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Probability of Two People Drawing Each Other’s Names in a Hat Game

July 12, 2025Socializing4098
Conclusion When a group of n people draw names from a hat, the probabi

Conclusion

When a group of n people draw names from a hat, the probability that two specific individuals draw each other’s names is a fascinating problem in probability theory. This situation can be analyzed through the lens of permutations and derangements. In this article, we will explore the probability of two specific people drawing each other's names and how this can be generalized to the study of permutation cycles.

Probability of Two Specific People Drawing Each Other's Names

Let’s denote the group by n people and the two specific individuals as A and B. The total number of ways to arrange the names is given by n! (n factorial). For A to draw B's name and B to draw A's name, the remaining n-2 people can be arranged in any of (n-2)! ways. Thus, the probability P that exactly A and B draw each other’s names is:

$$P frac{(n-2)!}{n!} frac{1}{n(n-1)}.

Therefore, the probability is simply the reciprocal of the product of n and n-1.

Generalizing to Permutation Cycles

The problem can be generalized to find the probability that no two specific people swap names. This is equivalent to finding the probability of having no 2-cycles in the permutation. We can use generating functions to solve this problem.

The exponential generating function for permutations by cycles is given by:

$$sum_{rgeq 0} frac{1}{r!} left(sum_{kgeq 0} (k-1)! frac{x^k}{k!} right)^r.

This can be simplified to:

$$sum_{rgeq 0} frac{1}{r!} left(sum_{kgeq 0} frac{x^k}{k} right)^r e^{ln(frac{1}{1-x})-frac{x^2}{2}} e^{-frac{x^2}{2}} (1-x)^{-1}.

The ordinary probability generating function for permutations with no 2-cycles is:

$$sum_{ngeq 0} (1-text{Prob}_{n}^{text{no 2-cycles}}) x^n e^{-frac{x^2}{2}} (1-x)^{-1} - frac{x^2}{2} e^{-frac{x^2}{2}} (1 x^{-1}).

Thus, the coefficient of x^n gives the probability:

$$text{Prob}_{n}^{text{no 2-cycles}} sum_{r0}^{lfloor n/2 rfloor} frac{(-2)^{-r}}{r!}.

For a group of 23 people, the probability is approximately:

$$1 - text{Prob}_{23}^{text{no 2-cycles}} 1 - left( sum_{r0}^{11} frac{(-2)^{-r}}{r!} right) frac{32165963699}{81749606400}.

Derangements and Applications

A derangement is a permutation where no element appears in its original position. This is relevant here as when no one draws their own name, we are dealing with derangements. The limit of the probability of derangements as n approaches infinity is given by:

$$lim_{n to infty} text{Prob}_{n}^{text{no 2-cycles}} e^{-1/2}..

This approximation is quite accurate even for moderate values of n, as shown by the examples provided.

Furthermore, the generating function method can be generalized to calculate the probability of a permutation having exactly s 2-cycles. This is given by:

$$text{Prob}_{n}^{text{exactly s 2-cycles}} sum_{rs}^{lfloor n/2 rfloor} frac{(-1)^{r-s}}{2^r r! (r-s)!}.

This formula allows us to calculate the exact number of permutations with any specified number of 2-cycles.

The calculations for the number of permutations with exactly s 2-cycles can be found in the OEIS sequence A114320, which verifies the correctness of the generating function approach.