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The question is: You have 2 fairly weighted dice. You and an opponent pick any integer one after the other. If your number is closer to the sum of the faces on the rolled dice, you win. Do you want to go first and what number do you choose? What's the probability of winning?

I want to pick 7 and be the first one to go because the expectation of the sum of two dice is 7. But what's the probability of winning the game ? the answer is 7/12, I don't understand how to arrive at this number. Thank you.

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Presumably your opponent picks either $6$ or $8$. Assuming it is $6$ ($8$ is symmetric) your opponent wins if the roll is $2,3,4,5,6$ and you win otherwise. Can you find the chance of each of those?

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  • $\begingroup$ I see. assuming the other player is rational he would pick either 6 or 8. Let's pick 6. The probability of the other player to win is when the sum of two dice is 2,3,4,5,6 which is 5/12, and the chance for me to win is 1- 5/12 = 7/12 $\endgroup$ – szd116 Feb 12 '16 at 21:29
  • $\begingroup$ Exactly. If he picks another number, his chance is worse. If he picks $5$, then $6$ is a tie. If he picks $4$ or less, you win $6$. Note that for three dice your ability to pick first doesn't help you. The expectation is $10.5$, so you pick $10$ or $11$. Your opponent picks the other and the game is fair, as the chance of $21-n$ is the same as the chance of $n$. $\endgroup$ – Ross Millikan Feb 12 '16 at 21:36
  • $\begingroup$ Great, thank you so much. Would you please help me with this one: math.stackexchange.com/questions/1652400/… $\endgroup$ – szd116 Feb 12 '16 at 23:16
  • $\begingroup$ the issue is subtle: The chance of getting $7-12$ combined rolls is $$ 1+2+3+4+5+6=21 $$ So $21/36=7/12$ still. But the reason is different. $\endgroup$ – Bombyx mori Apr 4 '16 at 4:15

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