For the normal die let be the value of the upper face. Then takes on the values
So . So Let be the value on the second die. Calculate the expected value for that. Then use the independence of the rolls to calculate
It's not that hard to do this directly: If the first roll is one, then the possible (and equally likely) totals are 1+ 7= 8, 1+ 8= 9, 1+ 9= 10, 1+ 10= 11, 1+ 11= 12, and 1+ 12= 13.
If the first roll is 2, the the possible totals are 2+ 7= 9, 2+ 8= 10, 2+ 9= 11, 2+ 10= 12, 2+ 11= 13, and 2+ 12= 14.
You can continue in the same way and see that there are 11 possible results, from 8 to 18. 8 occurs once, 9 occurs twice, 10 occurs 3 times, 11 occurs 4 times, 12 occurs 5 times, 13 occurs 6 times, 14 occurs 5 times, 15 occurs 4 times, 16 occurs 3 times, 17 occurs twice, and 18 occurs once.
That is a total of 36 outcomes so you can calculate the expected value.
Or you could do it the easy way: since "7, 8, 9, 10, 11, and 12" are just "1, 2, 3, 4, 5, 6", with 6 add. So find the expected value for the standard problem of rolling two dice and then add 6.
Hello, goldenrain!
Two dice are rolled. One is "normal" with the numbers 1, 2, 3, 4, 5, 6 on its faces.
The other is "speial with the numbers 7, 8, 9, 10, 11, 12 on its faces.
Find the expected value for the sum of the numbers on the upper faces of the dice.
The sum will range from 8 to 18.
You can construct a 6-by-6 table to see the distribution of these sums.
We find these facts:
. .
Now you can calculate the expected value.