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? Answer1 - 1/eor 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!) = 1/e Subtract this from 1 to get the probability that somebody draws their own name. Hide What Next?
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