I am trying to figure out probabilities for ten-sided dice for my RPG system just to make sure my stats are correct, but I thought I'd check my workings with better maths brains than mine.
- This system requires you to roll a pool of d10 of between 2 dice and 9 dice.
- You look for matching pairs and matching triples. A pair is a success, and so is a triple.
- (A triple is treated as a pair, with the third dice 'discarding' itself to give you a bonus effect of modifying its pair up or down by +1/-1, which is sometimes advantageous, as higher and lower pairs both have different advantages in system).
- More pairs mean more successes. So we want as many as we can.
- Naturally, with 2 dice rolled, the chance of a pair is 10%. There is an early biased curve of any success occuring, rising to 84.88% by the time you reach six dice rolled, and 99.64% at nine dice. This, I'm all good with.
Where I'm not sure I'm doing things right is working out the possibilites for a single triple, and then two pairs in a given pool.
What I've been doing for a triple is taking the chance of a pair, and then multiplying it by (1/10) + (1/10) for each extra dice besides the two dice already in the pair.
Hence, for a triple in six dice: $$0.8488 \times ((1/10)+(1/10)+(1/10)+(1/10))$$
And then, for the chance of two pairs in six dice, multiplying the chance of a pair in a pool of six dice by the chance of a pair in four dice.
So: $$(0.8488 \times 0.4960)$$
Which gives us this chart:
Now, assuming all this is correct, it means that, once we reach six dice rolled, the chance of there being two pairs exceeds that of there being a triple.
Which surprises me. It sounds like an anomoly. I may just be being thick about this, and that really is truly the case, but I just want to make sure I'm not doing something terribly wrong here!