Expected Values HELP!

Here is the question:

What is the expected value of a dollar bet?

There are three wheels, each with 20 spaces.

Reel one: 10 cherries, 6 blanks, 3 lemons, 1 money bag
Reel two: 6 cherries, 10 blanks, 3 lemons, 1 money bag
Reel three: 6 cherries, 6 blanks, 7 lemons, 1 money bag

It costs $1 to play. For a winning combination, the net payout is as follows:(in addition to the $1 bet):
money bag, money bag, money bag $1000
cherry, cherry, cherry $ 20
cherry, cherry, symbol (any except cherry or blank) $10
cherry, symbol, symbol $1


TS Contributor
The expected value is the same as a weighted average outcome. For each of the winning combinations you need to compute it's probability of occurring, and the same for the losing combinations, then multiply each by it's value. The expected value is the sum of all of these products.

In order to win $1000, you need to have all three wheels come up "money bag." The probability of 3 money bags is (1/20)^3 or .05^3 = .000125

In order to win $20, you need to have all three wheels come up "cherry." The probability of 3 cherries is (1/10)*(1/6)*(1/6) = 1/360

...and so on.

Hint: also - once you've found the probabilities of winning, add them up and subtract their sum from 1 to find the probability of merely losing your $1
I would like to point out a fine point of problems like this. in a game such as this you pay in $1 and then if you win $1000 you have a net gain of $999. as a result you have to take one of two approaches to solving this:

1. you can use the amount of the prizes ie p($1000)*1000 + p($20)*20... and allow a loss to be p($0)*0. in this case the expected return will be the number of dollars won for each dollar wagered.

2. you can use the "net" winnings so if you win the $1000 prize your winnings are $999 since you paid $1 to play and the net winnings of a loss will be -$1. when you do this the result is the "expected net profit(or loss)" for playing the game.

usually people use approach #1, but approach 2 can be useful at times.