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