Difficult calculation problem in statistics

[BelFish]

I am interested in the task with a starting stack of $10 (100BB at NL10) and stopping the game when the profit reaches $7 (BR 170BB). What exactly is the exact number for the risk of ruin under such conditions with a win rate of 20BB/100 and std.dev, for example, 80.

 

[BelFish]

The game continues either until the entire stack is lost (without additional purchases of chips), or until exactly $7 is won (as soon as we win, we immediately stop the game).

 

[BelFish]

It seems that according to the classical formula for the risk of ruin (with an integral) it will not be possible to calculate ...



R=exp(-2*m*Br/D)

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And it is necessary to use some other averaging of the type, as in molecular physics, integrals for the mean path length of molecules. Or similar from statistical mathematics.

P. S. Is it possible to calculate such tasks in some advanced poker calculator?

I know the "primedope" calculator, but it doesn't seem to be able to calculate...

 

[jordanbillie]

This is impossible to calculate, and anything you do calculate will just be a numeric "explanation" which you created and can fluctuate because of other external factors. Also, your "risk of ruin" should evolve (just like your win rate), making it more challenging to "pin down" or grasp.

I understand the interest in figuring something like this out, but honestly the "stopping at $17 or bust" is just a self inflicted filter you are placing on your own game. If you are at a table where you get to $17 within 50 hands, why stop?!?!?!? Conversely, if you are seated at an uber tough table, where you are struggling to stay above break even, continuing to play for the sake of "$17 or bust" seems like a mistake.

Conclusion: Many more variables besides 'win rate' should be considered, if an attempt to calculate "risk of ruin" is to be made. I would stop here (in the calculations), and simply play the tables that feel comfortable and stable (consistent 20BB/100 poker!).

 

[BelFish]

Profit is not $17, but $7. And at the same time, the game does not stop, but stops at NL10. This task is only part of another more complex task ))

It is clear that winrate and variance change over time, but we believe that there are quite a lot of statistics and these indicators are defined quite well (enough to assess the risk of ruin).

I think it can be calculated... You can calculate the risk of complete ruin for the case of an endless game but without the upper boundary condition for stopping the game when the amount of profit reaches $ 7. Then the total risk of ruin will be:

R=exp(-2*m*Br/D)=exp(-2*20*100/(80*80))=0.535 or 53.5%

But for our task, we need to subtract from this risk all the "trajectories" in which a profit of $ 7 was reached, and then ruin still occurred.

Or we can use another method: take an integral with profit margins of $7 and more. And somehow to average this integral...

 

[BelFish]

I would very much like to know how much the risk will decrease with the introduction of a boundary condition for profit. Will it drop to 20%? Maybe up to 10%. Or maybe only up to 30%-40% ...

P. S.

P(W7$) + P(L10$) = 1

P(L10$) - our risk of ruin.

 

[jordanbillie]

Profit is not $17, but $7. And at the same time, the game does not stop, but stops at NL10. This task is only part of another more complex task ))

It is clear that winrate and variance change over time, but we believe that there are quite a lot of statistics and these indicators are defined quite well (enough to assess the risk of ruin).

I think it can be calculated... You can calculate the risk of complete ruin for the case of an endless game but without the upper boundary condition for stopping the game when the amount of profit reaches $ 7. Then the total risk of ruin will be:

R=exp(-2*m*Br/D)=exp(-2*20*100/(80*80))=0.535 or 53.5%

But for our task, we need to subtract from this risk all the "trajectories" in which a profit of $ 7 was reached, and then ruin still occurred.

Or we can use another method: take an integral with profit margins of $7 and more. And somehow to average this integral...

I would very much like to know how much the risk will decrease with the introduction of a boundary condition for profit. Will it drop to 20%? Maybe up to 10%. Or maybe only up to 30%-40% ...

P. S.

P(W7$) + P(L10$) = 1

P(L10$) - our risk of ruin.

LOL, perhaps this is why I stick with MTTs. I have a very simple approach, and my statistics are calculated for me. LOL!

 

[BelFish]

Well, solving different problems is also fun...

After all, i generally play SnGs HU, and i can do not bother at all, because there all even more simple than in MTTs ))

But i'm just interested in solving...

 

[jordanbillie]

Well, solving different problems is also fun...

After all, i generally play SnGs HU, and i can do not bother at all, because there all even more simple than in MTTs ))

But i'm just interested in solving...

I know the sentiment, I just don't think anything is "solvable" here due to all of the assumptions that would have to be built in.

 

[BelFish]

I know the sentiment, I just don't think anything is "solvable" here due to all of the assumptions that would have to be built in.

In fact, there are no assumptions, but all the necessary statistical parameters (math. expectation and variance) are given. And it is known that the distribution is normal (this has been proven for poker). Accordingly, just as in molecular physics one calculates the mean free path of molecules, so here it should be possible to analytically obtain an exact solution.

As the 53.5% result for the infinite distance case, but without the boundary condition. After all, this is the exact result, and if you check it with simulations on a computer or in Excel, then exactly this percentage of ruin will turn out...

 

[BelFish]

In general, the "primedope" poker calculator is still similar to tasks from molecular physics. This is a problem of Brownian motion. A win rate is analogous to a steady flow or wind that drifts particles in one direction. And the profit is similar to the distance traveled by the particle. Start - at point 10 (there a source, particle sprayer). Point zero is one wall, and point 7 is the second wall. For example, there are filters there and they catch these particles, and eventually they will be distributed in some proportion. In my opinion, this is an elementary (in the sense of analytically solvable) problem of molecular physics. Maybe someone solved or met a similar task?