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.

__________________
1. There are two things you should never try to prove ...... the impossible and the obvious.

2. If you always do what you've always done, you'll always get what you've always got.

3. Lack of planning on your part does not constitute an emergency on my part.

4. Pressure makes diamonds.

I have a grid of all 36 possible outcomes set up. Now what should I do next?

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?

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.

__________________
1. There are two things you should never try to prove ...... the impossible and the obvious.

2. If you always do what you've always done, you'll always get what you've always got.

3. Lack of planning on your part does not constitute an emergency on my part.

4. Pressure makes diamonds.

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:

.

__________________
1. There are two things you should never try to prove ...... the impossible and the obvious.

2. If you always do what you've always done, you'll always get what you've always got.

3. Lack of planning on your part does not constitute an emergency on my part.

4. Pressure makes diamonds.

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.

__________________
1. There are two things you should never try to prove ...... the impossible and the obvious.

2. If you always do what you've always done, you'll always get what you've always got.

3. Lack of planning on your part does not constitute an emergency on my part.

4. Pressure makes diamonds.

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!)