Secret Santas IProbability puzzles require you to weigh all the possibilities and pick the most likely outcome.
A group of about twenty friends decide to exchange gifts as secret Santas. Each person writes their name on a piece of paper and puts it in a hat and then each person randomly draws a name from the hat to determine who has them as their secret Santa.
What is the probability that at least one person draws their own name?
HintThe probability converges on a single value as the number of people approaches infinity. At ten people the probability is accurate to better than 1 millionth for any number more than ten people.
Answer1 - 1/e
or approximately 0.63212
where e is the mathematical constant (e ~ 2.71828).
This comes from the number of derangements (permutations in which no element appears in its original position). There are n! ways to draw n names out of the hat. There are [n!/e + .5] derangements of n elements (where [x] is floor x--the integer portion of x). This gives:
[n!/e + .5] / n!
as the probability of nobody drawing their name.
Ignoring the .5 added for rounding (which becomes increasingly insignificant as n increases) this gives
(n!/e) / n!
= n!/e * (1/n!)
Subtract this from 1 to get the probability that somebody draws their own name.
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