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Finding the Break-Even Win Probability from Bet Payoffs

Article Quant Q&A · Author: dared
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Summary

The document explains how to find the minimum win probability needed for a wager to break even when the gain and loss amounts differ. Its example risks 12,000 to win 4,000. A winning outcome occurs with probability p, while the losing outcome occurs with probability one minus p.

The method is to write expected profit as the probability-weighted sum of the two outcomes, treating the loss as a negative payoff. Setting this expectation equal to zero gives the break-even probability; solving the equation yields 0.75 for the stated payoffs. A probability above that threshold gives positive expected value under the same payoff assumptions, while a lower probability gives negative expected value. This calculation depends on the win and loss amounts being accurate and on the outcomes being represented by those two possibilities. It does not account for fees, changing payouts, or uncertainty in estimating the true win rate, and break-even expected value does not guarantee profit over a finite sequence of bets.

Key ideas

  • Expected profit is the sum of each payoff multiplied by its probability.
  • Represent losses as negative payoffs when setting up the expectation.
  • The break-even win probability is found by setting expected profit to zero and solving for p.
  • The threshold applies only to the stated payoffs and excludes costs or uncertainty in the estimated win rate.

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StatisticsRisk management
Full text
# Calculate min. win ratio needed for a bet to be profitable


# Calculate min. win ratio needed for a bet to be profitable












If a bet 12000 to win 4000 my risk/reward ratio is .33 . How often must I win the bet to be profitable? I know it's 75% but have not found the formula yet.

## Answer by Alex C (score 0, accepted)

https://quant.stackexchange.com/a/39213

With probablility $p$ you win 4000, with probability $(1-p)$ you lose 12000 (or in other words you "win" -12000)

Write the expectation: $E=4000 p + (-12000)(1-p)$

Find the breakeven $p$ as the $p$ that sets the expectation to zero and solve for $p$:

$4000 p + (-12000)(1-p) = 0 \implies p=\frac{16000}{12000}=0.75$

Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)

This summary was written by Stratmill's research agent from the original; it is not a copy of the source.