Envoyé par Richard ALL-IN

> à partir de quelle trésorerie peut-on admettre que le risque est couvert pour 1 team de n joueurs ?

Envoyé par cutsss

shorthanded:
200 = nl10
500 = nl25
1,000 = nl50
2,000 = nl100
5,000 = nl200
16,000 = nl400
25,000= nl600
60,000 = nl1k

Envoyé par Richard ALL-IN

hé non ! reste le nombre de jours nécessaires pour absorber le risque (variance, bad beat,...) est 3, 4, 5 ou 10 ?

Envoyé par ffdpokerforum.com

Advanced bankroll requirements concepts
Those are typical numbers, but you may be curious as to how they are derived, or you may have atypical winrate, standard deviation or risk aversion and want something more specific for your needs. While you may not be interested in how the formula is derived (though I'm pretty sure at least two people on this board will be interested if they don't know it already), you may want to play with it a bit to see how things change if you increase or decrease a bit your winrate, etc.

The actual formula is :

B = -ln(r)*sigma²/(2w)
r = exp(-2wB/sigma²)

where
B = bankroll
r = risk of ruin
sigma = standard deviation (in some units)
w = winrate (in the same units as standard deviation)

This assumes you will play at a given limit forever. Although as soon as you start playing enough at a limit and don't withdraw you'll be fine : indeed, winning and thus adding money to your bankroll reduces your risk of ruin as a consequence. Doubling your roll makes your risk of ruin gets squared so it goes down very quickly. Also, if you are correctly disciplined and willing to go down in limits the risk of ruin or the bankroll requirements are significantly reduced.

Typically, you would get your winrate and standard deviation from PokerTracker (standard deviation is in the session tab, "more details...") in BB/100 hands. Standard deviation is, in simple terms, the amount off average something is on average. Let's take an example :

Desired risk of ruin : 0.5%
Winrate : 5BB/100
Standard deviation : 45BB/100
Bankroll = -ln(0.005)*45*45/(2*5) = 1073 BB = 2145 big blinds = 21.5 buyins

There you are, the 20 buyins guideline can be found as a direct application of the formula for some typical values.

Desired risk of ruin : 0.5%
Winrate : 10BB/100
Standard deviation : 45BB/100
Bankroll = -ln(0.005)*60*60/(2*10) = 536 BB = 1072 big blinds = 10.7 buyins

Here you can see the importance of winrate on the final result. The more you're crushing the game, the less buyins you need.

Desired risk of ruin : 0.5%
Winrate : 5BB/100
Standard deviation : 60BB/100
Bankroll = -ln(0.005)*60*60/(2*5) = 1907 BB = 3815 big blinds = 38.1 buyins

Standard deviation varies a bit from person to person, depending on the playing style and the number of players at the table. Here are typical SD values in BB/100 (courtesy Casper...) :

• weaktight players are ~<30, usually low 20s
• TAGs are anywhere from 30-60 depending on their game
• brilliant SH NLHE player is like 50
• brilliant fullring NLHE is 30-40
• a true good lag can have a sd over 50 in fullring NLHE, and over 60 in sh NLHE

So you can see the fullring players usually have a lower standard deviation than the shorthanded players, hence have lower swings and lower bankroll requirements. As you can see above when we changed the 45 s.d. to a 60 s.d., which could be considered typical numbers, we went from 20 buyins required to 38 buyins required. That's why it's generally recommended to have a bigger bankroll for 6-max games than for fullring.

It will also provide you with the results for slightly different winrates and risk of ruins, which is cool.


PS : Multi-tabling doesn't change anything to bankroll considerations mathematically. However, if you're prone to tilt, I can see having 12 tables open at the same time being potentially more harmful than 4, for example if you decide to go allin on all your tables at the same time But, essentially, it doesn't change anything to bankroll calculations.

Part II : Winrate considerations
As you should also notice, your winrate is key in determining your bankroll needs. For example, in a game you're crushing you can get away with a real low 10 buyins bankroll. However, how can you be sure you're crushing the game and not just being lucky ? And by the time you're sure you're crushing the game, you will have much more than 10 buyins anyway. We can see a posteriori why so many of us could build up easily from the $50 on party playing NL5 though

Determining your true winrate isn't obvious and always require a lot of hands in a game like poker where the variance is much higher than the winrate. It's always a hot topic and how many hands you really need is always somewhat mysterious. I remember hearing 10k hands when I first started as a good starting point for the long run. It's laughable now and anything under 100k hands will get people smiling at you. It's more subtle than that in reality and also directly related to your standard deviation.

So, as I was saying before, standard deviation is, in simple terms, the amount off average something is on average. It helps answer the question "how often will I be that far from the average ?" For example, if you flip a coin a hundred times, it is expected to land on heads 50 times. But how likely is it that it will land on heads more than 55 times, or 60 times ? How about less than 20 times ? Using basic statistics, we can answer all these questions.

Standard deviation is proportional to the square root of the sample size. That has a couple implications. First of all, when you have your SD in PT, it is per 100 hands. To convert it to different sample sizes you have to multiply by the square root of sample size / 100. So if you have an SD of 50 BB for 100 hands, on 100k hands you would have an SD of 50*sqrt(100000/100) = 1581 BB for 100k. It also has implications on the speed of convergence of your winrate to your true winrate. When you increase the number of hands played, your earnings increase linearly with respect to the number of hands, but at the same time, your standard deviation also increases, but in square root of the number of hands only. The earnings/deviation ratio, which is central in determining your confidence in your winrate, evolves only in square root of the number of hands, which is pretty slowly for those of you who don't see how the sqrt(x) function looks.

Let's get to that. The probability an event occurs decreases as it gets further from the mean, that's pretty intuitive. More quantitatively, an event wil occur within one SD of the mean 68% of the time, within two SDs 95% of the time (more exactly within 1.96 SDs), and within three SDs 99.7% of the time. To come back to our first example, when you flip a coin a hundred times, the standard deviation is 5. So we'll get 45-55 tails 68% of the time, 40-60 tails 95% of the time. Now, let's apply this to poker. Assume your winrate is 5BB/100 and your SD 60BB for 100. Where does your true winrate lie with 95% certainty after 10k hands ? What about after 100k hands ?

Estimating your winrate confidence inteval
After 10k hands, your SD is 60*sqrt(10000/100) = 600 BB. Your earnings are 5BB/100*10000 = 500BB. Now remember that the 95% confidence interval is within two SDs. Thus for 10k hands your true win rate is 500BB +/- 2*600BB, or per 100 hands 5BB +/- 12BB. Your true winrate is with 95% confidence between -7BB/100 and 17BB/100. See what I was talking about with the 10k sample size being laughable above What about 100k hands now ? Earnings are 5*1000BB +/- 2*60*sqrt(1000), which gives 5000BB +/- 3794BB, for a winrate per 100 hands of 5BB +/- 3.8BB, or between 1.2BB/100 and 8.8BB/100. Congratulations, you're a proven winner at 95% confidence. Notice how wide and inaccurate the confidence interval is, even after one hundred thousand hands ! So, the general formula would be :

Confidence interval = c*SD/sqrt(sample size/100)

where
c is a factor to decide the confidence you want. Typical values to use :
- 1 for 68% confidence (to NOT use would be more appropriate)
- 1.96 for 95% confidence
- 2.17 for 97% confidence
- 2.58 for 99% confidence
- 3 for 99.7% confidence
You can find c for your desired confidence level in a table of the standard normal distribution or in Excel with =NORMSINV((x%+1)/2).

You might be interested in the probability you're a winning player with a certain confidence. If so, use this formula in Excel :

=1-NORMDIST(0,winrate,SD/sqrt(hands/100),true)

where winrate is in BB/100, and SD is in BB for 100 hands. NORMDIST is the Excel function for the cumulative normal distribution from minus infinity to 0, with mean = winrate, and sigma = SD/sqrt(hands/100). For an observed winrate of 5BB/100 and an SD of 60 for 100 hands, you get approximately 80% after 100k hands and 99.6% after 100k hands (notice this formula is just for being a proven winner, not a winner at that specific winrate !).

How many hands would I need for a given accuracy
Another question you may ask, is how many hands must be played in order to accurately know one's winrate within +/- 0.5BB/100 for example. While it's somewhat more academic and less useful, it's pretty interesting to try it once for educative purposes.

It depends on what level of confidence you want of course. Let's see the requirements for 95% confidence and 99% confidence for typical values, even though, as you'll quickly see, you probably won't play enough hands at that level to ever reach this accuracy in your life. The actual winrate doesn’t enter into the calculation actually. It would if we had specified the accuracy as some percentage of the winrate of course.

The standard error (SE) of the win rate is always SD/sqrt(N/100), where SD is the standard deviation for 100 hands, and N is the number of hands. For 95% confidence, we need 0.5 bb to be about 1.96 standard errors. For 99% confidence, we need 0.5 bb to be about 2.58 standard errors. Call this number s, for the number of standard errors. Then we have :

0.5 = s*SD/sqrt(N/100)

N = 100*(s*SD/0.5)^2

We can use this equation to get the following results:

95% confidence:
For SD = 60 bb/100, N =~ 5.5 million hands.
For SD = 40 bb/100, N =~ 2.5 million hands.

99% confidence:
For SD = 60 bb/100, N =~ 9.6 million hands.
For SD = 40 bb/100, N =~ 4.3 million hands.

So as you can see, it takes a ridiculous amount of time to achieve 0.5 bb/100 accuracy. The bottom line is short term results are mostly meaningless. Even long term, comparing your 9BB/100 to someone else's 7BB/100 and bragging about how much you pwnzor him isn't very meaningful. Don't bother trying to pinpoint your winrate with 0.1BB/100 or even 0.5BB/100 accuracy, it's completely futile, even more so considering you will (hopefully) improve during the course of your play. Never forget that even if you are playing much better than your opponents, it will still be very likely for you to drop 1, 2, 5, 10, 15 buyins without even tilting. And you'll probably have an even worse downsing at some point assuming you play enough (one extremely good player on this site had a 40 buyins downsing once for example). Just the inherent luck factor in the game of poker.

Part III : Some variance
That brings up an interesting question. Given a winrate and standard deviation, how do you calculate the probability of winning/losing x BB in n hands ?

To do that, first compute your average win on n hands (winrate x n/100 if winrate is in BB/100). Compute your SD for n hands (SD * sqrt(n/100) if SD is in BB/100). Your target x will be some SDs away from the mean (below or above), find how many. Then just look the probability of that happening in a standard normal distribution table, or use excel and input =NORMSDIST(number of SDs). This will give you the probability of winning less than x BB in n hands. If you're interested in winning more than x, do 1 minus that number and you will have the probability of winning x BB or more.

Let's take an example. Winrate = 5BB/100, SD = 60BB/100. What's the probability you will lose 5 buyins or more in 10k hands. Your average win would be 500BB for 10k hands, and your SD 60*sqrt(100) = 600BB. Losing 500BB would be 1000BB below the mean, that is -1.67 SD. The probability you lose that amount or more is thus =NORMSDIST(-1.67) = 4.8%. That means that being what is considered a very good winner in that game, there is still roughly 5% chance you'll be down 5 buyins after 10k hands, while you should be up 5 buyins. If you're only a 2.5BB/100 winner, the probability shoots up to more than 10%. Yes, this game is tough.

Conclusion : combining winrate confidence and bankroll calculations
If you have made it to this point, you have either skipped to the conclusion, or you're pretty well versed in everything variance related. Notice how the winrate calculations are very useful for the bankroll requirements. Calculate the confidence interval around your winrate, at 98% confidence for example. There is 1% chance of a winrate below the minimum winrate you found to give you your observed winrate (1% and not 2% because there is also a 1% chance it was higher than the maximum winrate of the 98% confidence range). From there, you could say you're 99% confident your winrate is above that minium, and use that minimum winrate to calculate your bankroll requirements, for maximum safety.

Example :
Desired risk of ruin : 0.5%
Winrate in PT : 7BB/100
Standard deviation : 60BB/100
Sample size : 100,000 hands

Confidence interval at 98% = 2.33*SD/sqrt(sample size/100) = 4.4 BB/100
-> Winrate between 2.6 and 11.4 BB/100
-> Winrate higher than 2.6 BB/100 at 99% confidence
Bankroll at 0.5% RoR = -ln(0.005)*60*60/(2*2.6) = 3668 BB = 7336 big blinds = 73 buyins
Bankroll at 1% RoR = -ln(0.005)*60*60/(2*2.6) = 3188 BB = 6376 big blinds = 63 buyins

Same with a standard deviation of 45BB/100 instead :
Winrate higher than 3.7 BB/100 at 99% confidence
Bankroll (RoR 0.5%) = 29 buyins
Bankroll (RoR 1%) = 25 buyins

source : ffdpokerforum.com