Measuring best pools of probability My apologies if title sounds a little confusing, but the nature of the problem requires more than one line to explain.
I have a range of different probability pools (henceforth called games). In each game, a subject will roll N-faces dice where N is variable for each single roll. A "desired die result" is represented by a percental chance. So if a game has one 10% win chance roll, it means that the rule of that game is "you toss a 10-sided die and must get a 10". If a game has one 10% win chance roll and one 5% win chance roll, it means "you toss one 10-sided die where you win if you get a 10 and one 20-sided die where you win if you get a 20".
An example of list of available games to play in a pool:
Game 1:

*

*10% win chance roll;

*10% win chance roll;

Game 2:

*

*10% win chance roll;

*5% win chance roll;

*5% win chance roll;

Game 3:

*

*20% win chance roll;

I know from basic probabilities that the chances of scoring at least one win are, respectively: 19%, 18.775%, 20%.
Initially, it would be obvious that the best game to play is Game 3. However, there is another important factor to take into consideration: the only game where it's possible to score 3 wins is Game 2, and one must consider that 1 win has the same value among all three games.
Now I Believe the best game to play is Game 2 because the trade-off of a tiny 1.225% for the potential of winning 3 times as much sounds like a bargain (and please correct me if I'm wrong).
The core of the question is: how do I measure how much Game 2 is a better game to play than the others?
 A: Treat each win as having a value of 1, and each loss a value of 0. The expected value of each individual die roll is just the probability of winning expressed as a decimal (a 20% win chance has an expected value of 0.2). Since all the die rolls are independent, the expected value of the game as a whole is simply the sum of the expected values across rolls.
In game 1, for example, you have a 1% chance of winning twice, an 18% chance of winning once, and an 81% chance of not winning at all, for a total expected value of $0.01*2 + 0.18*1 + 0.81*0 = 0.2$, which is equivalent to the sum of the original 10% probabilities on both die rolls ($0.1+0.1=0.2$).
Since all three games have identical total win probability, the expected value of each game is also identical. All of the games are equivalent, none is better than any other. Any game with a sum probability of winning across trials of 20% is equivalent, whether it's one die roll with a 20% chance of winning, two dice rolls with a 10% chance of winning, twenty dice rolls with a 1% chance of winning, or any other combination resulting in a total of 20%.
