Question about finding minimum probability to play game 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. 
 A: You should solve inequality:
$$p\times 20\ge -10+(1-p)\times 20.$$
A: This is a classic disguised poker situation, where you must calculate pot odds as a ratio. Pot odds are a mathematical expression of risk and reward.
**EXPLANATION: **
Imagine you're told you have 3:1 pot odds, what does that mean??
This ratio means (reward:risk)
If the pot is \$80 and your opponent bets \$40, that means we must risk \$40 to win the new pot total (\$120). Your reward:risk ratio is 120:40, which simplifies to 3:1.
To turn the ratio into a percentage, we simply take RISK / RISK + REWARD. so, 40 / 40 + 120 = 40/160 = 1/4 = 25%. Another way to see it is we have 3:1 odds, so to get our percentage you just add one to the bottom fraction: its 1/(3 + 1) = 1/4 = 25%.
If we have 30% equity to make our hand and we are offered 25%, we should always make the call. If we have 20% equity to make our hand and are offered 25% equity, we should never call.
ANSWER:
In this situation, you each have put \$10 into the "pot", so the pot has \$20. Your opponent bets \$10 more, and the new pot total is \$30. You can either fold and lose your original bet $10, or call and risk \$10 to win whats in the pot (\$30). REWARD:RISK = 30:10 = 3:1 odds. % = RISK / REWARD + RISK = 1 / (3 + 1) = 1 / 4 = 25% needed.
A: 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
