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Risk-free gambling with math!

April 9, 2013

Disclaimer: I do not endorse gambling and I have no idea if this is in breach of any gambling websites’ terms and conditions.

I’ve recently picked my first ever professional sporting team to support – the Boston Celtics. In doing so, I’ve finally discovered the allure of watching sports; once you assign value to a specific team winning, the game is far more exciting. This then led me to making $5 bets on games that didn’t involve my team, so I could get excited over other games that I watched with friends. After making a few bets on successive games, I quickly got bored and stopped visiting the gambling website. This is how I discovered how quick they are to offer you free bets. They knew deep down that I was a chronic gambler in the making; they just needed to keep me coming back for a while.

From free bets, you can only keep the winnings. For example: If you place a $10 free bet on something paying $1.10 for a win, then you have used up the free bet, and either have $1 or empty pockets. Assume, for the moment, that we’re dealing with a rather altruistic bookie. In this case, $1.10 for team A to win implies $11.0 for team B to win. In general, if we have $X for team A and $Y for team B, then they will be related by Y-1=\frac{1}{X-1} and the bookie’s odds are based on a probability of A winning, P(A)=\frac{Y}{X+Y}. Let’s try to understand these equations now.

If a bookie thinks that team A will win 10 out of 11 times, then he will expect to keep your $1 (that you’ve bet on team A) once in 11 games. This means that he can only afford to give you one tenth of a dollar each of the 10 times that you win; that is, he’ll offer you $1.10 odds for team A. On the other hand, if you bet on team B then he’ll expect to receive $10 from you over the 11 games. So now he can afford to pay you $10 on the time that you do win; that is, he’ll offer $11 for team B. So we could say that the odds are 1.10 to 11, in favour of team A, or team A will win 11 games for every 1.10 games that team B wins. (Note: This is the same as 10 games to team A for 1 games team B wins, but we’ve phrased it in terms of the known odds.) So we could say the probability of team A winning, is \frac{11}{12.1}, which hopefully explains the formula given above for P(A).

Sadly I don’t know any altruistic bookies, so we’d better take their cut into account! Let’s imagine the same scenario, but now the bookie is keeping 5% of the money that comes in. Over the 11 games (and $11), the bookie will pocket $0.55 and so the $1 you would have previously won is now only $0.45. Since this $0.45 is to be distributed over 10 games, the odds he would give you for team A is now only $1.045. If you are betting on team B, he will still keep his $0.55 but this will come out of the $10 that he could afford to pay you for winning. This means the odds for team B would be given as $10.45. If Z is to be the bookie’s cut (in this case Z=0.05), then the astute reader should be able to verify the following relationship between X and Y: Y=\frac{(1-Z)X}{X-1+Z}. Note: The probabilities P(A) and P(B) are given by the same formulas.

Now, lets go back to that free bet I was talking about. Say we place a $100 free bet on team A to win, with odds given by $X. If team A wins, then we get $100(X-1) in our pocket and if team B wins then we’ve lost our free bet. Let’s now imagine that we bet something less than $100(X-1) on team B, so that if team A wins then we’re not losing anything and if team B wins then we’ve won our real bet and lost the free bet. It is actually possible to judiciously choose our bets in such a way that can be assured $63 cash from our free bet!

If we put an $a free bet on team A and $b of our own money on team B, then we have two possible outcomes:

  • Team A wins: You gain $((X-1)a-b)
  • Team B wins: You gain $(Yb-b)

Note that we have subtracted the $b that we’ve invested.
For now we are interested in a guaranteed win, so we would like both outcomes to result in the same gain for us. That is, we need (X-1)a=Yb. Now either way, we gain \frac{(X-1)(Y-1)}{Y}a. But we have a relationship between X and Y from before, so we can write this in terms of the single quantity Y.
Total gain =\frac{(1-Z)(Y^2-1)-Y(Y-1)}{Y(Y+Z-1)}a.
At this point we will simply state that the maximum total gain occurs when Y=\frac{(1-Z)(1+\sqrt{Z(1-Z)})}{1-Z+Z^2} and results in a gain of \frac{1-2\sqrt{Z(1-Z)}}{1-Z}a – the reader who remembers their calculus can easily verify this.

This is starting to get a little messy now, but using these equations with Z=0.04 (seems about right for the betting website I found), you can guarantee 63 cents on every dollar of free bets they offer. Simply look for a game with odds around $1.19 against $4.90; place your free bet on the team paying $4.90 to win, and place a bet 3.27 times larger than the free bet on the $1.19 team.

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8 Comments
  1. sean permalink

    make it work irl and ill follow you to the TAB

    • It does work!
      Although the free bets for new accounts on the gambling websites excludes Victoria. If you sign up with a NSW address then you could easily do this. I’m not sure if it’s in breach of some kind of fair use policies or something though.

  2. Spherical Cow permalink

    “xvi. This Bonus is for recreational gamblers only. tomwaterhouse.com reserves the right to suspend an account and all associated funds (including winning dividends) where the intent of the client may be considered as a direct attempt to fraudulently abuse the Bonus. Where this occurs, tomwaterouse.com will make all reasonable attempts to contact the client.”

    • Oh I had always assumed they’d have some kind of thing like that, although I’m sure you could easily get around that if you wanted to.

  3. Bogdan permalink

    What if the game ends draw?

    • This is assuming you’re betting on something that can’t end in a draw. If you bet on a sport that allows draws, then you would need to place a much smaller bet on the possibility of a draw too; the exact amount is left as an exercise to the reader 😉

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