The fist case is the so called exploding dice, i.e. if a die comes up with the (usually) maximum possible result, then another die is rolled and added to the dice pool. Eg. a player rolls 5 d6s, and gets 1,1,3,6,6. The player then rolls two more d6s, say 2,6. The second six again adds another die which gives 2. Given the target number of 5, the final result is then 3 hits. So how does this affect the probability distribution? Well, I’m glad that you asked! First consider the expected number of hits for each die. With standard dicepool, the probability of getting a hit is p. For the exploding dice, we need to account for the possibility of getting another roll. Let’s call $p_{max}$ the probability of exploding. So first there is a chance of getting a hit. Then there is a  $p_{max}$ chance of getting another roll that gives a hit with probability of p and again there is a probability $p_{max}$ of the additional die exploding and giving another roll with probability p of getting a hit and so on. This can be expressed as $$p + p_{max}\left(p + p_{max}(p + p_{max}(\ldots))\right)$$ This can be expanded as$$p +p \cdot p_{max} + p \cdot p_{max}^{2} + p \cdot p_{max}^{3} + \ldots$$ Continue reading “Dice pools – Addendum”