# THE “MEN & HATS” PROBABILITY PARADOX 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?”