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+a-1)] + [(x+a)/(2x+a)]*[x/(2x+a-1)]
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
. First, find a maximum value for a
by maximizing x = 25.
Thus, a<=6. Setting a=6
into the equation yields x=15
. 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