Pairs of Socks

  1. A sock drawer contains some number of red and blue socks (and no other color.) In each case below, find the number of socks of each color in the drawer. A pair of socks is removed at random.
    1. The probability they are both red is 1/3, and the probability they are both blue is 1/6.
    2. The probability they are both red is 1/2, and the probability they are both blue is 1/12.
    3. The probability they are both red is 1/2, and the probability they are both blue is 1/14.
  2. Some number of green socks are now also in the drawer. Again, find the number of socks of each color in the drawer. A pair of socks is removed at random.
    1. The probability they are both red is 1/3, and the probability they are both blue is 1/15.
    2. The probability they are both red is 1/5, and the probability they are both blue is 1/15.
    3. The probability they are both red is 1/6, and the probability they are both blue is 1/12.
    4. The probability they are both red is 1/8, and the probability they are both blue is 1/12.
Note: The problems above are not necessarily listed in order of difficulty.

Source: Mensa's Nearly Impossible Baffling Brainteasers #36, plus original extensions.


Solutions were received from Joseph DeVincentis, Sandy Thompson, Denis Borris, Philippe Fondanaiche, John Hewson, Allen Druze, Dave Bachtel, and Alan Zimmermann.

The answers are:

Part 1:
  1: 21 red, 15 blue
  2: 85 red, 35 blue
  3: 15 red, 6 blue
Part 2:
  1: 6 red, 3 blue, 1 green
  2: 3 red, 2 blue, 1 green
  3: 4 red, 3 blue, 2 green
  4: 6 red, 5 blue, 5 green
Here is Philippe's explanation for solving. Many people used a brute-force computer search.
We have to find a and b integers representing respectively the numbers of red 
and blue socks such as:
a*(a-1)/(a+b)*(a+b-1) = p   (1)
b*(b-1)/(a+b)*(a+b-1) = q   (2)
(1) and (2) ==> q*a(a-1) = p*b*(b-1)  (3)
So for each couple of probabilities (p,q), we determine easily all the 
couples (a,b) which satisfy the relation (3) and among them, we keep those 
which satisfy (1) (or (2) ).
His solution for the second part just added an additional variable for the green socks.
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