Probability of losing x big bets

[MarkD]

Given a known WR and SD what is the probability that a person will lose x big bets? Or, more importantly how do you calculate this?

I don't think the answer is ROR.

For purposes of this question if you need to use numbers let's use wr = 2 BB/100 hands and SD = 15 BB/100 hands and let x = 100, 200, 300.

- "Known" implies that there is no doubt at all in the numbers.

[NUReedy]

I think I may be doing this wrong, but:

P(X<-100) = P(((X-WR)/SD)<((-100-2)/15)) = P(Z<-6.8)
I don't have a Z-score table in front of me, but the probability is close to 0.

Actually, the more I think about this the calculation I just did doesn't make sense. In this case the variable X represents the number of BB won PER 100 hands, so I believe P(X<-100) represents the probability you lose 100BB/100. If someone would care to elaborate and correct me.

What I did though, the standardization, is the correct way to go about this one, I just can't get around the fact that X is a proportion.

[MarkD]

Let x be in units of BB/hr it's the same thing. What you are calculating though is not the probability of losing 100 BB's.

I've seen many people on multiple forum's spout about just doing this easy calculation but for the life of me I don't know how to do it.

[topspin]

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Given a known WR and SD what is the probability that a person will lose x big bets? Or, more importantly how do you calculate this?

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If you look down like 3 threads, Fianchetto posted the exact same question about 7 hours ago.

[MarkD]

Yah, I saw that after obviously. The question is slightly different though.

[topspin]

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Yah, I saw that after obviously. The question is slightly different though.

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If I have a known win rate and std dev. how do I calculate the probability of losing X BB's in a session?

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Given a known WR and SD what is the probability that a person will lose x big bets?

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Sorry if I'm being obtuse, but could you elaborate on the difference? Did you mean 100BB in a row or something?

[MarkD]

No. The difference is that he is talking about losing x BB's in one single session.

The question I am asking is what is the probability of dropping 100 BB's before you begin winning again? So you could lose 50, win 20, lose 30, win 10, lose 40, win 60, lose 70 and have a net loss of 100. To start winning again I mean that after dropping 100 you need to net +101 from that point forward before you are winning again.

Essentially if we can calculate this we can calculate the probabiilities of losing various different amounts and see how that correlates to ROR.

[MtnDave]

Its been a long time since I took probablity, but isn't this just the Gamblers ruin problem? Isn't it the same (or can be morphed into) the probablity of going broke given a certain bankroll?

[topspin]

[ QUOTE ]
Its been a long time since I took probablity, but isn't this just the Gamblers ruin problem? Isn't it the same (or can be morphed into) the probablity of going broke given a certain bankroll?

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It's actually somewhat harder than the typical gambler's ruin problem they usually start you off with. In the one I was taught, you would either win or lose a single dollar at a time, whereas in poker you can win or lose many different amounts on a single wager, with varying probabilities.

Another problem is that in the typical gambler's ruin problem, you win with probability q and lose with probability 1-q. You could choose q to equal your mean, but there's no other parameter to choose to fit the problem to your variance.

Having said that, there are some similarities. There's a nice description of the simplest form of the ruin problem here. In our case, the boundary conditions would be -100BB and 1BB, with an initial point of 0BB. The event of interest would be "ruin" at -100BB.

[MtnDave]

Thanks for the response. However if you had more information, I think you could make a reasonable approximation using the gamblers ruin formula. If you knew the average size of a pot, and the rake and betting sizes, that information plus your win rate and your SD could let you estimate p.