I am not a gambler, but I've noticed a lot of sports betting sites and horse racetracks now offer betting insurance - sometimes up to $1000. That means if you lose your first bet, they will allow you to make another bet for free and keep the proceeds if that second bet wins.
Their goal is obviously to addict new customers, but I wonder if there is an opportunity to play this game strategically and systematically across many sites and obtain a positive expected value by carefully selecting odds. This is similar to how credit cards are generally an awful deal, but can be played to obtain four-figures worth of bonuses per year while paying no interest. There are enough gambling sites offering first bet insurance that a person could run through dozens of them and probably make a decent net profit across them all IF they could find the mathematically optimal betting strategy.
With sports, casino games, and horses, one can generally choose the odds at which to bet. E.g. bet on an unlikely winner for either a big potential payout or small loss, or bet on a likely winner for either a small potential payout or big loss.
So I wonder, for your first bet is it optimal to bet on a team/horse/hand that has a high (e.g. 80%) chance of winning a small amount, cash out if you win, and if you lose the first bet then bet on another likely outcome for your second bet? Or is it more optimal to swing for the fences on that first bet, and if you lose make your second bet a high-probability one to probably recover your money?
At a superficial level, there are two ways to win and one way to lose:
1) Win first bet and cash out.
2) Lose first bet, win second bet, cash out.
3) Lose both first and second bets.
Yet one can vary the odds of both the first and second bets, which adds complexity.
This is where my memory of probabilistic math fails me.
If all bets were coin flips with 50/50 odds, double or nothing, then the possible outcomes would look like:
1) Probability of winning first bet and cashing out = 50%
2) Probability of losing first bet = 50% and the conditional probability of winning second bet = 50%
as a combination equals (0.5 * 0.5 =) 25%.
3) Probability of losing both coin toss bets equals (0.5 * 0.5 =) 25%
So in this example, the odds of doubling your money are scenario 1 + scenario 2 = 75% and the odds of losing all your money is scenario 3 or 25%. Based on this little exercise alone, it looks like there is profit to be made from these offers.
So we want to solve for the odds of winning at bets of various probabilities by simply continuing this process down our spreadsheet for bets with odds different than our 50/50 example.
However this method will simply tell us to take all the safest bets, with no accounting for the fact that the payouts are higher for risky bets than for safe bets, and not accounting for the house's cut. It is possible to win very frequently and still not make money if the payouts are low enough and you lose your entire bet some percentage of the time. E.g. if we make our 2 bets have an 80% chance of winning, the odds of winning one of them are 96%.
pW1 = probability of winning first bet
pL1 = probability of losing first bet
pW2 = probability of winning second bet
(pW1 + (pL1*pW2)) = (.8 + ((1-.8)*.8)) = 96%.
However if the house's cut on the payout is 5% for example, then we might have lost money by winning this game! We can envision the house's cut as a percentage taken off the payout if a win occurs.
If for example we lower the odds of each bet to 40%, we obtain a 64% chance of winning at least one of them. Then we need to figure out what the payout would be for a bet with 40% odds (or any odds). There is a 64% chance we receive this payout, and a 36% chance we lose our bet. Because I am not a gambler, this is where my grasp on the math gets fuzzy and I need your help.
According to my read of
this site, we should expect a true odds ratio of (p / (1-p)). So if there is a 40% chance of winning, the
Odds Ratio would be (0.4/(1-0.4)=) 0.67 or 2 to 3. That is to say, if 2 people put in a hundred dollars each betting this way and 3 other people put in a hundred dollars each betting against them, the people betting $200 could win the total payout of $500 dollars (getting their money back plus the other people's money) which is a profit of $300. Vice versa for the other side of the bet if they win.
So for our spreadsheet's purposes, the
Profit of a winning bet is the formula = 1/OddsRatio.
Total Payout is Profit+Bet, or Profit + 100%. Profit at various win probabilities with no house cut included is:
80%: 0.25x
60%: 0.67x
50%: 1x
40%: 1.5x
20%: 4x
So I built a spreadsheet to find the combination of 2 conditional bets with the highest expected value, with a given house cut. My formula for the EV of a winning bet looks like:
(OddsFirstBet*PayoutFirstBet)+(OddsSecondBetGivenLossOnFirstBet * PayoutSecondBet) - HouseCut
OddsFirstBet * PayoutFirstBet reduces to one because it's a fair bet with no house profit accounted for yet.
OddsSecondBetGivenLossOnFirstBet is ((1-OddsFirstBet)*OddsSecondBet).
The EV of a double losing bet is simply -100%.
The total EV of the game is the expected value of a win times the probability of winning either bet plus the EV of losing times the probability of losing.
Conclusions:Somebody please check my math because I have little confidence in my work.
Using a 10% house cut and 20% odds increments, initial results suggest that most combinations of odds have an EV less than 100% and are thus bad decisions. However, four combinations of two 20%, 40%, 60%, or 80% bets stand out as profitable decisions. For example, betting at 1:4 (20%) odds for your first bet and then 4:1 (80%) odds for your second bet yielded a total game EV of 126.8%. Thus if you repeated this play enough times you'd probably earn a 26.8% profit.
Of course the usual implementation caveats apply. What are the rules for withdrawing funds after gambling one or two rounds? Are the most profitable bet combinations even made available to betters doing the promotion? Are these games rigged? Will you get addicted and wreck your life instead of ripping off online casinos for a few hundred bucks? Will the bets have to be different sizes? Are the other players who set the odds rational experts? Is it worth the time? Etc.
For now I want to leave these concerns behind and focus on whether the mathematical system is right. Is it? Please check my attached spreadsheet and let me know your thoughts. If this is the correct approach, I'll expand it down to the percent to see what the optimal bet is.