# Average "Hits" on exploding dice

#### kryshen

##### New member
Hey folks,

I've been using a really simplistic equation for this and I'd like to get an actual table/equation/and standard deviations if possible.

Here are the rules.
roll xd6s (six-sided dice)
every die that rolls a 4 or greater is a "hit."
6s rolled count as a hit and then "explode," adding another die which is then rolled. That die hits on a 4+ and also explodes on 6s (this explosion chain can continue infinitely, theoretically).

What is the average number of hits for any value of x? Up til this point I've just assumed an average "hits per die" of 0.583 (50% of dice rolled "hit," 16.67% explode and of those 50% hit), but that only accounts for one explosion, not a 6 rolled into a 6 etc. For bonus points, how can I find the standard deviation for any given value of x?

#### Klaas van Aarsen

##### MHB Seeker
Staff member
Hey folks,

I've been using a really simplistic equation for this and I'd like to get an actual table/equation/and standard deviations if possible.

Here are the rules.
roll xd6s (six-sided dice)
every die that rolls a 4 or greater is a "hit."
6s rolled count as a hit and then "explode," adding another die which is then rolled. That die hits on a 4+ and also explodes on 6s (this explosion chain can continue infinitely, theoretically).

What is the average number of hits for any value of x? Up til this point I've just assumed an average "hits per die" of 0.583 (50% of dice rolled "hit," 16.67% explode and of those 50% hit), but that only accounts for one explosion, not a 6 rolled into a 6 etc. For bonus points, how can I find the standard deviation for any given value of x?
Hi kryshen, welcome to MHB! Let's start with $x=1$, and let $E$ be the corresponding average number of hits (the so called Expectation).
Then on average we have $0$ hits with probability $\frac 36$, $1$ hit with probability $\frac 26$, and $1$ plus the as yet unknown $E$ number of hits with probability $\frac 16$, yes?

So:
$$E=0\cdot \frac 36 + 1\cdot \frac 26 + (1+E)\cdot \frac 16 = \frac 12 + \frac 16 E \\ \Rightarrow\quad \frac 56E=\frac 12 \quad\Rightarrow\quad E= \frac 35$$
Now what would happen if we have $x$ dice that we throw completely independently from each other?

#### kryshen

##### New member
Thanks so much! Just to be clear, this means that the average number of hits on 10 dice would be 6. Inassumed that the numberbwould be hugher because of the dice explosions.

How would I find the standard deviation for any given value of x?

#### Klaas van Aarsen

##### MHB Seeker
Staff member
Thanks so much! Just to be clear, this means that the average number of hits on 10 dice would be 6. Inassumed that the numberbwould be hugher because of the dice explosions.

How would I find the standard deviation for any given value of x?
Let the average number of hits of 1 dice be $\mu=\frac 35$, which is what we found earlier.
Then the square of the standard deviation $\sigma(\text{1 dice})$ is equal to the what the squared deviation of $\mu$ is on average:
$$\sigma^2(\text{1 dice)} = (0 - \mu)^2 \cdot \frac 36 + (1 - \mu)^2 \cdot \frac 26 + ((1+\mu)-\mu) \cdot \frac 16$$

And the standard deviation $\sigma(x\text{ dice})$ is:
$$\sigma(x\text{ dice}) = \sqrt{ x\cdot \sigma^2(\text{1 dice})}$$