Roll until 5 or 6 is obtained on die without mid-game cash out

Question

This question is from QuantGuide(Busted 6 II):
Suppose you play a game where you continually roll a die until you obtain either a 5 or a 6. If you receive a 5, then you cash out the sum of all of your previous rolls (excluding the 5). If you receive a 6, then you receive no payout. You do not have the decision to cash out mid-game. What is your expected payout?
My Approach:
In the case when we can't cash out mid-game the expected value will be 2.5. Now in the case of stopping midgame, we calculate the expected value at each stage of the throw. For the $i^{th}$ throw the expected value will be: \begin{equation} \frac{2}{3}^i(2.5i)+\frac{2}{3}^{i-1}(2.5(i-1))\frac{1}{6} \end{equation} The 2.5 value is due to at each throw the average value of the dice roll will be $\frac{1+2+3+4}{4}$. The value I am getting is 2.59 but it doesn't pass. It would be great to get some help.

Answer 1

A faster method would be assuming 6 is also cashing out, then dividing by 2 (because 6 and 5 are identical other than the fact that 6 multiplies your score by 0). you have a 1/3 chance of cashing out each turn, meaning each outcome is expected to be rolled "half a time". the calculation $\frac{\frac{1+2+3+4}{2}}{2} =2.5$ is still correct, just this is an easier method.

One thing I don't understand. Where did you get 2.59?

Answer 2

Like I asked in the comments, I don't know why you have the first term, $\left( \frac{2}{3} \right)^i \cdot 2.5^i$, in your expression.

For the player to win anything, there must be at least two rolls and the game should end with a $5$. If the game ends at $i +1$ rolls, the payout would be $2.5i$. The probability that the game ends with a $5$ at $(i+1)$th roll is $\left( \frac{4}{6} \right)^i \cdot \frac{1}{6}$.

So, the expected payout is

$$\sum_{i=1}^{\infty} \left( \frac{4}{6} \right)^i \cdot \frac{1}{6} \cdot 2.5i \; = \; \frac{5}{12} \sum_{i=1}^{\infty} \left( \frac{4}{6} \right)^i \cdot i$$

I don't know how to get the sum but the answer would be $2.5$.

I believe there could be a simpler solution but it doesn't strike me at the moment.

Answer 3

The game ends if you roll a $5$ or $6$, with probability $\frac13$. The expected number of trials until a trial with probability $\frac13$ succeeds is $3$. So you expect to roll $3$ times, $2$ of which count towards the payout, each of which has an average of $2.5$. The probability that you actually get that payout is $\frac12$. So the expected payout is $2\cdot2.5\cdot\frac12=2.5$.

Answer 4

Ignore for a second that there's only a $\frac12$ chance of getting the payout. Then consider the first roll. There's a $\frac13$ chance of getting 5 or 6, so you end with $0$ and a $\frac23$ chance of getting any other roll, which puts you back in the same position except with on average $\frac52$ added to your final score. So for this 'game' we have $$\mathbb{E}S=\frac13\cdot0+\frac23(\frac52+\mathbb{E}S)$$ Which tells us $$\frac13\mathbb{E}S=\frac53$$ so $\mathbb{E}S=5$. The answer for your game will just be half of this.