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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 13. 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: 2,549 Thanks: 1399 
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 $P[1] = P[2]=P[3] = \dfrac 1 6 \dfrac 1 2 = \dfrac{1}{12}$ $P[4]=P[5]=P[6]=\dfrac 1 4$ $E[roll] = \dfrac{1}{12}+\dfrac{1}{6}+\dfrac{1}{4} + 1 + \dfrac{5}{4}+\dfrac{3}{2} = \dfrac{17}{4} = 4~\dfrac 1 4$ Last edited by romsek; June 27th, 2019 at 03:31 PM. 

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