Calculating the Average Throw in Backgammon: A Math Problem | Homework Statement

[Niles]

Homework Statement

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

 

[hgfalling]

Homework Statement

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

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.