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Old 09-12-2002, 12:18 PM
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Default ROR problem for stat weenies

Player X, is a good player, without any patience.

Use the following assumptions to determine the probability player X will go broke, vs. attaining a safe (300BB) bankroll.

Assumptions :
-Player X wins at a rate 1.05 BB per hour with a standard deviation of 13.11 BB/Hr.
-Player X's bankroll is 50 Big Bets.

Bonus ? : Using the same BB per Hour and Stdev, what size Bankroll would Player X need to have, in order to go broke less than 10% of the time?
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  #2  
Old 09-12-2002, 07:17 PM
irchans irchans is offline
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Default Re: ROR problem for stat weenies

I did some math with a lot of estimates and I got the following approximate formula

Prob of Winning = (1-R^n)/(1-R^u)

where

n = starting bank roll
u = winning bankroll
R = ( sigma^2 - e ) / ( sigma^2 + e )
sigma = 1 hour standard deviation
e = earn per hour

Plugging in the numbers

R = ( 13.11^2 - 1.05 ) / ( 13.11^2 + 1.05 ) = 0.987856
Pwin = ( 1 - R^50 ) / ( 1 - R^300 ) = 0.469158

Somebody posted a similar formula with u = Infinity a few weeks ago. (BruceZ maybe.)


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Old 09-13-2002, 02:08 AM
BruceZ BruceZ is offline
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Default Re: ROR problem for stat weenies

The formula I posted is from Blackjack Attack by Don Schlesinger. It is a highly respected formula used by the blackjack community, and I have demonstrated earlier that it gives the exact result when applied to a problem that can be done by a direct method. This formula gives the risk of ruin for a given bankroll if you play forever. We can use it for this problem by noting that if the risk of losing 50 bets before we get to 300 bets is p, and the risk of losing a 50 bet bankroll if we play forever is p(50) and the risk of losing a 300 bet bankroll if we play forever is p(300), then we have:

p(50) = p + (1-p)p(300)

That is, the probability of losing a 50 bet bankroll if we play forever is the probability of losing a 50 bet bankroll before we get to 300, plus the probability that we get to 300 (which is 1-p) times the probability that we lose a 300 bet bankroll before we play forever. Solving for p gives:

p = [p(50)-p(300)]/[(1-p(300)]

Schlesinger's forumla gives us p(50) and p(300) as follows:

P(50) = [(1 - 1.05/13.11)/(1 + 1.05/13.11)]^(50/13.11) = 54.2%.

p(300) = [(1 - 1.05/13.11)/(1 + 1.05/13.11)]^(300/13.11) = 2.53%.

So p = 53%.

So you will lose the 50 bets before you get to 300 bets 53% of the time, and you will get to 300 bets 47% of the time. Perfect agreement with irchans approximate formula (irchans: care to give your derivation?).

If you want a 10% risk of ruin for a bankroll of size B, then working backwards:

.1 = [p(B) - p(300)]/[1-p(300)]

since p(300) = .0253, p(B) = .123. This is the chance of going broke if you play forever with a bankroll of B. Now calculate what B is from the inverse ruin equation:

B = [-(13.11)^2/(2*1.05)]ln(.123) = 172 bets.
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Old 09-13-2002, 10:07 AM
irchans irchans is offline
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Default Re: ROR problem for stat weenies

Thanks for the interesting post BruceZ. I won't have time to post my derivation until Monday at earliest, but I would like to point out that the two formulas are very similar. (I will be travelling soon for the weekend.)

Notation

n = starting bank roll
B = winning bankroll
sigma = 1 hour standard deviation
e = earn per hour


BruceZ-Schlesinger Formula (with a little algebra):

pwin = (1 - r^(n/sigma)) / (1 - r^(B/sigma))
r = (sigma - e) / (sigma + e)

Irchans Formula :

pwin = (1 - R^n)) / (1 - R^B)
r = (sigma^2 - e) / (sigma^2 + e)


If {e/sigma, e/sigma^2} are small and {n, n/sigma} are large then our formulas produce very similar answers. For the example given ( sigma = 13.11, n = 50, B = 300, e = 1.05 ) we get the following results:

BruceZ-Schlesinger Formula pwin = 0.469788

Irchans Formula pwin = 0.469158


Both Formulas are similar to

Simplest Formula:

pwin = (1 - Exp[ - n q] ) / (1 - Exp[ - B q] )
q = 2 e / sigma^2

which gives

pwin = 0.469155







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