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There is $n$ pairs of socks that each sock having one mate,each pair is different from another pair .

There is $2$ drawers: left drawer and right drawer.

All the left socks are in the left drawer and all the right socks are in right drawer.

Every day I take $1$ sock from the left drawer and $1$ sock from the right drawer.

The socks are worn at that day and then thrown into a laundry basket,so after $n$ days the drawers will be empty.

What is the probability that in day number $k$ , I took a matching pair,when $1\le k \le n$ ?

I want to find the probability that a match occurs on day $k$?

How do I approach to that type of question?

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closed as off-topic by Did, Shailesh, C. Falcon, Zain Patel, Claude Leibovici Apr 23 '17 at 7:18

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    $\begingroup$ Are you asking for the probability that the FIRST match occurs on day $k$ or just the probability that a match occurs on day $k$? Also, are you replacing the socks each day? $\endgroup$ – lulu Apr 18 '17 at 12:33
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The chance you get a match on day $1$ is $\frac 1n$. Imagine you pull from the left first, then there are $n$ socks in the right drawer to choose from and $1$ matches.

The chance for any day is also $\frac 1n$. Imagine lining up all the socks in the order you will pull them. For a given day $k$, imagine swapping the first and $k$th stocks in both rows. Now day $k$ is a match if and only if the first day would have been a match before the swap. It is just confusing yourself to worry about whether day $1$ matched before you draw day $k$. You can do that, but it will still come out $\frac 1n$.

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  • $\begingroup$ But after $k$ days,I left with $n-k$ socks in each drawer. Why I don't need to check it? $\endgroup$ – Asaf Apr 21 '17 at 15:45
  • $\begingroup$ I showed why by the swap argument. The computation is not hard, either. On day $k$ you pull left sock. There is $\fraf{k-1}n$ chance you have pulled its mate, so $\frac {n-k}n$ its mate is available and $\frac 1{n-k}$ chance you pull it . Multiply them and you get $\frac 1n} $\endgroup$ – Ross Millikan Apr 21 '17 at 16:28
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Let's number the socks. We call them $\ell_1,\ell_2,\ldots, \ell_n$ for the left ones and $r_1,r_2,\ldots, r_n$ for the right ones. Now you pick one of each, so the set of possible outcomes is $\{(\ell_i,r_j) \mid 1 \leq i,j\leq n\}$, a set with $n^2$ elements. To take a matching pair, we have to take from the subset $\{(\ell_i,r_i) \mid 1 \leq i \leq n \}$, a set with $n$ elements.

As we are assuming that every pair has the same probability, can you compute the chance of picking a matching pair (not considering $k$ right now)? Let's for now call this probability $p$.

The second part of your questions asks: If we repeat the above experiment with chance to succeed $p$ every time, what is the chance that we first succeed after $k$ tries? Sounds familiar? (Hint: Bernoulli...)

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    $\begingroup$ depends if you put the socks back in to the drawers or not $\endgroup$ – JJR Apr 18 '17 at 12:32
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    $\begingroup$ Also the OP does not specify that the match on day $k$ be the first one. $\endgroup$ – lulu Apr 18 '17 at 12:32
  • $\begingroup$ And he also did not mention that all socks can be distinguished, so yes, I made some assumptions... :) $\endgroup$ – Dirk Apr 18 '17 at 12:34
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    $\begingroup$ @lulu - ignoring matches or not on other days, the probability of a match on day $k$ is the same as on day $1$. $\endgroup$ – Henry Apr 21 '17 at 15:17
  • $\begingroup$ @Henry Yes, but the probability of getting the first match on day $k$ is different than the probability of getting a match on day $k$. $\endgroup$ – lulu Apr 21 '17 at 15:18

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