Average "Hits" on exploding dice

In summary: So in summary, for a given value of x, the average number of hits is $E=\frac 12+\frac 16 E$, the standard deviation is $\sigma(x\text{ dice})=\sqrt{ x\cdot \sigma^2(\text{1 dice})}$, and the deviation from the average is $\sigma(\text{1 dice})-\mu=\sigma(\text{10 dice})$.
  • #1
kryshen1
2
0
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?
 
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  • #2
kryshen said:
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?
 
  • #3
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?
 
  • #4
kryshen said:
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:
$$\sigma(x\text{ dice}) = \sqrt{ x\cdot \sigma^2(\text{1 dice})}$$
 

Related to Average "Hits" on exploding dice

1. What are "average" hits on exploding dice?

Exploding dice are dice that have the potential to keep rolling and adding to their value when a certain number is rolled. "Average" hits on exploding dice refer to the expected number of times the dice will explode and add to their value in a given number of rolls.

2. How is the average number of hits on exploding dice calculated?

The average number of hits on exploding dice can be calculated by multiplying the probability of the dice exploding by the number of rolls. For example, if the dice have a 25% chance of exploding and you roll them 20 times, the average number of hits would be 5 (0.25 x 20 = 5).

3. What factors can affect the average number of hits on exploding dice?

The average number of hits on exploding dice can be affected by the probability of the dice exploding, the number of rolls, and any modifiers or bonuses that may be applied to the rolls. Additionally, the type of dice being used (e.g. 6-sided vs. 20-sided) can also impact the average number of hits.

4. Why is understanding the average number of hits on exploding dice important?

Understanding the average number of hits on exploding dice can help players make more informed decisions in tabletop games or simulations where exploding dice are used. It can also help game designers and developers balance the use of exploding dice in their games.

5. Can the average number of hits on exploding dice be manipulated?

No, the average number of hits on exploding dice cannot be manipulated. It is purely based on the probabilities and number of rolls, and cannot be altered by players or game designers.

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