Expected value of a recursive dice game

Question

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Context:

Consider a dice game that is played with a fair, six sided die. You are handed the die, and asked to roll.

• If you roll a 2, 3 or 5, then you take double the face value of the the roll, and are able to roll again.
• If you roll a 4 or 6, then you take your current winnings
• If you roll a 1, then you lose everything

Determine the expected value of playing the game.

For reference, I am currently refreshing my knowledge of probability theory before an upcoming course, and I am somewhat stuck on this problem.

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Working:

Intuitively, this seems like a problem that I can solve by conditioning on the first roll, and exploiting symmetry to solve for $\mathbb{E}(X)$. That is, if we let random variable $X$ represent the payoff from playing the game, then;

$$ \begin{align*} \mathbb{E}(X) &= \sum_{v \in \{1, 2, \dots,6 \}} \mathbb{P}(v) \mathbb{E}(X|v) \\ &= \frac{1}{6}\left( \sum_{v \in \{1, \dots, 6\}} \mathbb{E}(X|v)\right) \end{align*} $$

Now, if $v \in \{2, 3, 5 \}$ then $\mathbb{E}(X|v) = 2v + \mathbb{E}(X)$ (this follows from symmetry), so we can simplify the above to: $$ \mathbb{E}(X) =\frac{1}{6} \left( 20 + 3\mathbb{E}(X) + \sum_{v \in \{ 4, 5, 6\}} \mathbb{E}(X|v)\right) $$

I'm confident with what I have up to here, but now I struggle to proceed; I am unable to formulate $\mathbb{E}(X|4)$, $\mathbb{E}(X|6)$, or $\mathbb{E}(X|1)$.

I would like an explanation for how I can formulate these three expressions in terms of constants, and $\mathbb{E}(X)$.

Thanks alot!

Answer 1

Remarks

I saw your conversation with Lulu in the comment and I saw that you have obtained the correct answer. This is an alternative answer

Solution

Let $Y_{m}$ be a scenario where we roll $m$ times and the last roll is either $4$ or $6$. We have:

$$ \mathbb{P}\left(Y_{m}\right)=\frac{1}{2^{m-1}}\cdot\frac{1}{3} $$

The expected value of earning, when we roll $m$ times is:

$$ \mathbb{E}\left(X\mid Y_{m}\right)=\frac{4+6+10}{3}\cdot\left(m-1\right) $$

Therefore, the expected earning from the game is given by

$$ \begin{aligned} \mathbb{E}\left(X\right) &= \sum_{m=2}^{\infty}\mathbb{E}\left(X\mid Y_{m}\right)\mathbb{P}\left(Y_{m}\right)\\ &= \sum_{m=2}^{\infty} \frac{20}{3}\cdot\left(m-1\right)\cdot\frac{1}{2^{m-1}}\cdot\frac{1}{3}\\ &= \frac{20}{18}\cdot\sum_{m=2}^{\infty}\frac{m-1}{2^{m-2}}\\ &=\frac{20}{18}\cdot\left[\frac{d}{dx}\sum_{m=2}^{\infty}x^{m-1}\right]_{x=\frac{1}{2}}\\ &=\frac{20}{18}\cdot\left[\frac{d}{dx}\left(\frac{x}{1-x}\right)\right]_{x=\frac{1}{2}}\\ &=\frac{20}{18}\cdot\left[\frac{1}{\left(1-x\right)^{2}}\right]_{x=\frac{1}{2}}\\ &=\frac{40}{9} \end{aligned} $$

Answer 2

Let's define $Y$ as the event that a game ending value $(1, 4, 6)$ has just been thrown. Well, on any arbitrary throw we have a $3/6 = 1/2$ chance of the game ending, from which we can expect the average game to go for $2$ rolls.

On the first of these rolls, the average payout is $\$20/3$. Thus, the average cumulative total before the game is ended is $\$ 20/3$. Upon the game ending throw, we keep our reward $2/3$ of the time, and lose it $1/3$ of the time.

Therefore, the expected winnings, is $2/3 \times 20/3 + 1/3 \times 0 = \$40 / 9$

A rather simple answer.