All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
perplexus dot info

Home > Probability
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 | 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)]


P = [2x+2xa]/[4x+4xa+a-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.

a+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
Please log in:
Remember me:
Sign up! | Forgot password

Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (2)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Copyright © 2002 - 2018 by Animus Pactum Consulting. All rights reserved. Privacy Information