What is the average amount of dice rolled when you add another die when a 5 or 6 appears?

Question

• Initially we roll 3 fair dice at the same time.
• Every time we roll a 5 or 6 on any of the dice we roll we will roll another dice.
• On average how many dice will be rolled in total? (including the initial 3)

How would you go about calculating this? If you can point me in a direction where I can find out how to solve this myself that will be more than welcomed. I'm stuck on the fact that I don't know how many dices will be rolled considering the fact that there's a chance that more will be added apart from the initial 3. I have no idea how I should even continue or where to start. I have little knowledge on statistics in general but if you can point me to the right direction where I can start learning or at least start a rabbit hole of learning I can take it from there. I made a program that conducted the question 15,000 times and it comes up to 4.5 rounded up. I would like to know how to solve it on paper rather than just brute forcing it.

Answer 1

You can treat each die separately; the result is just three times the result for one die. For one die, let $x$ be the expected number of rolls. This is $1$ with probability $\frac46$ and $1+x$ with probability $\frac56$. Thus

$$ x=\frac46\cdot1+\frac26(1+x)\;. $$

The solution is $x=\frac32$, so for three dice the expected number of rolls is $\frac92=4\frac12$. Since Ross' approach yields the same answer, it looks like there's a bug in your code that suggested the result was $4$.

Answer 2

Hint: When you roll one die you have $\frac 13$ chance to roll another. Each roll multiplies by another $\frac 13$, which makes it a geometric series to sum. The three dice are independent, so you can multiply the result for one die by $3$