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%?
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+a1)] + [(x+a)/(2x+a)]*[x/(2x+a1)]
or
P = [2x²+2xa]/[4x²+4xa+a²2xa]
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(a1)
with boundaries: x>0, x=26, a>0, a<=(26x).
Now, the trick is to find integers that satisfy
x &
a. First, find a maximum value for
a by maximizing x = 25.
a²+a50=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 20070306 13:28:51 