# Recurring Event, Probability of Totals

#### joepinion

##### New Member
Hi.

I'm actually not working on homework, but trying to make odds tables for a card game called Summoner Wars to practice/learn math for fun.

It's a typical dice battle game, where various characters roll varying amounts of dice to attack. For each die, if they role a 3, 4, 5, or 6, it counts as a "hit." 1 or 2 is a miss. So for each die, the probability of a hit is 2/3.

So I did some research and learned about binomial distribution. Fun!! I used it to make tables that showed, given how many dice you were rolling, how many attempts you were making, and a couple other variables in the game, what your odds are of getting at least 3 hits, 4 hits, 5 hits, etc., with an attack.

Very cool! I had never even heard of binomial distribution before today.

The only "odds" I have yet to solve in this game is a certain character who rolls 1 die to attack. Then, whether you hit or miss, you roll a die and if you roll a 5 or 6, you get to attack again. This could theoretically go on forever if you kept rolling 5 or 6 on those rolls between each attack.

So the guy has a 2/3 chance at a hit for each attack, and after the first attack, gets a 1/3 chance to attack again. IF that attack happens (regardless of whether it's a hit or miss), there's a 1/3 chance to attack AGAIN, etc., until the 5 or 6 isn't rolled.

So is there some direction you can all point me in to figure out how to calculate a percentage odds of getting 1, 2, 3, 4, 5, etc., hits with this character in one turn, given his unique recurring attack? Is there even a way to figure this out without doing 1,000,000 simulations?

Joe

ps If these types of games interest you, this seems like one of the best ever. Summoner Wars. Check it out.

#### BGM

##### TS Contributor
Note

You have 0 hit is equivalent to getting 1,2 in the first roll, so the probability is 1/3

You have 1 hit is equivalent to getting 3,4 in the first roll, or getting 5,6 in the first roll and 1,2 in the 2nd roll, so the probability is 1/3 + 1/3^2 = 4/3^2

Inductively, you have n hit is equivalent to getting 5,6 in the first (n-1) rolls (and thus
have (n-1) hit) and getting 1 more hit. So the probability is (1/3^(n-1))(4/3^2) = 4/3^(n+1), n = 1, 2, 3, ...