Strategy and Expected Value of Dice GameDate: 09/03/2005 at 13:19:57 From: Matt Subject: Dice game You have the option to throw a die up to three times. You will earn the face value of the die. You have the option to stop after each throw and walk away with the money earned. For example, if I throw the die once and a 2 shows up, I earn 2 dollars or throw again. If a 3 comes up after the second throw, I earn 3 dollars and if I want to stop I can or I can throw a third time. The earnings are not additive, so my total earning is just the 3 dollars I got on the last roll. What is the expected payoff of this game? I know if I throw it once, the expected value is 3.5. I just don't know how to incorporate the stopping option. I was thinking of approaching this with a risk aversion idea, meaning that if the throw gives 4 or above, the user should stop. Date: 09/04/2005 at 01:59:47 From: Doctor Julien Subject: Re: Dice game Hey Matt, Thanks for writing in. I love these kinds of questions! Here's how I would approach it. Let's say that we've rolled two consecutive ones. We have one throw left. What is our expected payoff? As you correctly stated, it would be 3.5. Let's call that the payoff at stage 3. So that tells us that if our second roll is 1, 2, or 3, we should reroll. Otherwise we should stop. So our stage two payoff will be the average of 4,5, and 6 half the time, and half the time it will be the payoff for stage 3. Stage 2 payoff = 1/2 [(4+5+6)/3] + 1/2 [3.5] = 1/2 [5] + 1/2 [3.5] = 1/2 [5 + 3.5] = 1/2 [8.5] = 4.25 With me so far? Now stop for a moment and think what you would do for the next step in this problem. When you think you have an answer, read on! What we want to do is now calculate the function for stage 1 based on stage 2. If we roll a 5 or a 6 on the first throw, we'll want to keep that. Otherwise, we'll go to stage two where our expected payoff is 4.25. Stage 1 payoff = 2/6 [(5+6)/2] + 4/6 [4.25] = 2/6 [5.5] + 4/6 [4.25] = 11/6 + 17/6 = 28/6 = 4.666... So that's our expected payoff! Can you see how to generalize this for any number of throws? - Doctor Julien, The Math Forum http://mathforum.org/dr.math/ Date: 09/04/2005 at 15:55:32 From: Matt Subject: Dice game Thank you for your quick reply. I think my answer was wrong in the sense that I was assuming a constant risk aversion strategy. Meaning I was assuming that at each throw I would replay only if I was getting a 1, 2 or 3. But your approach is more accurate. Thanks again. Matt |
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