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Math Help - Rigged Dice - Expected Sum

  1. #1
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    Unhappy Rigged Dice - Expected Sum

    1. A pair of dice (A & B) are rigged to have a higher than normal chance of getting snake eyes (double ones).
    • For each dice there is 3 times the probability of getting a 1 as the other numbers. So, for dice A, the probability of getting a 1 is 3/8 while the probability of getting each other number is 1/8.
    E.g., there are 4 ways to get a 9 (3;6, 6;3, 4;5, 5;4), so the probability of getting a 9 is 4 x 1/8 x 1/8 = 1/16. There are 3 ways to get a 4 (1;3, 3;1, 2;2), so the probability of getting a 4 is 3/8 x 1/8 + 1/8 x 3/8 + 1/8 x 1/8 = 7/64.

    Question:
    What is the expected sum of Dice A & B, i.e., the mean? [Use the ∑x p(x) formula. The expected sum of two fair dice is 7. Because the rigged dice have a greater chance of giving ones, the mean will be less than 7.]
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    Quote Originally Posted by ccdelia7 View Post
    1. A pair of dice (A & B) are rigged to have a higher than normal chance of getting snake eyes (double ones).
    • For each dice there is 3 times the probability of getting a 1 as the other numbers. So, for dice A, the probability of getting a 1 is 3/8 while the probability of getting each other number is 1/8.
    E.g., there are 4 ways to get a 9 (3;6, 6;3, 4;5, 5;4), so the probability of getting a 9 is 4 x 1/8 x 1/8 = 1/16. There are 3 ways to get a 4 (1;3, 3;1, 2;2), so the probability of getting a 4 is 3/8 x 1/8 + 1/8 x 3/8 + 1/8 x 1/8 = 7/64.

    Question:
    What is the expected sum of Dice A & B, i.e., the mean? [Use the ∑x p(x) formula. The expected sum of two fair dice is 7. Because the rigged dice have a greater chance of giving ones, the mean will be less than 7.]
    I suggest that you first construct a grid showing the probability of all possible pairs of numbers from the two dice.
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  3. #3
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    I have a grid of all 36 possible outcomes set up. Now what should I do next?
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    I got 5. Since the median of the set of possible rolls is 2.5, the sum of 2 2.5's is 2 x 2.5 = 5. Is that a correct approach?
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    Quote Originally Posted by ccdelia7 View Post
    I have a grid of all 36 possible outcomes set up. Now what should I do next?
    Now calculate the probability of each pair (i, j) of numbers (i is the number of spots on die A and j is the number of spots on die B).

    eg. The probability of (1, 1) is \left( \frac{3}{8}\right) \cdot \left( \frac{3}{8}\right) = \frac{9}{64}.

    eg. The probability of (1, 2) is \left( \frac{3}{8}\right) \cdot \left( \frac{1}{8}\right) = \frac{3}{64}.

    Now use the grid to calculate the probability of the sum of spots, that is, Pr(S = 2), Pr(S = 3), .... Pr(S = 12).

    eg. \Pr(S = 2) = \frac{9}{64}.

    eg. \Pr(S = 3) = \frac{3}{64} + \frac{3}{64} = \frac{6}{64}.

    Now calculate the mean value of S in the usual way.
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  6. #6
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    Are you saying i should sum all probabilities and take a mean value?

    i.e. (sum of probabilities)/36
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    Quote Originally Posted by ccdelia7 View Post
    Are you saying i should sum all probabilities and take a mean value?

    i.e. (sum of probabilities)/36
    What I've suggested you should do is very clear. I can't be any plainer except to add (what you should already know) the following:

    E(S) = 2 \cdot \Pr(S = 2) + 3 \cdot \Pr(S = 3) + \, ... \, + 12 \cdot \Pr(S = 12).
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    Did you get 5.75?
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    Quote Originally Posted by ccdelia7 View Post
    Did you get 5.75?
    If you have done the baisc arithmetic correctly then you will get the correct answer. I don't check basic arithmetic - I always make careless mistakes.

    However, if you post full details of your calculations I will have a look.
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  10. #10
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    2*(9/64) + 3*(3/32) + 4*(7/64) + 5*(1/8) + 6*(9/64) + 7*(5/32) + 8*(5/64) + 9*(1/16) + 10*(3/64) + 11*(1/32) + 12*(1/64) = 5.75

    (If the given probabilities are correct!)
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  11. #11
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    Quote Originally Posted by ccdelia7 View Post
    2*(9/64) + 3*(3/32) + 4*(7/64) + 5*(1/8) + 6*(9/64) + 7*(5/32) + 8*(5/64) + 9*(1/16) + 10*(3/64) + 11*(1/32) + 12*(1/64) = 5.75

    (If the given probabilities are correct!)
    Well, the probabilities add up to 1 and the first couple are correct so I'll give them a provisional tick. And I suppose you correctly entered the sum into a calculator (since you didn't keep a common denominator of 64) so I'll give the answer a provisional tick.

    However, someone much more diligent than me and with way too much time on their hands might check it more thoroughly and, perhaps, disagree with me.
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