A box contains a number of socks, the number not exceeding 1993. Some socks are Red, some are Blue. The probability of randomly picking two socks of the same color is known to be 1/2.
What is the maximum number of Red socks that the box can contain?
Socks in a box

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 Joined: Tue Jun 08, 2004 8:55 pm
 Location: That Good Night
Say there are R red socks and B blue socks.
The odds of first picking a red sock is R / R+B,
and likewise, the odds of the first sock being blue is B / R+B.
The odds of the first two socks being red is (R/(R+B)) ((R1)/(R+B1))
and (B/(R+B)) ((B1)/(R+B1)) for the two socks being blue.
Since they are mutually exclusive, we can just add them and get a probablility of 0.5.
((R/(R+B)) ((R1)/(R+B1))) + ((B/(R+B)) ((B1)/(R+B1))) = 0.5
Manipulating and dropping the obvious parentheses for clarity:
(R/R+B)(R1) + (B/R+B)(B1) = 0.5(R+B1)
R(R1) + B(B1) = 0.5 (R+B1)(R+B)
2(R^2  R + B^2  B) = R^2 + BR  R + BR + B^2  B
2R^2  2R + 2B^2  2B = R^2 + 2BR + B^2  R  B
R^2  R + B^2 B = 2BR
R^2  2BR + B^2  R  B = 0
Help Me Mister Diophant!
The odds of first picking a red sock is R / R+B,
and likewise, the odds of the first sock being blue is B / R+B.
The odds of the first two socks being red is (R/(R+B)) ((R1)/(R+B1))
and (B/(R+B)) ((B1)/(R+B1)) for the two socks being blue.
Since they are mutually exclusive, we can just add them and get a probablility of 0.5.
((R/(R+B)) ((R1)/(R+B1))) + ((B/(R+B)) ((B1)/(R+B1))) = 0.5
Manipulating and dropping the obvious parentheses for clarity:
(R/R+B)(R1) + (B/R+B)(B1) = 0.5(R+B1)
R(R1) + B(B1) = 0.5 (R+B1)(R+B)
2(R^2  R + B^2  B) = R^2 + BR  R + BR + B^2  B
2R^2  2R + 2B^2  2B = R^2 + 2BR + B^2  R  B
R^2  R + B^2 B = 2BR
R^2  2BR + B^2  R  B = 0
Help Me Mister Diophant!

 Posts: 1462
 Joined: Wed Jun 02, 2004 11:04 pm
 Location: UK
Beleth's equation can be rewritten:
(R  B) (R  B) = R + B
This says that the difference in numbers between the colours, squared, is the total number of socks, so the total number must be a square, and the largest square less than 1993 is 1936.
The square root of 1936 is 44, so there must be (1936 + 44) / 2 = 990 socks of one colour and (1936  44) / 2 = 946 of the other.
So the maximum number of red socks is 990.
(R  B) (R  B) = R + B
This says that the difference in numbers between the colours, squared, is the total number of socks, so the total number must be a square, and the largest square less than 1993 is 1936.
The square root of 1936 is 44, so there must be (1936 + 44) / 2 = 990 socks of one colour and (1936  44) / 2 = 946 of the other.
So the maximum number of red socks is 990.

 Posts: 131
 Joined: Thu Jun 10, 2004 9:14 pm
 Location: Vancouver
Taking this one step farther...Beleth wrote:R^2  2BR + B^2  R  B = 0
R^2  2BR + B^2 = R + B
RB = sqrt(R+B)
Since the difference between red and blue socks is an integer, the total number of socks must be a perfect square. The largest square less that 1993 is 44^2 = 1936.
RB = 44 and R+B = 1936, so R=990 and B=946.
I believe that an equal number of socks of each colour doesn't work because once you have withdrawn a sock (say), you are now more less likely to draw a matching sock, so the odds are slightly less that 50%.
That's gotta be a damn big box of sox though. ;)
EDIT: Really, ceptimus's solution wasn't there when I wrote this post. Honest! :P

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 Location: Copenhagen