Optimal BRM shot taking and Kelly criterion in poker
I want to share some things I learned recently about optimal BRM and shoot taking. The First question is at what point a winning player should start to play higher games in order to maximize the rate of growth of his BR. It will depend on his WR on current stake and on higher stake as well as on variance on both limits. Formula for this is
C=((SIGMA1)^2-(SIGMA2)^2))/(2*WR1-2*WR2)
Where SIGMA1 and SIGMA2 are standard deviations in dollars on higher and lower limits, WR1 and WR2 are win rates in dollars and C is BR at which player should start playing higher limits. I'll call this critical bank roll formula. This formula is from book Mathematics of poker page 301 and it applies Kelly criteria to poker.
Example
Let's say you have 4BB WR on NL50 and you'll keep the same WR on NL100 and standard deviation is 100BB on both limits. Then for C we get C=(100^2-50^2)/(2*4-2*2)=1875$. So this player should take his first 1-2buyin shot right around 2k BR, if BR drops down below 1875$ or so he should go back to NL50.
Definition of win rate
When you use formula for critical BR, it's important to point out that the WR you should put in the formula is not exactly the same as WR you see in holdem menage. WR in that formula is all the money you put back in BR. If you don't intend to spend any money from poker winnings, then your WR is your WR from HM+ rakeback you get, if you do spend your poker winnings then you need somehow incorporate that in calculation. One way is just to subtract your average spending form WR and then use the formula.
What is better: 10 shots of 1buyin or one shot of 10buyins?
To answer this question will need one more formula. This is formula for risk of ruin in poker which is
R=e^(-2wr*b/SIGMA^2)
Where R-risk of ruin, wr is win rate, SIGMA is standard deviation. b is bank roll for a specific game(so if you are taking 5buyin shot your BR is 5BI) and e is euler's number which is 2.72.To understand whether 1bi or 10bi shots are better we get back to the example of NL50 player with WR of 4bb. Lets say he has a BR of 1k and wants to find the quickest path to NL100.
Strategy A
He'll grind up to 2900 and take 10BI shot on NL100 if loses 10BI on NL100 he'll move back down to NL50 regrind those 10BI until one of those shots stucks. First let's calculate how likely it is to have unsuccessful shot take with 10BI and 4bb wr. Using previews formula R=0.45. Now let's calculate how many hands our hero will play on average on NL50 until he moves up finally to NL100. Well first he has to win 1900$ to take his first shot with 4bb wr on average this will take him 95k hands, after that there is 45% chance shot wont stuck and hell be forced to move down and grind another 1k$ or 50k hands then there is 45% next shot wont stuck, so he moves down and so on... We get this sum
95000+0.45*50k+0.45^2*50k+0.45^3*50k+...
This is sum of the geometric series and sum of it comes as
45k+50k*(1/(1-0.45))=45k+50k*1.18=135k
So on average he will play 135k on NL50 until he finally moves up to NL100.
Strategy B
He will take a 1BI shot when he reaches 2k. Now using the formula for Risk of ruin for this kind of shot we get R=0.92. So there is only an 8% chance that this kind of shot will succeed, sounds pretty bad but let's do the math. Its similar to strategy A, but this time he needs 50k until the first shot and then he only plays on average 5k hands to regrind and take next shot, so we have
50k+0.92*5k+0.92^2*5k+0.92^3*5k...=45k+5k*1/(1-0.92)=45k+62500=107500
Surprisingly strategy B saved almost 30k hands! This is a very big difference and unintuitive result.
How to apply this to real poker?
Once you reach critical BR, my advice would be to start by introducing 1-2 tables of higher stake. Be very selective which tables you choose. This way often even if a shot fails you already will regradin some of the money back on other tables. It's important to be honest and realistic with what your WR is. We all have a tendency to overestimate what is true WR and when doing this kind of calculation it's better to be more conservative than to overestimate your EV. Also focus more on quality then quantity. Someone who has 6bb wr can move up to higher stakes much faster then 2bb winner, he needs to play less hands and he can take shots with lower BR, also 2bb winner often becomes a loser or BE player on higher stakes.
It's also important to be disciplined and move down when you have to and be prepared to handle bigger swings. This kind of strategy can take you very quickly 2 even 3 limits up if you start running hot but it also requires you to move down when its time to do so.
How quickly can you move from NL50 to NL400?
If you have 4bb WR at every stake (of course you need to work super hard to have 4bb on NL400). According to the previous calculation you'll play 107k on average until you finally move up to the next one. We need to beat 3 limits before NL400, that's NL50/100/200 so we can expect to be NL400 reg after 321k hands. If you happen to be a 6bb winner you can get there even faster within 153k hands! This shows how much more important it is to high WR over high volume when you move up the stakes.
Also here is table for critical BR for various wr
[url=https://ibb.co/5LjfqS4]
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I hope this helps someone, there isn't much good information about BRM and its very important topic and it can save you a lot of time when you move up the stakes if you use a wise BRM strategy. Be sure to double check math. It should be fine, but it never hurts to check it. If you have any comments questions let me know.
6 Replies
Let's say you have 4BB WR on NL50 and you'll keep the same WR on NL100 and standard deviation is 100BB on both limits. Then for C we get C=(100^2-50^2)/(2*4-2*2)=1875$. So this player should take his first 1-2buyin shot right around 2k BR, if BR drops down below 1875$ or so he should go back to NL50.
Once you reach critical BR, my advice would be to start by introducing 1-2 tables of higher stake. Be very selective which tables you choose. This way often even if a shot fails you already will reg
I've been thinking about this. If we start by introducing just 1-2 tables, can't we move up even faster?
Say we introduce 1 table of NL100 and play 3 tables of NL50.
Then winrate and standard deviation increase by 25% and we get:
Kelly = (62.5^-50^2)/(2*(2.5-2)) = $1406
2 tables of NL100 and 2 tables of NL50:
Kelly = (75^2-50^2)/(2*(3-2)) = $1563
Kelly just gives you the fastest way to grow you roll. If you have enough roll for nl100 you play that, if not you play lower. This turn if you have same WR on all tables, ofc this is not the case so for very soft tables you need smaller roll. In practice there is usually 0-2 soft tables, so you just start with those.
I dont think you cant put average WR and stnd dev in formula if you are mixing limits. Playing 2k in one session where half of them is on nl50 and other half is on nl100 is same like playing 2 sessions of 1k hands first on nl100 and second on nl50.
I think Zamadi's idea is valid.
There comes a point where the Kelly Optimal stake is between the current one and the next one up. Mixing tables allows you to achieve that.
After reading chapter 25 of MoP, I think I was wrong.
Expected value obviously still increases with 25% in my example above.
However, I don't think you can just add 25% to the standard deviation.
Instead, to calculate the standard deviation of a "portfolio" of uncorrelated bets it seems we have to raise the standard deviation of
every bet to the power of two and then take the square root of that sum.
So if we have a std.dev of 10bb/hand at both NL50 and NL100.
Simultaneously playing 1 hand of NL100 and 3 hands of NL50 gives us a std.dev of: SQRT(3 * $5^2 + $10^2) = $13.2 per 4 hands.
$13.2 per 4 hands = $13.2/SQRT(4) = $6.61 per hand
$6.61/$5 = 1.32
So our std.dev increases not by 25%, but by 32% when playing 1 table of NL100 and 3 tables of NL50, compared to playing only NL50.
This gives us a kelly value of: (6.61^2-5^2)/(2*(0.025-0.02)) = $1875 = exactly the same as if we changed all the tables at once
But introducing just 1 or 2 tables to begin with still makes practical sense, for the sake of table selection and mindset.
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However, if we had to choose between playing 2 tables of NL50 or 1 table of NL100 for the rest of our life, we would prefer 2 tables of NL50, because it allows to achieve the same $/hour, but at a lower standard deviation.
1 table of NL100: 100 hands per hour at a std.dev of $10/hand = $10 * SQRT(100) = $100/hour std.dev
2 tables of NL50: 200 hands per hour at a std.dev of $5/hand = $5*SQRT(200) = $71/hour std.dev
Simultaneously playing 1 hand of NL100 and 3 hands of NL50 gives us a std.dev of: SQRT(3 * $5^2 + $10^2) = $13.2 per 4 hands.
$13.2 per 4 hands = $13.2/SQRT(4) = $6.61 per hand
I think you have to take the weighted av...
This formula doesn't work for one hand. But we can model it as playing 75 hands at NL50 and 25 hands at NL100:
Converting back to a single hand: $55.32 / sqrt(sample size) = $5.532
So the average standard deviation when mixing stakes like this is $5.532 per hand, or $55.32 per 100 hands.
Converting back to a single hand: $55.32 / sqrt(sample size) = $5.532
So the average standard deviation when mixing stakes like this is $5.532 per hand, or $55.32 per 100 hands.
I haven't studied math at this level and I don't really I understand the formula... but the answer feels wrong.
Unless I am missing something, this would give us a kelly value of: (5.532^-5^2)/(2*(0.025-0.02)) = $560
That's lower than the kelly of going from NL25 to NL50 ($938).
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Also, I am able to replicate the answer in the sales example in the link, but I am not able to replicate your answer.
Trying to do use that formula I get: SQRT((74*43.3^2 + 24*50^2)/98) = $45 per 100 hands
Not sure where I went wrong...
