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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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What precisely does it mean to withdraw? – Charles May 2 '12 at 18:39
I'm not a native English speaker as you've certainly noticed. Withdrawal = A removal of funds from an account. ( I don't really know if exists the verb 'withdraw' lol. – Tom Dwan May 3 '12 at 19:00
So if you bet $\$$20 and win $\$$30, you can take your $\$$30 and leave the site for good. If you bet $\$$20 and lose, you get a $\$$20 credit with which you can place a second bet (either losing, in which case you get nothing, or winning, in which case you can take your winnings home). Is that right? – Charles May 3 '12 at 19:20
Yes sir ;) Perfect :) – Tom Dwan May 4 '12 at 12:20
Judging by your question, you are interested in betting. The proposal for a SE site about Sports Betting might be interesting for you. Two weeks ago, it was suggested to merge the proposal with Sports. – Martin Sleziak Sep 11 '13 at 5:48
up vote 2 down vote accepted

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.

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