# Help on Expected Value

1. Jul 1, 2010

Thanks everyone!:)

If I understand it right, since there are 3 dice, the probability that any of the numbers 1 to 6 appears after the throwing the dice is 36/216 which gets simplified to 1/6. so the probability of winning back your dollar is 1/6 and losing it would be 5/6. so intially, i have this expected value >> 1/6- 5/6 = -4/6. Bu then it said there that if you win, you will have additional k dollars depending... so i tried to compute the different winning combinations... if i am correct, then there 56 different possible outcomes.. of those 56, 20 of them composed of numbers without replacement i.e. (a number only appear once in each combination), 30 of them contains those combination which some numbers appear twice i.e {112, 334, 551, ...} and 6 combinations of the numbers appearing three times i.e. {111, 222, ...} so the probability that you get an additional dollar to your winning is 20/56, 2 dollars is 30/56 and 3 dollars is 6/56.

Is it correct to say that the expected gain from playing the game once is equal to E[X] = 1(1/6) - 1 (5/6) + 1(20/56) + 2(30/56) + 3 (6/56)?

Or is it equal to,

E[X] = (1(20/56) + 2(30/56) + 3 (6/56) + 1)(1/6) - 5/6?

Please, I just want to be clarified... Thanks a lot!

Last edited: Jul 1, 2010
2. Jul 1, 2010

### mathman

I haven't worked out your arithmetic, but my calculation is a net of -17/216.

P(losing)=125/216 net=-1
P(winning 1)=75/216 net=2
P(winning 2)=15/216 net=3
P(winning 3)=1/216 net=4