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Deal or No Deal

  1. Feb 21, 2009 #1
    The contestant has 6 suitcases unopened. Two of them are disadvantages and contain large numbers--which would count against him. The remaining 4 contain small numbers to his advantage.

    The contestant may quit, or if he chooses to go on, he must choose to open two suitcases. (It is disadvantages to pick even 1 large number out of his two choices.)

    The contestant sees that four of the cases are to his advantage and only two against him, and so he decides to play on and open two cases. He remarks to the audience, "The numbers favor me!"

    But do they? What is the most probable outcome?
    Last edited: Feb 22, 2009
  2. jcsd
  3. Feb 22, 2009 #2


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    4*3 ways to choose both small; 4 * 2 * 2 ways to choose one small and one big; 2 * 1 ways to choose both big. Overall, it's a 40% chance for 'good stuff', 53% chance for 'bad stuff', and a 7% chance for 'very bad stuff'.
  4. Feb 22, 2009 #3
    In other words, yeah, the odds probably don't favor his playing.
  5. Feb 22, 2009 #4
    csprof2000 In other words, yeah, the odds probably don't favor his playing.

    IT CERTAINLY WORKED THAT WAY! He came on with about $109,000, hit one small and one big, and the next thing I remember he left the stage with only $66,000. In other words his outcome was in the majority case, that is, the CRGreenhouse case of 53%.

    I have noticed that, mostly, if the contestant can reach $100,000 he is well advised to quit, at least in my book, and those unfortunate big cases have a way of happening that might not be expected, at least if we go individually case by case.

    If we went further with this and supposed he did three cases, we would have (4/6)(3/5)(2/4) = 24/120 = 20% probability of winning all three. Yet the contestant, we know, had a individual chance of 50% to 67% in all three cases.

    The most likely outcome in three cases would be to hit two small and one big = 60%. The other case of two big and one small would be the remaining 20%.
    Last edited: Feb 22, 2009
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