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Suppose that instead of having one drawer, you have two drawers. Each drawer has some socks that are white and some that are black. Drawer 1 has w black socks and x white socks. Drawer 2 has y black socks and z white socks. w+x=y+z. If you take out all the socks randomly, 1 each from each drawer to make pairs, until both drawers are empty, what is expected (or average) number of times you pulled out a black sock from drawer 1 and 2? Is it [{w/(w+x)}{y/(y+z)}](w+x)?

mark
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2 Answers2

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We change notation a little. Let $n$ be the number of socks in each drawer, and suppose the first drawer has $a$ black and the second has $b$.

On the $i$-th draw we pull out a sock from each drawer. For $i=1$ to $n$, let $X_1=1$ if both are black, and let $X_i=0$ otherwise. We want $E(Y)$, where $Y=X_1+\cdots +X_n$. By the linearity of expectation this is $E(X_1)+\cdots+E(X_n)$.

Finally, $E(X_i)=\Pr(X_i=1)=\frac{a}{n}\cdot\frac{b}{n}$. So $E(Y)=\frac{ab}{n}$.

Remark: If I am reading your formula correctly, you came to the same conclusion.

André Nicolas
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  • The socks aren't being put back in the drawer, correct? In that case, it would seem that the expectation of $X_1$ is what you gave for $X_i$, but not necessarily for $i>1$. – cxseven May 30 '15 at 21:40
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    I assumed socks were not being put back. All the $X_i$ have the same distribution. Imagine that the socks have secret ID numbers. All sequences of ID numbers in the draw from the left drawer are equally likely, as are all sequences of the draw from the right drawer. So we are just as likely to get the pair $(i,j)$ on the $k$-th (double) draw as on the first draw. Note that the $X_i$ are not independent. However, linearity of expectation holds in general, independence not required. – André Nicolas May 30 '15 at 21:45
  • That is correct, the socks are not being put back. I am trying to see if for any large n (your notation) and given enough a and b, if a black sock pair can exist. Essentially this is a Goldbach problem, to see if an even number k has two prime pairs that when added together equal that even number. The first drawer are the odd numbers from 3 to k/2 and the second drawer has odd numbers from k/2 to k. And the primes are spread throughout. – mark May 30 '15 at 21:49
  • The kind of reasoning you mention can be used to find out how likely Goldbach is to hold, for large enough even numbers, and objects distributed "like" the primes. – André Nicolas May 30 '15 at 21:58
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A drawer contains 4 red socks and 4 blue socks. Find the lowest number of socks that must be drawn from the drawer to be assured of having a pair of red socks.

This is not a probability question so I'm not sure how to approach it. I am thinking that I would need to pick at most 6 because the first 4 could all the blue. Then the only socks left are red. After two more pulls, I am guaranteed to have a pair of red. I am wondering if I am overthinking this though.

Linda Lofton
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