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I've posted a similar question some days ago: Maximize a welcome bonus. , but I didn't specify some details that turn this in a different question (maybe). (If this has the same answer of the previous one I'll edit that question).

A betting website offers to refund your first bet in case of losing. In case of winning you can keep all the money and have the possibility to withdraw, in case of losing the site gives you the same amount of money to bet in another event. In case you win this second bet you can withdraw.

I don't want to risk any money so my intention is to bet on the other possible outcomes of the match with another bookmaker to have a little margin of profit.

In this scenario how can I maximize my profit?

Consider the probability of winning a bet is $p$ and the payoff is $r$ (both as ratios of your bet) and $pr = k$.

Should I choose a bet with low $p$ or high $p$ in the new website? Will this influence the profit?

Thanks in advance!

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    BTW this is sometimes called [bonus harvesting](http://www.google.com/search?q=bonus+harvesting), see [Wikipedia](http://en.wikipedia.org/wiki/Arbitrage_betting). When trying things like this in real life, you should pay attention to the rules of the bookmaker. Often, they require you to wager the amount of bonus several times before withdrawing it; some have even the requirement that this has to be done during one month. (Personally, I had problems to fulfill this with a bookmaker who made my limit for betting very low after I won a few bets.)2013-09-11

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If you lose your first bet, your second bet has value $k$ (or $k\cdot \$20$ if you prefer to monitize). So your first bet should maximize $p \cdot r + (1-p) \cdot k$. This equals $2k - pk, so the lower the probability the higher your expected winnings.

I think that in practice low-probability bets are overpriced, so you may not actually go with the longest bet but one of the long ones would be a good choice. You can redo the calculations for a few of the long bets substituting the actual r$ value for the approximation $r = k/p$ if you want to find the actual best bet.