Advanced Dice Probability: 3d6

Question

I know how to calculate basic probabilities: 1 in 6 x 1 in 6 = 1 in 36, for example; and 1 in 36 x 1 in 6 = 1 in 216. However, I have forgotten, and cannot find instructions for, how to calculate for multiple-possibility-of-success measures.

My specific problem is this: I want to determine the odds of rolling 2 6s on 3 six-sided dice (the kind you find in Monopoly, craps, etc) for a tabletop RPG rewrite that I'm working on. So long as you roll 2 6s, the third die doesn't matter. After that, I can calculate the odds for explosion dice (every time you roll 2 6s, you can roll another die; if that one gets a 6, roll another, and so on), as that part's easy: just x6 the probability each time thereafter.

I don't want to just know what the odds are, of course; I want to know how to calculate those odds.

On that note, give me the same method and an answer for "odds of rolling an 8, 9, or 10 on 2 dice of 3 10-siders", so I can better organize the system. Thanks!

Answer 1

Call the dice $A, B$ and $C$. There are $6 \times 6 \times 6 = 216$ possible outcomes, so we need to count how many of those attend your event.

The result may be $666$, which I believe also attends your condition. Now, assume $A = 6$, $B = 6$ and $C \neq 6$. There are $5$ possibilities for that: $661, 662, 663, 664$ and $665$. The same could be said if $A$ or $B$ were the ones different of $6$.

Therefore, the probability that you are looking for is $16/216$, where these sixteen possibilities com from the scenarios described above $$16 = 1 (666) + 5(66?) + 5(6?6) + 5(?66).$$

This process can be analogously extended for dice with $10$ sides.

Answer 2

So there are a number of ways to do this, but the way I find easiest is (since this is a small calculation) to find the exact combinations for which your favourable result is achieved. This is $$ 6*6 \\ *66\\ 6 6 *\\ 666$$ from which you can easily calculate your desired result. There's one small thing you need to keep in mind, however; none of the *'s can have a 6; if it does, you essentially overcount the last case of 3 sixes.

So essentially the key steps are

1. Find number of ways in which your desired result is achieved.
2. Make sure that none of the ways you count in one 'step' appears again in another one, i.e., avoid overcounting. We did this by making sure that none of the blanks have a 6.
3. Divide by the total number of outcomes.

Now step 2 is usually where it can get confusing. The confusion usually increases with the number of favourable outcomes. You'll see this if you try and compute the second one yourself. And undercounting can be just as common as overcounting for sufficiently taxing examples! If you're starting out, actually writing them down by hand might help you make fewer mistakes.

This entire business of carefully leaving things out is known as the principle of inclusion and exclusion, which you might have seen in some form or another.