Probability of A given B & C?

Question

I would like to sum the probability that given 3 (potentially biased) die rolls, all 3 rolls will be different-- what is the correct way to do this?

So far, I have: $$ 1 - \sum_{i=1}^{6}\sum_{j=1}^{6} \left( P(d_i \mid d_i\cap d_j) + P(d_j\mid d_i\cap d_j) \right) $$ I am pretty sure this is not correct--any suggestions?

Answer 1

For a fair die with $n$ sides, each side has an equal probability of $1/n$.

For a biased die, each side has a probability $p_i$, with $\sum_{i=1}^np_i=1$.

For two biased dice $A$ and $B$, the probability that they will roll the same number is:

$$\sum_{i=1}^np_{Ai}p_{Bi}$$

For 3 dice, the chance that they are the same is the sum of the chance of each pair are the same less twice the sum that all 3 are the same. So the chance they are different is:

$$1 - \sum_{i=1}^np_{Ai}p_{Bi} - \sum_{i=1}^np_{Ai}p_{Ci} - \sum_{i=1}^np_{Bi}p_{Ci} + 2\sum_{i=1}^np_{Ai}p_{Bi}p_{Ci}$$

Note that this expression also works with a fair die (and gets much simpler).