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 June 27th, 2019, 03:00 PM #1 Newbie   Joined: Jun 2019 From: California Posts: 1 Thanks: 0 maximize dice roll Hi, I'm trying to learn probability with maximization in mind (with decision choice as a twist). Here is a sample question. If you roll a regular 6 sided dice, and then depend on the result decide if you will roll a second time. Only the latest outcome counts, the point is to maximize the number. What would be the expected numeric outcome? I feel the answer should be somewhere between 3.5 and 6 since if you just roll the dice once, 3.5 is the average and the outcome answer. I would only want to roll a second time for another chance if my first roll was between 1-3. But beyond that, I'm not too sure how to frame the concept to get the final answer. Any help is appreciated. Thanks!
 June 27th, 2019, 03:25 PM #2 Senior Member     Joined: Sep 2015 From: USA Posts: 2,494 Thanks: 1369 If we roll 1 or 2 we roll again. Else we take first roll $P[1] = \dfrac 1 6 \dfrac 1 3 = \dfrac{1}{18}$ $P[2] = P[1] = \dfrac{1}{18}$ $P[k] = \dfrac 1 6 + \dfrac 1 6 \dfrac 1 3 = \dfrac 2 9,~k\in \{3,4,5,6\}$ $E[roll] = \dfrac{1}{18}+\dfrac{2}{18}+\dfrac 6 9 + \dfrac 8 9 + \dfrac{10}{9}+\dfrac{12}{9} = \dfrac{25}{6} = 4~\dfrac 1 6$ Actually it looks like we can do better by rolling a 2nd time when we roll a 3. Here $P[1] = P[2]=P[3] = \dfrac 1 6 \dfrac 1 2 = \dfrac{1}{12}$ $P[4]=P[5]=P[6]=\dfrac 1 4$ $E[roll] = \dfrac{1}{12}+\dfrac{1}{6}+\dfrac{1}{4} + 1 + \dfrac{5}{4}+\dfrac{3}{2} = \dfrac{17}{4} = 4~\dfrac 1 4$ Thanks from topsquark Last edited by romsek; June 27th, 2019 at 03:31 PM.

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