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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: 2,549 Thanks: 1399 If we roll 1 or 2 we roll again. Else we take first roll $P = \dfrac 1 6 \dfrac 1 3 = \dfrac{1}{18}$ $P = P = \dfrac{1}{18}$ $P[k] = \dfrac 1 6 + \dfrac 1 6 \dfrac 1 3 = \dfrac 2 9,~k\in \{3,4,5,6\}$ $E[roll] = \dfrac{1}{18}+\dfrac{2}{18}+\dfrac 6 9 + \dfrac 8 9 + \dfrac{10}{9}+\dfrac{12}{9} = \dfrac{25}{6} = 4~\dfrac 1 6$ Actually it looks like we can do better by rolling a 2nd time when we roll a 3. Here $P = P=P = \dfrac 1 6 \dfrac 1 2 = \dfrac{1}{12}$ $P=P=P=\dfrac 1 4$ $E[roll] = \dfrac{1}{12}+\dfrac{1}{6}+\dfrac{1}{4} + 1 + \dfrac{5}{4}+\dfrac{3}{2} = \dfrac{17}{4} = 4~\dfrac 1 4$ Thanks from topsquark Last edited by romsek; June 27th, 2019 at 03:31 PM. Tags dice, maximize, roll Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post mepoom Probability and Statistics 2 September 12th, 2014 05:05 PM techcrium Advanced Statistics 0 February 28th, 2014 05:49 PM Kinroh Probability and Statistics 9 May 19th, 2012 11:24 AM coolguy29 Algebra 1 May 19th, 2012 06:49 AM maxj Advanced Statistics 2 July 22nd, 2010 12:18 PM

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