A friend told me this math prob:

I'm asked to calculate the probability of every postman put the letter in the wrong mailbox. There are infinite postmans and infinite mailboxes (only one is right for each letter and all the others are wrong).

I'm not a Mathematician but I told him the probability is 0.

  • $\begingroup$ How many letters each postman has to deliver? $\endgroup$
    – jimjim
    Jan 29 '12 at 9:20
  • 3
    $\begingroup$ Possibly related: If there were $n$ letters and $n$ mailboxes, then the probability approaches $\frac{1}{e}$ as $n\to\infty$. en.wikipedia.org/wiki/… I'm not sure what this question means. $\endgroup$ Jan 29 '12 at 9:21
  • $\begingroup$ I think each postman has only one letter to deliver. $\endgroup$
    – KingBOB
    Jan 29 '12 at 9:27
  • $\begingroup$ @Jonas's answer refers to $n$ letters put in $n$ mailboxes with the condition that no letter lands in the proper mailbox and with the additional condition that each mailbox receives exactly one letter. If one omits this additional condition, the probabilities for some given finite $n$ are different but their limit when $n\to\infty$ is $1/\mathrm e$ as well. $\endgroup$
    – Did
    Jan 29 '12 at 9:51

You have to treat this a limit.

If there are $n$ postmen delivering one letter each, and $m$ mailboxes, so if the postmen deliver at random then the probability one postman puts the letter in the wrong mail box is $\left(1-\frac{1}{m}\right)$, and so (assuming independence) the probability all the letters go in the wrong mail boxes is $$\left(1-\frac{1}{m}\right)^n.$$

You want the limit of this as $n \to \infty$ and $m \to \infty$. Sadly that limit is not well defined.

If $n = m $ then you can use $$\lim_{n \to \infty} \left(1-\frac{1}{n}\right)^n = e^{-1} \approx 0.367879$$ and this may be what your friend was thinking.

But if $\lim n/m = k$ then $\lim_{n \to \infty} \left(1-\frac{1}{m}\right)^n = e^{-k}$ and this can take any value between $0$ and $1$, achieving the extremes if $k$ is infinite or zero.


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