1. A pair of dice (A & B) are rigged to have a higher than normal chance of getting snake eyes (double ones).
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.
- 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.
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.]
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 .
eg. The probability of (1, 2) is .
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. .
eg. .
Now calculate the mean value of S in the usual way.
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.