My shoe is missing!

Probability Level 4

A group of N N people with distinct pairs of shoes attended a party. But when they left, they randomly took a pair of one left shoe and one right shoe. Let the probability of none of them getting both shoes of the same pair be P P .

Let lim ⁡ N → ∞ P = X \displaystyle \lim_{N\rightarrow \infty} P = X .

Find the value of ln ⁡ ( X ) \ln(X) .


The answer is -1.

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Zawad Abdullah by Zawad Abdullah , 23, Bangladesh

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1 solution

Once everybody (of the N N people) chooses a left shoe, the number of derangements of the right shoe is given as

D N = N ! ∑ r = 0 N ( − 1 ) r r ! D_N =N! \sum_{r=0}^{N} \frac{(-1)^r}{r!}

The total number of permutations of N N object would be N ! N! .

Thus, the probability that none of them bets both shoes of the same pair is

P N = ∑ r = 0 N ( − 1 ) r r ! P_N = \sum_{r=0}^{N} \frac{(-1)^r}{r!}

and

lim ⁡ N → ∞ P N = ∑ r = 0 ∞ ( − 1 ) r r ! = e − 1 = X \lim_{N \to \infty} P_N = \sum_{r=0}^{\infty}\frac{(-1)^r}{r!} = e^{-1}=X

Thus, ln ⁡ ( X ) = − 1 \ln(X) = \boxed{-1}

5 Helpful 0 Interesting 0 Brilliant 0 Confused

Sir Can you please tell us how did you got the expression for P N P_N

Syed Baqir - 5 years, 9 months ago

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