Tuesday, June 19, 2007

A Closet Full of Shoes

A closet contains 10 pairs of shoes (20 shoes altogether). If 6 shoes are chosen at random, what is the probability that there will be no matching pair in the sample?


  1. You know I love math problems, Mike. I thought about you yesterday when arguing with a friend that the chance of getting a bad hand in cards is no better or worse as a result of previous hands -- statistical independence, if you will.

    Any shoe will do for your first pick. For your second pick, there are 18 out of the 19 remaining shoes that will keep your streak alive. For your third pick, there are 16 out of the remaining 18, and so forth, so the probability looks something like:

    20/20 * 18/19 * 16/18 * 14/17 * 12/16 * 10/15 = 112/323

    That comes to about 34.67%.

  2. I wish the other folks had left some work; it would be interesting to see what went wrong.

    I agree 100% with Abe, and prefer the clarity of his conditional probability method (and his reduction of fractions.. no calculator until - maybe - the long division)

  3. Are each of the shoes unique? Or just consider that they are left and right?

    This is considering the latter..
    100% you don't get a pair first time.
    9/19 that you don't get a pair, 2nd time
    8/18 that you don't get a pair, 3rd time
    7/17 that you don't get a pair, 4th time
    6/16 that you don't get a pair, 5th time
    5/15 that you don't get a pair with the sixth pick.
    Therefore, 1.08% that you don't choose a left and right.
    And I like abe's calculations.

  4. And I like there are several different ways to think about this problem. I'm glad I asked it. ;-)


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