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You may toss a fair dice up to 3 times. You are trying to maximize the value of the final dice roll. What is the expected value of your last dice roll?

I'm not sure how to approach this. I think intuitively, I think it's 4.5 (because 3 dice rolls, so expected value at 1.5, 3, and 4.5) but I'm not sure. I thought maybe it would be similar to this question about expected max value of 3 dice but I couldn't put it together.

2easy
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  • Have you figured out the optimal strategy for maximize the expected last role? – Pax Aug 25 '21 at 05:12
  • @pax - Optimal strategy would be to stop after some cutoff. I think this is as a function of the number of dice rolls, but not sure exactly. I think last dice roll has E=3, so 2nd dice roll you would take any roll with value > 3. This leads back to the idea in the question where I think the second roll has E=4 by even split of probability. Thus, we can extend the same logic and get E=4.5 for all 3 dice rolls – 2easy Aug 25 '21 at 14:43

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The optimal strategy when you have a choice to select an independant random variable $Y$, after seeing a random variable $X$, with the goal to maximize the expected value, is take $Y$ if $X<\mathbb{E}[Y]$, and otherwise take $X$. If we let $Z$ denote the final value after doing this, we have that $$\mathbb{E}[Z]=\mathbb{E}[XI(X\ge \mathbb{E}[Y])]+\mathbb{P}(X<\mathbb{E}[Y])\mathbb{E}[Y]. $$

We have that the expected value of a single dice roll is $3.5$. Using the above we, we compute the expected return of atmost two dice roles is $$\frac{1}{6}(4+5+6)+\frac{3}{6}3.5=4.25$$

Thus, when deciding whether to keep the first roll, or to try with the other two dice, its optimal to keep it if its greater than $4.5$. Thus the result for three rolls is $$\frac{1}{6}(5+6)+\frac{4}{6}4.25=4.66...$$

Pax
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  • I think in your computation you are also assuming that the expected value of the second role is greater or equal to the expected value of the first role which might not be the case. Please feel free to correct me if I'm wrong. Thanks. – Anil Bagchi. Aug 25 '21 at 05:57
  • I don't thing this is the case? Do you mean in the statement that it is optimal to try again when $X<\mathbb{E}[Y]$? – Pax Aug 25 '21 at 06:01
  • When you have computed expected return after two dice rolls you are using the formula for expectation which you have derived in such a way that it is maximum in the final turn which in this case the second turn. Though I may misinterpret your argument. – Anil Bagchi. Aug 25 '21 at 06:05
  • I am a bit confused, how does knowing the maximum of the first two rolls help? – Pax Aug 25 '21 at 06:11
  • When determining whether to make a 3rd roll can only depend on the 2nd roll. – Pax Aug 25 '21 at 06:13
  • That's right. But it is not assumed in the question that the expected value of the second roll is greater than the first role. When determining the expected value of second turn didn't you assume that? – Anil Bagchi. Aug 25 '21 at 06:16
  • I mean, the expected value of each role is the same cause its a fair dice. – Pax Aug 25 '21 at 06:19
  • Is there any closed form solution for n dice rolls? Seems like it would be difficult as we have to reason about the cutoff as dice rolls are discrete – 2easy Aug 25 '21 at 14:46
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    @2easy I think for a general random variable, this would be hard, but in this case for more than 5 dice, the expected is greater than 5, so all the cut-offs are the same after this point. – Pax Aug 25 '21 at 18:03