(-2/3) to the power of 2 minus 3/4 times (2 1/3)
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Since the dice are fair and the rolling are independent, each single outcome has probability 1/15. Every time we choose
We have and
, because the dice are fair.
Now we use the assumption of independence to claim that
Now, we simply have to count in how many ways we can obtain every possible outcome for the sum. Consider the attached table: we can see that we can obtain:
- 2 in a unique way (1+1)
- 3 in two possible ways (1+2, 2+1)
- 4 in three possible ways
- 5 in three possible ways
- 6 in three possible ways
- 7 in two possible ways
- 8 in a unique way
This implies that the probabilities of the outcomes of are the number of possible ways divided by 15: we can obtain 2 and 8 with probability 1/15, 3 and 7 with probability 2/15, and 4, 5 and 6 with probabilities 3/15=1/5
