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.

  • $\begingroup$ great thanks. That is what I was looking for. $\endgroup$ – dared Apr 10 '18 at 12:26

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 $p$:

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

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