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Suppose you and I are playing a game, and it could be any game. We have bet 10 dollars which would go to the winner. At some point, I offer to double the bet to 20 dollars. If you accept, the game continues with the new bet. If you refuse. you lose the game, along with the original ten dollars. What is the minimum probability of winning the game that you would need to accept the increased bet?

My knowledge I know I want to set the expected value of winnings to $0$ and solve for p.

What I have (This assumes you accept the game):

X (winnings) can be 20 or -20 if I accept the game. $p$ = probability of winning $$0 = -20(1-p) + 20(p)$$ $$ p = .5$$

My issue is X can be -10 if I refuse the game. Not sure how to implement this into the expected probability.

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You should solve inequality: $$p\times 20\ge -10+(1-p)\times 20.$$

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  • $\begingroup$ How did you get to this? The answer is 1/4 but that is 3/4. $\endgroup$ – the boy 88 Nov 12 '19 at 23:09
  • $\begingroup$ I guess this is the question about whose probability to win is meant: yours or mine. If I pay 10$, I need larger probability to win. It is logical, is not it? $\endgroup$ – user Nov 12 '19 at 23:14
  • $\begingroup$ I just know the answer is 1/4 but I'm not sure about the solution. You're only paying 10 if you do not accept to play the game. If you accept the game, then you either make or lose 20. $\endgroup$ – the boy 88 Nov 12 '19 at 23:18
  • $\begingroup$ I see. I wrongly understood the problem. Now corrected. $\endgroup$ – user Nov 12 '19 at 23:24
  • $\begingroup$ That should be (1-p)*-20 because that is the loss scenario. $\endgroup$ – the boy 88 Nov 12 '19 at 23:29
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If you both pay 10 to play the game, there is now 20 in the "pot";

if opponent proposes to double his bet to 20, pot is now 20 + 10 = 30;

you can choose to pay additional 10 to win 30, you have 3 to 1 pot odds

thus you need 1/4 chance of winning to break-even/continue playing

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