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You throw a die. Each time you get a 4,5, or 6, you get the value on the face of the die. When you get a 1, 2, or 3, you quit the game (but keep any winnings). What's the expected value of this game?

Here's how I'm approaching this. I tried to first write an equation for the EV, as so:

$EV = (1/2)\cdot (0) + (1/2) \cdot (5+EV)$ - however, this assumes that you don't get to keep your winnings; how do I account for the winnings?

Next, I tried approaching it like this, where I sum up all of the possibility; you can get a $\{1,2,3\}$ on your first roll, second roll, third role, etc. where the EV of a non-$\{1,2,3\}$ roll is $5$.

$(1/2)(0) + (1/2)(1/2)(5) + (1/2)(1/2)(1/2)(10) + (1/2)(1/2)(1/2)(1/2)\cdot 15 ....$ how do I find a closed for expression for this series (it looks both like an arithmetic and geometric series)

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Both approaches look correct! The $5+{}$in the first equation does account for the winnings you get to keep.

The series you found in the second case is $$ \frac52 \sum_{k=1}^\infty \bigg( \frac12 \bigg)^k k. $$ This is indeed a notch more complicated than a simple geometric series, but we can still handle it! Starting from $$ \sum_{k=0}^\infty x^k = \frac1{1-x}, $$ valid for $|x|<1$, we can take the derivative of both sides to get $$ \sum_{k=0}^\infty kx^{k-1} = \frac1{(1-x)^2}, $$ which implies $$ c\sum_{k=1}^\infty kx^k = \frac{cx}{(1-x)^2}. $$ This is exactly the series you want, with $c=\frac52$ and $x=\frac12$, and therefore your series equals $$ \frac{(5/2)(1/2)}{(1-1/2)^2} = 5, $$ which is the correct expected value.

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  • $\begingroup$ Thanks so much Greg! I guess now I want to know how I would represent not being able to keep the money, using the first method I used above. EV = (1/2)*0 + (1/2)(5+EV). It seems to me that this would assume you don't keep the money, so I want to see how you would correctly represent that. $\endgroup$ Jul 26, 2020 at 21:30
  • $\begingroup$ Why don't you keep money here? Consider the first roll - with the probability 1/2 you get nothing and the other half goes to you winning something. Now obviously the game itself doesn't care if you won the first roll or not so if we forget the 1 roll and assume we are just starting, the Exp of this "tail" should be the same. But then we remember that actually won the 1 roll already so we should add it to our result. Hence your equation which leads to a correct answer $\endgroup$ Jul 26, 2020 at 21:55
  • $\begingroup$ @AlexeyKubarev Yup, I agree with that. My question now though is how would I represent NOT keeping the money? $\endgroup$ Jul 26, 2020 at 22:21
  • $\begingroup$ @JamesFlanagin well, that's pretty easy - you lose at some point with the probability equal to 1, therefore you always go home with nothing :-) $\endgroup$ Jul 26, 2020 at 22:32
  • $\begingroup$ @JamesFlanagin if you want to have some formula that gives this result, the expected win in that case is equal to $\frac{1}{2}\times 0 +\frac{1}{2}\cdot \frac{1}{2}\times 0 +...$ $\endgroup$ Jul 26, 2020 at 23:04
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Here is another approach.

Let's say your expected payoff is $E$. On your first throw, you roll a 4,5,or 6 with probability $1/2$ and an initial expected gain of $$\frac{1}{6}(4+5+6) = \frac{15}{6}$$ and keep going, or you roll a 1,2,or 3 and stop. If you get to keep going, then you are in exactly the same state you were in before the first throw, with an expected gain of $E$. So $$E = \frac{15}{6} + \frac{1}{2} E$$ Solve for $E$.

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