Probability Problem from The Game "Dice & Fold"

Question

I'm wondering if someone can give the analytical solution to this problem:

1. You start with a single die.
2. You roll the die, if the result is <5 then you lost the die while if the result is >= 5 you gain an extra die such that you now have 2 dice.
3. You continue rolling all your dice one by one until you have no die left.

For clarity: Say you have 2 dice now, you roll one of them and get 4. Since 4<5 you lost that die and are left with a single die. If you were to get a 6, you will get an extra die such that you now have 3 dice. The game ends when you are left with no die.

All the dice have uniform probabilities of getting 1-6. Let $X$ be the random variable representing how many rolls you did. What is $E[X]$ and $Var[X]$?

Two possible sequence of the number of dice that you have is: 1 2 1 0 (you rolled 3 times) and 1 2 3 2 1 0 (you rolled 5 times).

I've done simulation in R which suggests that $E[X]=3$ and $Var[X]=24$, but I can't seem to prove this analytically. My simulation code is as follows:

library(purrr)

set.seed(102)

number_of_simulation <- 10^6
roll <- 0
number_of_dice <- 1
simulation_number <- 1
simulation_result <- c()

for(simulation_number in 1:number_of_simulation){
  while (number_of_dice>0){
    roll <- roll + number_of_dice
    roll_result <- rdunif(number_of_dice,1,6)
    number_of_dice <- 2*sum(roll_result>4)
  }
  simulation_result[simulation_number] <- roll
  simulation_number <- simulation_number + 1
  roll <- 0
  number_of_dice <- 1
}

mean(simulation_result)
var(simulation_result)

Note

This problem arises from "Dice & Fold", a game released in 2Q24 on Steam. The starting character is called Jack. His skill is to roll 2 dice given that you can provide him with a die greater than 4. There exists a trinket in the game that gives you +2 heal whenever you use a skill. The expected value of the heal amount we get if we use all our dice for this skill is $2E[X]$ times the number of dice that we have at the beginning of the round. Thus, this problem.

Answer 1

This is a left-boundary biased random walk or a ruin game against an adversary with infinite resources. With each die roll, the number of dice increases by 1 with a probability $p=1/3$ and decreases by 1 with a probability $q=2/3$. The distribution of stoppings times is given by equation 4.14 on page 352 (page 370 of the pdf) of this book. The probability of stopping on roll $n$ from starting point $z$ is

$$u_{z,n}=\frac{z}{n}\binom{n}{(n+z)/2}p^{(n-z)/2}q^{(n+z)/2}$$

Note that $n$ and $z$ must have the same parity. The expected number of rolls is $z/(q-p)=1/(2/3-1/3)=3$.

Plugging it into Mathematica further confirms your simulation results:

In[1]:= 
u = ProbabilityDistribution[(Binomial[k, (k + 1)/2] 2^((k + 1)/2))/(
   k 3^k), {k, 1, \[Infinity], 2}];
Mean[u]
Variance[u]

Out[2]= 3

Out[3]= 24