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January 25, 2015

January 25, 2015

Posted by **Jaime** on Wednesday, February 1, 2012 at 4:01pm.

So far I found the probability distribution of X, where X is the sum of the dice:

2: P(X=x)= .0278

3: P(X=x)= .0556

4: P(X=x)= .0833

5: P(X=x)= .1111

6: P(X=x)= .1389

7: P(X=x)= .1667

8: P(X=x)= .1389

9: P(X=x)= .1111

10: P(X=x)= .0833

11: P(X=x)= .0556

12: P(X=x)= .0278

I also found the expected value of X: 7.

So from this I know that probability of each sum differs, with rolling a sum of 7 being more probable than rolling a sum of 2 or 12.

I'm having some trouble figuring out the expected value of the game, however. Say that to play my game, a person has to pay $1. If he wins, he wins $6, making the net outcome of a win $5. So the two outcomes are: -1 and 5. The way we found the expected value in class was we found the outcomes and the probability of each outcome, multiplied the outcome by the probability, and summed the products. The part that is confusing me is, doesn't the probability of winning depend on the sum that a player guesses? Isn't someone who guesses 7 as their sum going to be more likely to win than someone who guesses a 2?

Am I overthinking this or going about this in the wrong way? Can someone help me figure this out?

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