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!
Best Answer
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.