Fictional Prop Bet

[AaronBrown]

I agree you should take this bet. The only way to construct a scenario in which the bet is negative EV is to make your ROI very close to zero, and set the payoffs so there is a high probability of ending slightly negative.

For one unrealistic example, assume a 10% rake, and payoffs of $300 for 1, 2 and 3 on a $100 buy-in. If you win 3 or fewer times, you are negative for the 10 tournaments. You can have a 35% chance of winning, which gives you an EV of +5%, and still have more than 50% chance of winning 3 or fewer times.

But this example has all payouts the same size (which increases the chance of ending up at a slight loss), a payout calculated to leave you down slightly and a very low ROI. Any realistic example makes this a good bet.

[Siegmund]

A few quick test cases via simulation (10,000 trials = margin of error of 1% on the chance of winning the bet.)
$11 buyin, $50/30/20 payout structure. I assumed that winning exactly $110 in 10 tourneys was a push.

Each line below gives P(1st),P(2nd),P(3rd) as inputs, and ROR and chance of winning the bet afterward.

10,10,10: -10%, 39%
11,10,10: -5%, 43%
12,10,10: 0%, 47%
13,10,10: 5%, 51%
14,10,10: 10%, 54%

14,8,8: 0%, 46%
15,8,8: 5%, 49%
16,8,8: 10%, 53%

15,7,7: 0%, 46%
16,7,7: 5%, 49%
17,7,7: 10%, 53%

17,5,5: 0%, 45%
18,5,5: 5%, 48%
19,5,5: 10%, 52%

There are a wide variety of ways to lose money on this bet if your ROI is in the 5% range, and ITM up to about 33%.

Subject to the constraints P1+P2+P3>39 and P1>P2>P3, however, your ROI will always be over 20%, and your chances for the prop bet will be 61% at worst.

Edit: my ROIs were dividing by 100, not 110... so, for 5, 10,20%, read 4.5,9.1,18.2%.

[AaronBrown]

I did an exact calculation with these numbers. The best result for disproving the claim was your 18%, 5%, 5%. That gives an ROI of 4.5%, as you say. The chance of losing money after 10 tournaments is 0.5006, the chance of breaking even is 0.0197.

But I don't think this is reasonable. If you have an 18% chance of winning, and you don't win, I expect you to have close to an 18% chance of coming in first among the remaining players. So your chance of second place should be close to 18%*(100% - 18%) = 15%. It might be 12% or 20%, but 5% is a stretch. Also, you've got a small ROI and a tiny negative edge in the bet.

If you make all probabilities 11%, you get exactly 0% ROI, and a 50.27% chance of losing money. If you introduce even small variation among the probabilities and positive ROI, your chance goes down. For example, 12% of winning, 11% of 2nd and 10% of 3rd gives 2.7% ROI and 48.25% chance of losing. 10% chance of winning, 12% chance of 2nd and 13% chance of 3rd gives 1.8% ROI and 48.38% chance of losing money.

The reason you can get greater than 50% chance of losing by making the probabilities unequal is you get a high probability of being just below the breakeven point. So in your 18%, 5%, 5% case, out of the 50.06% chance of losing money, 13.89% of it comes from earning exactly $100, 10.53% of that comes from the chance of coming in first twice with no other payouts. With 9 tournaments instead of 10, 2 firsts is only slightly less likely, and it becomes a win instead of a loss. So I consider the 18%, 5%, 5% situation carefully constructed to make the 10-tournament bet negative EV.