## Wednesday, June 21, 2017

### Reroll damage - the math behind it

If you had been reading the internet these past few days you might have seen that Cygnar is getting some new sweet models, because we deserve it.

In particular, one unit that called my attention was the Trencher Longguners, without going into too much detail, they can reroll damage rolls. This is a primer, and I asked myself: "How does this change the math of damage rolls?"

Always reroll this!

Currently, we have ways of rerolling to hit, but that math is different, if your odds of hitting something are p, the odds of hitting it when you can reroll are p + (1-p) * p, meaning that you either hit first, or you miss and you hit it on your reroll.

However, damage is different, you roll damage and you have to decide if you want to keep said roll or you get greedy and try to do more. In this article I will address the math behind this: "Should I reroll or should I keep it?" Intuitively, I think you should reroll everything that is under average.

This is an easy question when you are shooting a single wound model because it will become binary again: "Did I kill it?" if the answer is yes, do not reroll it, if the answer is not, reroll. But it becomes more complicated when you try to maximize against multiple box models.

For instance, a person could be greedy and say: "I will reroll anything that is not box cars!!", well let's see how this changes the math: the odds of rolling 2 sixes is 1/36, the odds of not rolling 2 sixes is 35/36, so you just doubled the odds of rolling box cars! (or almost: 1/36+35/36*1/36 = .055 instead of 0.28 ). However, does this maximize the damage that you can do?

The algorithm to estimate the expected roll when you get to reroll is easy, first you need to decide what's the magical number that will force you to reroll, let's call that R. If R=12, means that you reroll everything but 12s, or if R=9 it means that you reroll anything that is 8 or lower.

Therefore, the probability of rolling x, if x is higher or equal than R, means that you either rolled x on the first roll, or you rolled less than R and then you rolled x on the second try:

now, if x is lower than R, it is not possible that the roll is the result of the first roll, which means that:

$\large&space;P[\mathbf{x}]&space;=&space;Pr[\mathbf{r}

with this new function, we can compute the expected damage of this roll:

Table with the expected damage when you get to reroll if the first roll is under R
R Expected Damage
3 7.138889
4 7.361111
5 7.611111
6 7.833333
7 7.972222
8 7.972222
9 7.833333
10 7.611111
11 7.361111
12 7.138889

It is clear that if you only reroll when you roll less than a 3, the expected damage does not increase that much, however if you are only rerolling when you roll anything less than box cars it has the same effect, because you are keeping the same number of combinations (only 1 out of 36).

From this table, the optimal number seems to be either 7 or 8, meaning that it is the same to reroll anything that is under 7 or 8 and it is the optimal value.

My question to you is: if you have rolled a 7 (meaning that you have less than 8) are you going to risk it? I do not think so...