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Fair Pairs (Posted on 2007-03-05) Difficulty: 2 of 5
The probability of drawing a pair of cards from a standard deck of 52 and having those cards be different colors is 26/51, slightly over 50%.

What is the smallest number of cards that need to be removed from the deck so that the probability of drawing a mismatched pair is exactly 50%?

See The Solution Submitted by Brian Smith    
Rating: 4.6667 (3 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution Triangles (solution) | Comment 1 of 5

a=number of red card, b= number of black cards

One can either get a red then black or black then red.  The probablity of each is the same so the probability is 1/4.  We have:

(a/(a+b))*(b/(a+b-1))=1/4 which simplifies to

a^2 +(-1-2b)a + (b^2-b) =0 using the quadratic equation

a= (2b+1 } (8b+1))/2

8b+1 is a perfect square whenever b is a triangular number:
Solutions

b=0, a=1 or 0
b=1, a=3 or 0
b=3, a=6 or 1
b=6, a=10 or 3
b=10, a=15 or 6
b=15, a=21 or 10
b=21, a=28 or 15
b=28 (We can stop here.  Too many cards or one color)

The solution removing the least cards is 15 of one color and 21 of the other.  This is 36 cards so 16 have to be removed.

It is very interesting to me that all of the solutions are consecutive triangular numbers.


  Posted by Jer on 2007-03-05 12:35:26
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