I have the below scenario.
Suppose you are playing a game where you and your opponent have put $\$10$ in the pot each. Your opponent bets another $\$10$. What is the minimum probability of you winning to call this $\$10$ bet?
Attempt
There is $\$30$ in the pot: $\$20$ from the opponent and $\$10$ from me. If I call, I increase the pot to $\$40$ and the game continues. Is the minimum probability $10/30=1/3$? This is the ratio of my risk to reward before I decide to call.
On
To determine the minimum probability of winning necessary to justify calling a $10 bet in a game where both players have already contributed $10 to the pot, we need to consider the expected value of the game. Calculate the expected value by multiplying the probability of winning by the amount won, and subtracting the probability of losing multiplied by the amount lost. Let x be the probability of winning the game. The expected value is then: (x * $10) + ((1 - x) * -$10). Solve for x by setting the expected value equal to zero and solving for x: (x * $10) + ((1 - x) * -$10) = 0. Simplify the equation: 10x - 10 + 10x = 0, 20x - 10 = 0, 20x = 10, x = 0.5. Therefore, the minimum probability of winning necessary to justify calling a $10 bet in this game is 0.5, or 50%.
Suppose that we win nothing if we do not call. Let $p$ be our probability of winning. Then when we call the expectation of our gain is $$(30p-10(1-p))\$=(40p-10)\$ .$$ So when $p>1/4$ the expectation is a positive number of dollars.