Having bumps in the road with expected value.

[gardenia]

Two people roll a pair of dice. If the sum on the dice is 7, Bob pays Jane $25.00. If the sum is not 7, Jane pays Bob the number of dollars represented on the dice.

I got Jane's expected value as $8.33


Did I solve this correctly?

 

[JeffM]

Two people roll a pair of dice. If the sum on the dice is 7, Bob pays Jane $25.00. If the sum is not 7, Jane pays Bob the number of dollars represented on the dice.

I got Jane's expected value as $8.33


Did I solve this correctly?

I get a very different answer. How did you get yours?

 

[gardenia]

P(2) 1/36 + P(3) 2/36 + P(4) 3/36+ P(5) 4.36 + P(6) 5/36 + P (6) 6/36 + P (8) 5/36) + P(9)6/36 + P(10) 3/36+ P(11) 2/36 + P(12) 1/36

There are 6 ways to roll a 7 with 2 dice. I'm all turned around on this.

 

[JeffM]

P(2) 1/36 + P(3) 2/36 + P(4) 3/36+ P(5) 4.36 + P(6) 5/36 + P (6) 6/36 + P (8) 5/36) + P(9)6/36 + P(10) 3/36+ P(11) 2/36 + P(12) 1/36

There are 6 ways to roll a 7 with 2 dice. I'm all turned around on this.

Your probabilities look OK except for P(5) and P(9), but I am guessing that those are typos.

So then what did you do?

 

[HallsofIvy]

P(2) 1/36 + P(3) 2/36 + P(4) 3/36+ P(5) 4.36 + P(6) 5/36 + P (6) 6/36 + P (8) 5/36) + P(9)6/36 + P(10) 3/36+ P(11) 2/36 + P(12) 1/36

There are 6 ways to roll a 7 with 2 dice. I'm all turned around on this.

As JeffM said, you have typos at P(5), which should be 4/36 rather than "4.36", "P(6) 6/36" which should be P(7), and P(9) which should be 4/36 rather than "6/36". But I don't understand why you have written those as a sum (their sum is, of course, 1). To find the "expected" value, you multiply each probability by the value to the person before summing. Here, Jane wins $25.00 if the dice sum 7 (which has, as you say, probability 6/36= 1/6) but loses $2 with probability 1/36, $3 with probability 2/36, etc. Her expected value is \(\displaystyle -2(1/36)- 3(2/36)- 4(3/36)- 5(4/36)- 6(5/36)+ 25(6/36)- 8(5/36)- 9(4/36)- 10(3/36)- 11(3/36)- (12)(1/36)\).