Multiple dice throws with "probability gates"

Question

You're throwing 6 sided dice and there is 5 "levels" you need to pass. You start at level 1 and to advance to the next level you will need to roll a 6. Once you're on level 2 if you roll a 6 on your first try you will advance to level 3, but if you roll anything other than 6 you will go back to level 1. So in essence you need to pass four 1/6 probability gates to get to level 5. Now the question is, how many throws in total do you need at level 1 to get to level 5? A friend and I have been arguing what is the right way to solve it and we seem to be unable to convince the other that our way is correct.

I deposit that it's a simple as $1/\left(\frac{1}{6}\right)/\left(\frac{1}{6}\right)/\left(\frac{1}{6}\right)/\left(\frac{1}{6}\right)$ ~ 1296

My friend argues that it is $\left(1-\left(\frac{1}{6}\right)\cdot \left(\frac{1}{6}\right)\cdot \left(\frac{1}{6}\right)\cdot \left(\frac{1}{6}\right)\right)^x=0.5$ so $^x$ ~ 898

Now to the best of my understanding I am calculating the mean value and he is calculating the median value. I know that median can be better because it "cuts out" outliers, but I'd argue that in this case it should not be used since the data points further away from the median are not random.

Am I right? Is he right? Is neither one of us right? Please explain why mean or median should be used.

Answer 1

This can be described as a markov chain with the following transition matrix with states written in reverse (stage 5 corresponds to first column, stage 4 corresponds to second column, etc...):

$\begin{bmatrix} 1&\frac{1}{6}&0&0&0\\ 0&0 &\frac{1}{6} &0 &0\\ 0&0 &0 &\frac{1}{6} &0\\ 0&0 &0 &0 &\frac{1}{6} \\ 0&\frac{5}{6} &\frac{5}{6} &\frac{5}{6} &\frac{5}{6} \end{bmatrix}$

This is in standard form $\left[\begin{array}{c|c}I&S\\\hline 0&R\end{array}\right]$

We look to the fundamental matrix $(I-R)^{-1}$ for the information about the expected time until reaching a stationary point.

$I-R = \begin{bmatrix}1&-\frac{1}{6}&0&0\\0&1&-\frac{1}{6}&0\\0&0&1&-\frac{1}{6}\\-\frac{5}{6}&-\frac{5}{6}&-\frac{5}{6}&\frac{1}{6}\end{bmatrix}$

Using a calculator, $(I-R)^{-1}=\begin{bmatrix}6&6&6&6\\30&36&36&36\\180&210&216&216\\1080&1260&1290&1296\end{bmatrix}$

We look to the final column since we begin the "game" in stage 1 and add the entries.

The expected number of rolls of the die then will be $6+36+216+1296=1554$

Answer 2

You seek the expected number of rolls at level 1 before reaching level 5; call this number $E$. Now, starting from level 1:

• $\frac56$ of the time you don't roll a 6 and you stay on level 1. Now you need an expected $E$ rolls – in addition to the roll you just did – to get to level 5.
• $\frac16\cdot\frac56$ of the time, your first roll is a 6 but your second isn't. Again you return to level 1, and you need an expected $E$ more rolls in addition to the roll at level 1 you just did. This pattern continues for three rolls and four rolls.
• $(\frac16)^4$ of the time, you roll four sixes out of four and reach level 5. You still need a roll at level 1 for this.

Hence we have the following equation for $E$: $$E=\left(\frac56+\frac5{6^2}+\frac5{6^3}+\frac5{6^4}\right)(1+E)+\frac1{6^4}$$ Solving, we find $E=1296$, i.e. it takes 1296 rolls on average at level 1 to get to level 5, and you are correct.