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Probability Problem

Posted By: joe freedman
Date: Thursday, 27 August 2009, at 8:26 p.m.

In Response To: Probability Problem (Gregg Cattanach)

Let's say you deal me two cards from a "deck" of 4 aces (each of a different suit) and 4 kings (same). (1) At that point, the odds of 2 aces are 4/8 x 3/7 = 3/14. (2) You then ask me to tell you if at least one of the cards is an Ace. I truthfully say, "at least one of the cards is an Ace." At that point, the odds of two aces is 3/11, because the combinations involving only two kings have been eliminated. (3) You then tell me: "show me your ace. If you have two aces, select one randomly and show it to me." I show you an ace.

How can my showing you an ace change the odds that the unseen card is an ace? I've already told you that one of the cards is an ace, so my showing you an ace should not reveal any new information. I don't think this is the same as the Monty Hall.

Messages In This Thread

• Probability Problem
  poseidon -- Thursday, 27 August 2009, at 1:26 p.m.
  • Probability Problem
    misja -- Thursday, 27 August 2009, at 1:42 p.m.
    • Probability Problem
      Stick -- Thursday, 27 August 2009, at 3:31 p.m.
    • Probability Problem
      Petter Bengtsson -- Thursday, 27 August 2009, at 3:46 p.m.
  • Probability Problem
    Gregg Cattanach -- Thursday, 27 August 2009, at 3:49 p.m.
    • Probability Problem
      Bob Koca -- Thursday, 27 August 2009, at 4:19 p.m.
    • Probability Problem
      Chuck Bower -- Thursday, 27 August 2009, at 4:46 p.m.
    • Probability Problem
      Jason Lee -- Thursday, 27 August 2009, at 7:54 p.m.
  • Probability Problem
    Neil Robins -- Thursday, 27 August 2009, at 4:14 p.m.
    • Probability Problem
      Gregg Cattanach -- Thursday, 27 August 2009, at 5:37 p.m.
      • Probability Problem
        Bob Koca -- Thursday, 27 August 2009, at 5:56 p.m.
        • Probability Problem
          Gregg Cattanach -- Thursday, 27 August 2009, at 6:07 p.m.
          • Probability Problem
            Bob Koca -- Thursday, 27 August 2009, at 7:42 p.m.
            • Probability Problem
              Gregg Cattanach -- Thursday, 27 August 2009, at 8:09 p.m.
              • Probability Problem
                joe freedman -- Thursday, 27 August 2009, at 8:26 p.m.
                • Probability Problem
                  poseidon -- Thursday, 27 August 2009, at 8:37 p.m.
                • Probability Problem
                  Jason Lee -- Thursday, 27 August 2009, at 8:47 p.m.
              • Probability Problem
                Gregg Cattanach -- Friday, 28 August 2009, at 1:58 a.m.
                • Probability Problem
                  Paul Vianna -- Friday, 28 August 2009, at 2:11 a.m.
                  • Probability Problem
                    Chuck Bower -- Friday, 28 August 2009, at 4:06 p.m.
            • monte hall called this
              garyo -- Saturday, 29 August 2009, at 7:51 a.m.
  • Probability Problem
    Chuck Bower -- Thursday, 27 August 2009, at 5:16 p.m.
  • Probability Problem
    Timothy Chow -- Thursday, 27 August 2009, at 6:03 p.m.
    • Probability Problem
      Bob Koca -- Thursday, 27 August 2009, at 6:06 p.m.
      • Probability Problem
        poseidon -- Thursday, 27 August 2009, at 6:18 p.m.
  • Probability Problem
    joe freedman -- Thursday, 27 August 2009, at 6:35 p.m.
    • Probability Problem
      joe freedman -- Thursday, 27 August 2009, at 6:39 p.m.
  • Probability Problem
    Jason Lee -- Thursday, 27 August 2009, at 7:57 p.m.
  • Probability Problem
    Bob Koca -- Friday, 28 August 2009, at 2:13 a.m.
    • Probability Problem
      poseidon -- Friday, 28 August 2009, at 10:29 a.m.