# Homework Help: Backgammon math problem

1. Aug 12, 2010

### Niles

1. The problem statement, all variables and given/known data
Hi

I am trying to find the average of a throw in backgammon with two dice. Recall that two dice with the same amount of eyes (is that how one say it in English?) count double. What I have is

$$\frac{{2 \cdot 3 + 3 \cdot 4 + 4 \cdot 5 + 4 \cdot 6 + 6 \cdot 7 + 5 \cdot 8 + 4 \cdot 9 + 2 \cdot 10 + 2 \cdot 11 + 1 \cdot 16 + 1 \cdot 20 + 1 \cdot 24}}{{36}}$$

The integer to the right of the multiplication-sign is the throw, and the integer to the left is the different ways of achieving it. I get 8.17

Can you confirm my answer? It seems high

2. Aug 12, 2010

### hgfalling

Re: Backgammon

Well...I think you probably didn't copy your counts right, because your answer is right, but doesn't equal what you wrote above. (It's missing a 12).

But think of it this way. If you just threw two dice and took the average, it would be 7, right? So now 1/6 of the time, you get twice as much. So your new average should be 7 + (1/6)7 = 8.17.