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## A Game of Dice

Probability puzzles require you to weigh all the possibilities and pick the most likely outcome.

 Puzzle ID: #43901 Fun: (2.45) Difficulty: (2.66) Category: Probability Submitted By: javaguru Corrected By: krishnan

Your friend offers to play a game of dice with you. He explains the game to you.

"There are two dice and the goal is to have the highest total of two dice. You roll first. If you don't like your roll, then you can roll both dice again. If you roll a second time, you must keep the second roll. I'll roll second. I have to keep my first roll, but if we both roll the same number, I win."

On your first roll you roll a 7. Should you roll again or keep the 7?

Bonus question: if you play the optimal strategy, what is the probability that you will win?

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 melzerh Jan 13, 2009 Its ok but you won't always beat the odds. Malika123 Jan 15, 2009 It's nice. here2 Jan 22, 2009 very nice and thanks for the full explanation. mumbojumbo1234 Nov 15, 2009 First part was simple enough but couldn't follow the chance of winning playing optimally. I get .58.. BOBdaMAN May 16, 2010 This is way to confusing. this is me, reading the answer :p opqpop Sep 28, 2010 There is an error. The probability you win when you roll second time should be 575/1296. There are two arithmetic errors, when you do 4*6 and 4*26. You can also get 575/1296 from (1-2[(1/36)^2-(2/36)^2 - ... - (5/36)^2] - 6/36) / 2 via symmetry. The probability you win is [1 - P(both of you roll same sum)] / 2. Subsequently, the error carries over to answer for the bonus question. 398/1926 should be 400/1926 and 12159 should be 12075. Luckily, the final answer still comes out to .57 if you round. This is an interesting question because at first glance, EV(sum of 2 rolls) = 7, and since you want the highest sum, you'd think one would be indifferent to rolling again, but this turns out to be false. I'm having trouble figuring out why this is so. Anybody want to try and give a good explanation? opqpop Sep 28, 2010 I suppose if one is familiar w/ the game where you roll a dice and win the number of dollars equal to your roll, or can choose to roll again and win the number of dollars equal to your new roll, one's intuition would be more likely to trick him here. In that problem, we know EV(roll) = 3.5, so you roll again if you get 3 or lower because 3 < 3.5. Here, we should not be concerned with EV, but instead the probability of winning, which is why we can't use the same approach.