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 Fair Pairs (Posted on 2007-03-05)
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)

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 Solution | Comment 3 of 5 |
Let x=# of one color
Let x+a = # of other color

The probability of drawing two different colors is:

P = [x/(2x+a)]*[(x+a)/(2x+a-1)] + [(x+a)/(2x+a)]*[x/(2x+a-1)]

or

P = [2x²+2xa]/[4x²+4xa+a²-2x-a]

For a standard deck with no cards removed (i.e. x=26, a=0), P=26/51.

Setting P=0.5, the equation can be rewritten as follows:

2x = a(a-1)

with boundaries:  x>0, x=26, a>0, a<=(26-x).

Now, the trick is to find integers that satisfy x & a.  First, find a maximum value for a by  maximizing x = 25.

a²+a-50=0;  a≈6.6

Thus, a<=6.  Setting a=6 into the equation yields x=15, x+a=21.  Thus, 16 cards need to be removed (five of one color, eleven of another) to ensure that the probability of drawing a mismatched pair is exactly 50%.

[(15/36)*(21/35)] + [(21/36)*(15/35)] = 0.5

 Posted by hoodat on 2007-03-06 13:28:51

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