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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?

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  • $\begingroup$ From where do you have the numbers of the win chance rolls? $\endgroup$ Commented Aug 12, 2022 at 18:46
  • $\begingroup$ @callculus42 I determine them. Will edit the question with more details on it. $\endgroup$ Commented Aug 12, 2022 at 18:47
  • $\begingroup$ It would be nice. I don't get the link between the dice and the numbers of the win chance rolls. $\endgroup$ Commented Aug 12, 2022 at 18:48
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    $\begingroup$ @callculus42 I edited it. Dice are just a physical representation. But in reality we are working with virtual random numbers that can go from 1 to N. The percentages used in my example can be represented by existing physical dice, though. $\endgroup$ Commented Aug 12, 2022 at 18:53
  • $\begingroup$ OK. Now for me the question comes up, what is the goal of the game? $\endgroup$ Commented Aug 12, 2022 at 19:00

1 Answer 1

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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%.

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    $\begingroup$ +1 for the linearity of expectation. $\endgroup$ Commented Aug 12, 2022 at 19:05
  • $\begingroup$ If I make Game 1 roll 1 60% and roll 2 also 60%, I get 0.36 * 2 + 0.84 * 1 + 0.16 * 0 = 1.56. What is this 1.56? $\endgroup$ Commented Aug 12, 2022 at 19:19
  • $\begingroup$ @AlexandreSeverino The middle term's probability of exactly one win should be 0.48, not 0.84 - note that the sum of the probabilities of your exhaustive list of mutually exclusive outcomes is 0.36+0.84+0.16, which is greater than 1 and therefore indicates a mistake. With the correct value of 0.48, the sum is 1.2, giving the correct expected number of wins for the game as a whole. $\endgroup$ Commented Aug 12, 2022 at 19:23

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