Red socks and Blue socks

Probability Level 5

A drawer contains a mixture of red socks and blue socks, at most 1950 in all. It so happens that, when two socks are selected randomly without replacement, there is a probability of exactly 1 2 \frac{1}{2} that both are red or both are blue. What is the largest possible number of red socks in the drawer that is consistent with this data?

Image Credit: Flickr Alison Clayton .


The answer is 990.

Alan Yan by Alan Yan , 20, USA

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2 solutions

Alan Yan
Nov 4, 2015

Let there be r r and b b red and blue socks respectively. Since we want to find the maximum number of red socks, we can assume WLOG that r ≥ b r \geq b . Thus, the probability of selecting both red or both blue is ( r 2 ) + ( b 2 ) ( r + b 2 ) = 1 2 ⟹ ( r − b ) 2 = r + b \frac{\binom{r}{2} + \binom{b}{2} }{ \binom{r+b}{2}} = \frac{1}{2} \implies (r-b)^2 = r + b

This implies that r + b r+b is a perfect square, let this be n 2 n^2 . Thus we have r + b = n 2 r − b = n r = n 2 + n 2 \begin{aligned} r+b & = n^2 \\ r - b & = n \\ r & = \frac{n^2 + n}{2} \end{aligned}

Now by the condition, we have that n 2 ≤ 1950 n^2 \leq 1950 . The greatest integer n n can be is 44 44 . Plugging this in, we find that r = 990 r = \boxed{990} . We check to see that this is indeed a solution.

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I did a recent problem to do this with my University. This is a second way you can do it just like Alan's: Proof: If there is a 1/2 chance of picking a matching pair, the must be an equal number of mismatching pairs. Then, ( r 1 ) ( b 1 ) = ( r 2 ) + ( b 2 ) r b = r ( r − 1 ) 2 + b ( b − 1 ) 2 2 r b = r 2 − r + b 2 − b r + b = r 2 − 2 r b + b 2 r + b = ( r − b ) 2 \begin{aligned} \binom{r}{1}\binom{b}{1} & = \binom{r}{2} + \binom{b}{2} \\ rb & = \frac{r(r-1)}{2} + \frac{b(b-1)}{2} \\ 2rb & = r^2-r+b^2-b \\ r+b & = r^2-2rb+b^2 \\ r+b & = (r-b)^2 \end{aligned} Using the rest of Alan's proof finds the number of socks to be: r = ( n + 1 2 ) b = ( n 2 ) \begin{aligned} r& =\binom{n+1}{2} \\ b& =\binom{n}{2} \\ \end{aligned}

Jaleb Jay - 5 years, 7 months ago

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I also started in a slightly different way: I changed the 1/2 statement to, equivalently, saying that the probability of first picking a red and then picking a black is 1/4. That is to say, 4 r b = ( r + b ) ( r + b − 1 ) 4rb = (r+b)(r+b-1) , etc.

Peter Byers - 5 years, 7 months ago

Oh my god... I answered with the maximum TOTAL. hitting myself right now... Nice solution though. Mine was much less elegant.

Brandon Monsen - 5 years, 6 months ago
Edgar Wang
Nov 21, 2015

There was a similar problem to this one on the 2015 COMC (it was B4) There are n sheep in a barn, where n lies between 2000 and 2100. We split the sheep into 2 barns. We randomly pick two sheep: the probability that both of them are in different barns is exactly 1/2. Determine n.

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