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View Full Version : Slight Edge Question

MEbenhoe
11-23-2004, 04:15 AM
Lets say you can bet on the outcome of a random event in which one outcome has a 50.5% chance of happening and the other outcome has a 49.5% chance of happening, but you would be able to get even odds on this bet from anyone because almost anyone would think this is clearly a 50:50 proposition.

If you could make this bet 300 times per hour you should on average make 3 bets per hour. What would an acceptable risk of ruin be for a bet like this? and Given a \$1000 Bankroll for this best what would be the highest amount you could bet per outcome while maintaining an acceptable risk of ruin?

BruceZ
11-23-2004, 05:13 AM
[ QUOTE ]
Lets say you can bet on the outcome of a random event in which one outcome has a 50.5% chance of happening and the other outcome has a 49.5% chance of happening, but you would be able to get even odds on this bet from anyone because almost anyone would think this is clearly a 50:50 proposition.

If you could make this bet 300 times per hour you should on average make 3 bets per hour. What would an acceptable risk of ruin be for a bet like this? and Given a \$1000 Bankroll for this best what would be the highest amount you could bet per outcome while maintaining an acceptable risk of ruin?

[/ QUOTE ]

Your risk of ruin ror for constant bets is given exactly by:

ror = (49.5/50.5)^B

where B is your bankroll in bets. Solving for B gives:

B = log(ror) / log(49.5/50.5)

You decide what risk of ruin you are willing to accept. For example, if you want a 1% risk of ruin, then B = log(0.01) / log(49.5/50.5) = 230 bets. So if you have \$1000, you would make bets of \$4.34.

For a derivation of these formulas, see the first part of the derivation in this post (http://forumserver.twoplustwo.com/showthreaded.php?Cat=&amp;Number=683150&amp;page=0&amp;view=ex panded&amp;sb=5&amp;o=14&amp;fpart=2#Post682045683150). I also recently answered a similar question here (http://forumserver.twoplustwo.com/showthreaded.php?Cat=&amp;Board=probability&amp;Number=117 0950&amp;Forum=,,All_Forums,,&amp;Words=&amp;Searchpage=1&amp;Limi t=25&amp;Main=1166221&amp;Search=true&amp;where=&amp;Name=197&amp;date range=&amp;newerval=&amp;newertype=&amp;olderval=&amp;oldertype=&amp;b odyprev=#Post1170950).

pzhon
11-23-2004, 05:41 AM
If you are allowed to change the size of your wagers depending on your bankroll, you may want to use the Kelly Criterion to determine what fraction of your bankroll to bet.

The Kelly Criterion for rational gambling with an advantage says to maximize the expected logarithm of your bankroll. This will maximize the median growth of your bankroll. It recommend betting roughly so that your bankroll equals variance/edge. Here, the variance is about 1 square bet, and the edge is .01 bets, so the Kelly Criterion recommends that you choose your bet size so that your bankroll is about 100 bets.

If you bet according to the Kelly Criterion, you will never go broke because you are always betting only a proportion of your bankroll. However, there will still be fluctuations. Your bankroll will be cut in half before it doubles 1/3 of the time. Your chance of ever falling to a fraction x of your initial bankroll is x. The probability that your bankroll would ever reach \$100 is 1/10.

If you bet a fixed fraction of what the Kelly Criterion recommends, you can dramatically cut your downswings. If you bet 1/k of what the Kelly Criterion recommends, your chance ever to fall to x times your initial bankroll is x^(2k-1). If you bet 1/2 of what the Kelly Criterion recommends, k=2, and you will fall to \$100 with probability (1/10)^(2*2-1)=1/1000. Your bankroll won't grow as quickly, but you may find these fluctuations more acceptable. See this paper (http://www.bjmath.com/bjmath/proport/riskpaper1.pdf).

If you bet more than twice as much as the Kelly Criterion recommends, your median result will be to lose. You will eventually fall past any fixed positive threshold with probability 1.