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Probability Question has my stumped! Can you help

  1. Oct 12, 2012 #1
    Hello so here is the question:

    You have a deck of 97 cards and I will pay you $10 if I draw 4 cards and they are in ascending order (not necessarily consecutive order) and you pay me $1 if they are not. Would you play?

    I am thinking a few different ways:

    A the easiest: its symetric so equal chances higher or lower so the chance to get 5 cards each one higher than the next is 1/1 * 1/2 * 1/2 * 1/2 = 1/8

    or b. the average pick out of the cards is 49 so the next has to be 50 or higher the average of 50-97 is the second choice(that has a probability of 47/96).. the third has to be higher than that average second pick which is 73 average of 73-95 is 85 and so on

    so the prob is 1*47/96*26/95*10/94

    Thank you
     
  2. jcsd
  3. Oct 12, 2012 #2

    Ray Vickson

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    Your expression above is 13/912 ≈ 0.01425, but I get something very different: I get p{win} = 4465/18624 ≈ 0.2397.

    The point is that when the first number is small, there are lots of ways to choose the second number (to get a win), but when the first number is large the number of choices for the second number is much more limited.

    RGV
     
  4. Oct 12, 2012 #3
    Are the 97 cards all distinct? If so, the probability of drawing 4 in ascending order is simply 1/4!, approximately 0.04167, since all 4! orderings are equally likely.
     
  5. Oct 13, 2012 #4

    Ray Vickson

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    No fair, you did it the easy way! My previous post was nonsense, as I had made the stupid mistake of forgetting some factors, etc. When I correct the results (but using my lengthy method) I end up with P{win} = 1/24, as you say.

    RGV
     
  6. Oct 13, 2012 #5
    I didn't necessarily do it the easy way the first time, heh heh.
     
  7. Oct 13, 2012 #6

    Ray Vickson

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    There must be something a bit deeper going on here. I used
    [tex]P_{\text{win}}= \sum_{i=1}^{94} \frac{1}{97} \sum_{j=i+1}^{95} \frac{1}{96}
    \sum_{k=j+1}^{96} \frac{1}{95} \sum_{l=k+1}^{97}\frac{1}{94} = \frac{1}{24} = \frac{1}{4!}.[/tex]
    Of course, 1/4! is the volume of the region
    [tex] R = \{(x_1,x_2,x_3,x_4) : 0 \leq x_1 \leq x_2 \leq x_3 \leq x_4 \leq 1 \}.[/tex]
    The volume of R can be computed by nested integrations; somehow, we get the same value by replacing the integrations by summations, and with slightly changing denominators (1/97, 1/96, 1/95, ... ). Surely this cannot be coincidental.

    RGV
     
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