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Old 12-27-2005, 05:54 PM
NoSoup4U NoSoup4U is offline
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Join Date: Apr 2004
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Default Risk of Ruin and sportsbetting

I have taken to using my poker bankroll to make sports bets so that it is earning me some money while I play poker. I find this profitable for two reasons: 1) I get a bunch of deposit bonuses that are free money and 2) I can almost always locate bets where I bet on both sides and make anywhere from 1/2% to 4% no matter who wins. But there is a catch (isn't there always?). I started with 5k in three different books (to max out the bonus) and 5k in reserve. One book won every wager and the other two lost every wager. Now I have 15K in one book, 2K in one book and 5K in reserve (the extra money is a result of bonuses earned). Having all my money in one book would be a disaster, because my entire strategy is predicated on making opposing wagers at two different books. So, for all practical purposes, if I lose my remaining 7K into the book with 15K, I am ruined. I could always cashout and give up my bonus, but I don't want to do that.

So my question at the end of all that long-winded stuff is something like this:

If I place wagers that are exactly 50/50 between two books, one with 15k and one with 7k, how do I calculate the risk of one book winning all the bets as I vary bet size? Obviously, if I placed a single $7,000 vs $7,000 this would happen 50% of the time. If I bet $100 vs $100, I assume the risk would be very, very remote. I'd like to be able to calculate what bet size will give me a 5% risk of ending up with all my money in one book or a 10% risk of same.

There also exists the situation where I could bet $10 to win $100 against $100 to win $10 where I assume the true odds are 90/10. If I want to minimize the risk of one book winning all my money, should I prefer wagers with unbalanced likelihoods or ones where the risk is 50/50? Intuitively, I think it shouldn't matter, but I've learned that my intuition is mostly crap.
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