# Probability of Exactly X Successes with Variable Success Dice

I am interested in finding the probability that on any single roll of dice you can have an exact number of successes. However, depending on the particular dice success is a different probability and I can mix dice types. Example for colors of six sided dice:

• Green Dice
• 6/6 sides are success
• Yellow Dice
• 4/6 Sides are success
• White Dice
• 2/6 sides are success

Looking around its easy to find information for rolling all successes with the dice being independent (like Yahtzee with 5 dice all rolling a 1) can be referenced as:

1/6 * 1/6 * 1/6 * 1/6 * 1/6 * 1/6 = 1/7776

So, if I was to roll one of each color and wanted to know what the probability of rolling 2 successes is rather than all, could I reference it as something like this but do chance to fail on later dice possibly? Example of my thinking:

6/6 * 4/6 * 4/6 (white fail chance) = 96/216 = 4/9

The order of the dice rolling for the problem doesn't matter aas they all go at once, but if I reverse this for white to be sucess and yellow to fail it becomes

6/6 * 2/6 * 2/6 (yellow fail chance) = 24/216 = 1/9

So by ordering it like this I have given the value weight which I do not intend, but the different probability's complicating this makes me thing that I am going about it the wrong way or missing something else. Mathematics has never been my best subject even if it is interesting. The order would matter with the green dice though since if I rolled 2 of them I could never get only 1 success because there is a 100% probability to get 2 successes.

Is there a better way to look at this or compensate for the fact that the order of success doesn't matter only the number of success?

In general, whether or not you consider the order of rolling when you look at your dice to determine whether the "successful" event has occurred, any actual set of dice, rolled all at once, will satisfy an event with the same probability as if you had rolled them one at a time in a predefined sequence.

Therefore you generally cannot go wrong if you assume a predetermined sequence of rolling the dice. You might say the green die is rolled first, then the yellow die, then the white die.

But since the event you described (exactly two successes) does not specify the order in which the successes occur, now you must consider all possible orders. As you say, fail-success-success is not possible, since every roll of the green die is a success, but success-fail-success and success-success-fail each can still occur.

You correctly computed that success-fail-success occurs with probability $\frac19$ and that success-success-fail occurs with probability $\frac49$. These two events are mutually exclusive (you cannot have the second roll fail and succeed on the same roll), so the probability that at least one of them occurs is the sum of their individual probabilities. And that sum is the probability of exactly two successes.

If you had not engineered the green die so that every roll is a "success", you would also have a non-zero probability of fail-success-success, which also would have to be included in the sum.

• Ok, so I have the right idea I just have an equal chance of each probability and need to consider each permutation of win order and add them. With the green being a zero probability to fail it would be considered but in this case it would be zero then? So for my one of each example instance the overall probability would be 0/9 + 1/9 + 4/9 = 5/9? – Piskie Oct 29 '15 at 3:59
• "Equal chance" has a special meaning in probability--but yes, you need to add all the permutations, and the result is 5/9. – David K Oct 29 '15 at 7:19
• OK thanks for clearing that up (sorry if I am using the incorrect terminology). If I took 2 of each dice then and wanted 3 successes i could set the order and just need to consider every possible success combination (assuming dice 1/2 are the always win greens then it would be dice 1/2/3, 1/2/4, 1/2/5, 1/2/6) and add those probabilities then like before? – Piskie Oct 29 '15 at 23:19
• Yes, because 2 of your successes will always be green, and you want just one success from one of the other four dice, each of the four dice has a separate chance to be the third success. – David K Oct 30 '15 at 2:01

Even if you don't consider the order in which dice are rolled, you still have to specify which of the dice will come up as success when you are counting successes. For example, to compute the prob that there are exactly two successes when you roll Die 1, Die 2, and Die 3, you should think of the event "exactly two successes" as the union of three mutually exclusive events:

1. Only Die 1 failed.
2. Only Die 2 failed.
3. Only Die 3 failed.

Since events (1), (2), (3) are mutually exclusive, adding the probabilities of events (1), (2), (3) will give the prob of the desired event. In the case where Die 1 is green, Die 2 is yellow, and Die 3 is white, the prob of event (1) is $(0/6)\times(4/6)\times(2/6)$. In the case where all dice are white, the prob of event (1) is $(4/6)\times(2/6)\times(2/6)$.