# Calculating dice probabilities with multiple, overlapping success criteria

Alright, so I play a tabletop game where models get in fights and roll dice to see if they deal damage to the other models. I'm wondering how to calculate the probabilities of dealing 0, 1, and 2 damage in a 2v2 fight.

Here is the situation:
- We have Model A and Model B rolling to deal damage to Model C and Model D
- A needs to roll a 5 or 6 to damage C, and a 6 to damage D
- B needs to roll a 4, 5, or 6 to damage C, and a 5 or 6 to damage D
- C and D can only be damaged once each (i.e., they die when they get damaged)

So for starters, I know the probabilities for this scenario just by listing out the possibilities:

  A | 111111 222222 333333 444444 555555 666666
B | 123456 123456 123456 123456 123456 123456
----+------------------------------------------
Dmg | 000111 000111 000111 000111 111122 111222


So the probability of dealing 0 damage is $$\frac{12}{36}$$, (exactly) 1 damage is $$\frac{19}{36}$$, and 2 damage is $$\frac{5}{36}$$. One tricky outcome here is when A rolls a 5 and B rolls a 4. This only results in 1 damage since both rolls are only enough to damage C and C can only be damaged once.

I am hopelessly stumped on how to model this so that I can then apply it to other scenarios such as 3v3s or when some models use more than 1 die.

It is pretty easy to calculate the probabilities of doing damage in a 1v1 fight. I've used binomial distribution statistics to calculate those. Conveniently it works for multiple dice per model as well. However, when it comes to combining these probabilities ... I am at a loss. They don't seem to follow the rules for multiplication and addition.

For example, I naïvely expected the probability of doing 0 damage to be the probability of A doing 0 damage and the probability of B doing no damage. That sounds right intuitively, but no matter how I try to do it ... it never makes sense:

$$P(A\;miss\;C) = \binom{1}{0} \bullet 2^0 \bullet 4^1 = 4$$
$$P(A\;miss\;D) = \binom{1}{0} \bullet 1^0 \bullet 5^1 = 5$$
$$P(B\;miss\;C) = \binom{1}{0} \bullet 3^0 \bullet 3^1 = 3$$
$$P(B\;miss\;D) = \binom{1}{0} \bullet 2^0 \bullet 4^1 = 4$$

Therefore,

$$P(A\;miss\;C) \bigcap P(A\;miss\;D) \bigcap P(B\;miss\;C) \bigcap P(B\;miss\;D) = 4 \bullet 5 \bullet 3 \bullet 4 = 240$$

Wait what?!?! I expect the answer to be 12. Clearly there is something else going on here that I do not understand. :/

• You can't say $\Pr(X\cap Y)=\Pr(X)\Pr(Y)$ unless $X$ and $Y$ are independent events, and these are clearly not. When A doesn't damage C it's likely that he doesn't damage D, either, because he rolled a low number. Mar 16 '20 at 20:56
• Also, you're multiplying counts instead of probabilities. The product of the probabilities (even if you could just multiply them, i.e. if they were independent, which they aren't) would be $\frac46\cdot\frac56\cdot\frac36\cdot\frac46=\frac{240}{6^4}=\frac5{27}\approx0.185$, which is at least a lot closer to the $\frac{12}{6^2}\approx0.333$ that you were apparently expecting than $\frac{240}{6^2}\approx6.67$ would have been. Mar 16 '20 at 21:44

First I'd like to rephrase your question:

A and B each roll a die. The results of the die determine how much damage is dealt. The maximum{damage dealt by A, damage dealt by B} is the number we are interested in.

A Deals 1 damage by rolling a 5 or 2 damage by rolling a 6. B deals 1 damage by rolling a 4 or 2 damage by rolling a 5 or 6. Now take the maximum of the damage dealt by A or B.

I think it may be easier to frame it this way:

A misses and B Misses simultaneously:

P(0 Damage) = P(A Rolls 4 or less) * P(B rolls 3 or less) = 2/3*1/2= 1/3 = 3/9

A Hits both or B hits Both (don't double count):

P(2 Damage) = P(A Rolls 6) + P(B Rolls 5 or 6) - P(A Rolls 6 AND B rolls 5 or 6) = 1/6 + 1/3 - 1/6*1/3 = 4/9

P(1 Damage) = 1-P(0)-P(2) = 2/9

Alternatively,

A hits 1 (rolls 5) or B hits 1 (rolls 4)

P(1 Damage) = P(B Rolls 4)*P(A Rolls 5 or less) + P(B Rolls 3 or less)*P(A Rolls 5) = 1/6*5/6+ 3/6*1/6= 2/9

How I verified my answer: The table below shows how much damage occurs when (A rolls, B rolls) happens. Count the number of 0s and divide by 36 to determine how often 0 damage occurs. Repeat for 1s and 2s.

## | 6 | 2 | 2 | 2 | 2 | 2 | 2 |

Summary:

P( 0 damage ) = 3/9

P( 1 damage ) = 2/9

P( 2 damage ) = 4/9