = 7 p = .367857

N = 8 p = .367882

N = 9 p = .367879

N = 10 p = .367879

It should be noted that this puzzle question is sometimes referred to as “Montmort’s Problem.” Montmort was a Frenchman who studied the probability behind a game called “Treize.” (*Treize* is the French word for 13.) In its original form, the puzzle question dealt with a jar containing identical balls numbered 1, 2, 3, … , 13. If balls are randomly pulled out of the jar, one at a time, the puzzle question was stated like this: “What’s the probability that the 1st ball taken from the jar will *not* be the ball numbered 1, that the 2nd ball will *not* be the ball numbered 2, and so on, with the end result being that no number on any ball matches the order in which the ball is removed from the jar?”