I'm wondering how I'd go about calculating the probability required to make a one-time wager "worth it".

First Example:

"I'll bet you 20 that I can eat this entire burger" with no previous data on the subject's burger-eating habits. I could estimate the odds, but how high would those odds have to be to accept that simple wager? (20 bets, 40 payouts, no information on odds).

I'd guess "50/50" intuitively, but probability often defies intuition.

Second Example:

  • Wager Amount: 1,000
  • Expected Payout: 20,000
  • Odds: Unknown.

How do I know what probability I would require to make this bet "worth doing"? When does the expected value "break-even".

What I've done already:

I've tried rearranging the Kelly Criterion, but this hypothetical is a one-time wager.

I've tried searching for scenarios regarding an "infinite bankroll" - such as the burger example. (Where 20 is likely far below the maximum wager)

I've tried searching for scenarios of "all in" such as the second example, where 1000 is the maximum someone is willing to put up for the bet, but also their entire "bankroll" since it's a "one time".

  • $\begingroup$ As long as your expected winnings is greater than 0, then I guess a one time wager would be worth it? $\endgroup$
    – Igor
    Sep 19, 2021 at 4:06
  • $\begingroup$ Expected winnings being P_win * Payout - P_lose * Loss, right? The issue is, in these scenarios, the probability is unknown. The question is, "What probability would make it worth it", and I can't find a formula for that. Is it as simple as "Anything above zero"? That seems intuitively wrong. $\endgroup$ Sep 19, 2021 at 4:10

1 Answer 1


In response to my comment:

Let $P_w, P_l, W_{amount}, L_{amount}$ denote the probability of winning, losing and winning/losing amount respectively.

I argue that if your expected winnings is greater than 0, then it is a "safe bet" I will take your example 1.

$P_w = x, P_l = 1-x, W_{amount} = 20000, L_{amount} = -1000$, we need $P_w \cdot W_{amount} + P_l \cdot L_{amount} > 0 \rightarrow 20000x - 1000(1-x) > 0$ which is true when $x > \frac{1}{21}$, therefore if your odds are anything higher than $\frac{1}{21}$ you are expected to make money.


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