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This is still a work in progress, so bear with me. Instead of trying to write everything down coherently, I am going to ramble about here, and clean it up later.
Until then, please post you comments/questions in this thread which I unashamedly hijacked:
http://www.flopturnriver.com/phpBB2/...ic.php?t=22300
var=SD^2
B = -(sigma^2/2u)ln(r)
r = exp(-2uB/sigma^2)
where u is your hourly rate
sigma is your hourly standard deviation
r is your desired risk of ruin
B is your bankroll
Thank you to BruceZ on 2+2 for his explanation here.
Derivation of bankroll and risk of ruin formula
Let’s start with something really simple.
A Coin flip game,
P(win) = p,
P(lose) = q = 1-p
$1/flip, bankroll = $1, opponent is infinitely wealthy.
What is risk of ruin r?
r = q + (1 – q)*r^2.
The above formula says that we can go bust one of two different ways.. We can lose on the first flip with probability q and be finished, or we can win the first flip with probability 1-q, and then proceed to lose a $2 bankroll with probability r^2 since this is the probability of losing $1 twice.
Solving for r:
r = q/(1-q) = q/p.
What if our bankroll were B? Then to bust we would have to lose $1 B times, so
r = (q/p)^B
Now what if the bet size were b? Then our bankroll would be B/b bets, and
r = (q/p)^(B/b)
eq. 1
Now here is the main idea: We want to set p and the bet size b so that the EV and variance of the coin flip game is the same as the win rate WR and sigma^2 (Variance) of our poker game.
Let:
bet size b = sqrt(WR^2 + sigma^2)
p = ½ + WR/2b
q = ½ - WR/2b.
eq. 2
We can verify that
EV = p* b + q*(-b) = WR
and
Variance = p*b^2 + q*(-b)^2 – EV^2 = b^2 –WR^2 = sigma^2
Substituting p and q into (eq. 1)
r = [ (1 – WR/b) / (1 + WR/b) ]^(B/b)
This is our formula for the risk of ruin for any game with win rate WR and standard deviation sigma. Remember that b is function of both WR and sigma from (eq. 2). This can be put into a simpler form. Take the log of both sides.
ln(r) = (B/b)[ ln(1-WR/b) – ln(1 + WR/b) ]
Since WR is much smaller than b, we can use the approximation ln(1 + epsilon) is approximately equal to 1 + epsilon, when epsilon is a small number. Then we have:
ln(r) = (B/b)(1 – WR/b -1 –WR/b) = (B/b)(-2WR/b)
r = exp(-2WR*B/b^2).
Rearrangement gives the bankroll B needed for a risk of ruin r.
B = -b^2/2WR*ln(r).
Replacement of b by sigma for sigma >> E gives the formula in its familiar form. (in the post above)
Assume you have a u=3BB/hr (mean) and a var=225BB/hr per table, and consequently a StdDev=15BB/hr.
If you are two-tabling, You would have a u=6BB/hr, var=450BB/hr, andStdDev=21.21BB/hr
Note that by playing 2 tables at half the limit, you can reduce your standard deviation per hour without reducing your win rate, and hence greatly reduce your risk of ruin for a given bankroll, or cut your bankroll requirement in half for a given risk of ruin. This assumes that you can make the same number of BB/hr at each table at the lower limit as you did at the higher limit.
Why do we reccommend 300BBs for a limit player, and 15-30 buy ins for a NL player?
Using the BR equation above:
A good winrate for a limit player would be 3BB/100 with a StdDev of about 17BB/100Quote:
var=SD^2
B = -(sigma^2/2u)ln(r)
r = exp(-2uB/sigma^2)
where u is your hourly rate
sigma is your hourly standard deviation
r is your desired risk of ruin
B is your bankroll
substituting in to find the risk of ruin
we have
r = exp(-2uB/sigma^2)
= exp[ -2(3)(300)/17^2]
= 0.002
or a risk of ruin of 0.2%
The equation works for NL also.
If we want to determine the bankroll required to achieve the same risk of ruin in NL
Assuming a player with a 10BB/100 winrate and StdDev of 50BB/100
B = -(sigma^2/2u)ln(r)
= -[50^2/2(10)ln(0.002)
= 776BBs
Given that BBs in this example is 2xbig blind, this would equate to 1552 times the big blind, or just over 15 buy ins.
Note how much of an effect your winrate and Standard deviation have on your bankroll requirements.
Increasing your standard deviation, or decreasing your winrate would require your bankroll to be much larger.
Mason Malmuth's essay
The following calculation does not produce your true standard deviation, but rather an estimate of it. 3 sessions is obviously not enough for this estimate to be very accurate, and in reality you would compute it over many more sessions. (Malmuth reccomends at least 30) The more sessions you use, the more accurate this estimate will become.
But realize it takes relatively few sessions to determine an accurate estimate for your standard deviation compared to the number of sessions required to determine an accurate estimate of your win rate. This is a good reason to compute this statistic. While your average win may be somewhat uncertain, there is really no good reason you cannot have an accurate estimate of your standard deviation after a relatively small number of sessions.
Here we will assume that all sessions are the same length. The formula in the essay above allows you to adjust for variable length sessions. You simply need to log your win and the number of hours played for each session. We will assume constant length sessions here for the purpose of clearly illustrating what standard deviation means. Namely, it is the "square root of the average of the squares of the differences from your average", or sometimes called a "root mean square" or rms average. Perhaps you have heard of rms voltage. The 120 volts AC you hear about is an rms average of the voltage used in the US. It is actually the standard deviation of the voltage, which is a sine wave with peaks at +/- 170 volts.
Here is a simple example of how to estimate standard deviation:
Suppose you play three 4 hour sessions.
In the first session you win $200.
In the second session you win $400.
In the third session you lose $300.
Your average win or EV for these sessions is ($200+$400-$300)/3 = $100/session or $25/hour.
In the first session, you won $100 more than average.
In the second session you won $300 more than average.
In the third session you won $400 less than average.
Now take the SQUARE of these 3 differences from your average (100, 300, -400):
100^2 = 10,000
300^2 = 90,000
(-400)^2 = 160,000.
Note that it doesn't matter whether your differences are positive or negative since we are squaring them.
Now average these numbers to get your variance per session.
session variance = (10,000+90,000+160,000)/3 = 86,667.
Take the square root of this to get your standard deviation per session denoted by the Greek letter sigma.
session sigma = sqrt(86,667) = 294
Now how do we find standard deviation for 1 hour?
A session here is 4 hours, but you cannot divide 294 by 4 to get your standard deviation for 1 hour.
Your standard deviation increases as the square root of the number of hours you play. Therefore, you have to divide it by the square root of the number of hours.
sqrt(4)=2
So your standard deviation for 1 hour is:
sigma = 294/2 = $147 for 1 hour.
You could also compute variance by taking the average the square of our actual wins, instead of the square of our differences from our average, and then subtract from this the square of our average win.
session variance = (200^2 +400^2 + 300^2)/3 - 100^2 = 8667
Notice that variance has units of dollars^2/hr so it follows that standard deviation actually has units of dollars/sqrt(hr).
Standard deviation is usually referred to in units of bb/hr. This is incorrect, and what is actually meant is standard deviation for exactly 1 hour as we computed above.
If standard deviation actually had units of bb/hr, you could simply multiply this number by the number of hours played to get your standard deviation for any number of hours. You actually must multiply it by the square root of the number of hours, so it has units of bb/sqrt(hr) or bb/hr^.5.
So if our true standard deviation were $147, and if we are going to play for 100 hours, and we want to know our standard deviation for that period of time, it is sqrt(100)*147 = 10*147 = $1470.
You should be able to see how this relates to multitabling. You are effectively playing more "hours" in the same amount of time, so your standard deviation will increase by the square root of the number of tables you play.
Note in the above example that the standard deviation only increases by a factor of 10 in 100 hours, but our average win increases by a factor of 100 to 100*$25 = $2500. So our average win increases faster than our standard deviation.
This is why gambling "works" when you have an edge. In the beginning, your average win will be small compared to fluctuations caused by luck. Over time, your results will be determined mostly by your edge, and the effect of luck will be proportionally smaller. The effect of luck will still be larger in absolute dollars, but it will be a smaller in proportion to your win.
-Thanks again to BruceZ and his great explanations of gambling math.
As a quick note for Pokertracker users:
You can find your standard deviation by going to Session notes and clicking on "more detail"
Note in the example above that your average win for 100 hours is already greater than your standard deviation for 100 hours. When your average win becomes exactly equal to your standard deviation, you will be ahead more than 84% of the time.
This is because your results will lie within +/- 1 standard deviation from average 68% of the time, so 32% of the time they will lie outside this +/- interval. 16% of the time they will lie more than 1 standard deviation below the average, and 16% of the time they will lie more than 1 standard deviation above average.
This is based on a normal distribution or "Gaussian curve"
normal distribution: n. theoretical frequency distribution for a set of variable data, usually represented by a bell-shaped curve symmetrical about the mean. Also called Gaussian distribution.
Assuming $147 represents your true standard deviation for 1 hour, after 100 hours your average win will be $2500/$1470 = 1.7 standard deviations.
From a table of the standard normal distribution (I have posted one below), we can determine that you will be ahead more than 95% of the time at the end of this period. Note that with a different standard deviation, your probability of winning over X #of hands is much different.
Below is a "z" table which shows the probabilities of being x standard deviations from your expected win.
Code:-4.0 0.00003 -3.0 0.00135 -2.0 0.02275 -1.0 0.15865
-3.9 0.00005 -2.9 0.00187 -1.9 0.02872 -0.9 0.18406
-3.8 0.00007 -2.8 0.00256 -1.8 0.03593 -0.8 0.21185
-3.7 0.00011 -2.7 0.00347 -1.7 0.04456 -0.7 0.24196
-3.6 0.00016 -2.6 0.00466 -1.6 0.05480 -0.6 0.27425
-3.5 0.00023 -2.5 0.00621 -1.5 0.06681 -0.5 0.30853
-3.4 0.00034 -2.4 0.00820 -1.4 0.08076 -0.4 0.34457
-3.3 0.00048 -2.3 0.01072 -1.3 0.09680 -0.3 0.38209
-3.2 0.00069 -2.2 0.01390 -1.2 0.11507 -0.2 0.42074
-3.1 0.00097 -2.1 0.01786 -1.1 0.13566 -0.1 0.46017
The probability of -0.0 is exactly 0.50000.
This means half of the time you will be above, and half of the time you will be below your expectation.
This table also shows the probability that you will be "down" after playing an amount of time where your expected win/stddev is one of the above numbers.
To determine how long it will take for your average win to be 1 standard deviation, divide the square of your standard deviation for 1 hour by the square of your hourly rate.
hours to be ahead 84% of the time = (sigma/ev)^2.
*84% = the 68% of the time you will be +/- 1SD and the 16% you will be up more than 1SD.
This is one way to define the "long run". To find out how long it will take for your hourly rate to equal let's say 1.6 standard deviations, this is simply
hours to at least break even 95% of the time = (1.6*sigma/ev)^2.
At this point you have a 95% probability of being ahead.
Lets say we have a limit player with a 2BB/100 "true" win rate, and a 20BB/100 standard deviation.
How long will it take for his actual win to be GREATER THAN ZERO 99% of the time?
(2.3*20/2)^2 = 529*100 = 52,900 hands
So there is still a 1% chance that he will be down money at the end of FIFTY THREE THOUSAND hands.
All of us who have played for some time have had runs >2000 hands where we were down a SIGNIFICANT amount.
When you use your standard deviation for 1 hour to compute your swings for longer lengths of time, the results will be more accurate than when you only use it to estimate your swings for 1 hour. The reason is that in the long term, your results will closely resemble a normal or Gaussian distribution (bell curve), but in the short term this is not exactly the case.
For example, a true normal distribution has tails that go off to infinity and negative infinity. Since you can't actually win or lose infinity in 1 hour, the result is that the extra probability that would normally be in the tails of the curve get pushed in, making the tails thicker. This means that your swings for short periods of time are likely to be a little larger than what your standard deviation would suggest.
If your results were truly normal, your swings would lie within +/-1 standard deviation of your average 68% of the time, and within +/- 2 standard deviations 95% of the time.
Your average swing will be +/- 0.8 standard deviations.
Your median swing, which is the swing you exceed exactly half the time, will be +/- .67 standard deviations.
These estimates can give you a rough sense of how you are doing without a lot of calculations. Just remember that results in the short term are just crude estimates, and they are less reliable for reasons that have to do partly with statistics, and partly with your particular circumstances, such as being in a particularly wild game.
A couple links:
These are some very in depth papers on gambler's ruin and bankroll requirements.
http://www.math.utah.edu/~davar/PPT/ARCHIVES/ruin.pdf
http://www.bjmath.com/bjmath/sileo/sileo.pdf
Your bankroll is simply defined as the amount of money which if you lost, would prevent you from continuing to play poker.
If you can not deposit more, it is obviously the amount in your account.
If you can deposit, it is your account plus the amount you are willing/able to deposit to continue playing.
If you can't beat the game, bankroll management rules do not apply.
They are to prevent a winning player from going broke.
If you are not a winning player you should play the lowest limits possible or quit playing altogether.
What other factors do we need to consider when determining a bankroll for tournaments?
How does ROI and ITM% affect this?
The obvious answer is that it affects your variance.
So how do we determine variance and standard deviation for tournaments?
Variance is still simply the average of the squares of all results, and winrate total $won/#tournaments.
$109 buy-in, 100-player MTTs, same prize structure (35%/21%/14%/10%/6%/4%/3%/3%/2%/2%), same opponents, and neither you nor your opponents ever get any better
Let's suppose you finish in every place with equal probability. Then the standard deviation would be $440, or 4.04 buy-ins.
The standard deviation of your observed ROI after n tournaments would be 404%/sqrt(n). If you want to have a 95% confidence interval of +- 10%, you need your standard deviation to be 5%, so n=(404/5)^2=6,531 tournaments.
That is more than most individuals play, so it is almost impossible to get a high confidence of your winrate (ROI)
If you are a winning player, your distribution of places is not uniform. This will tend to increase the standard deviation. If you make k times as many final tables, with an even distribution at the final table, your standard deviation will increase by almost a factor of sqrt(k). for k=2, the standard deviation would be $606, or 5.57 buy-ins.
Here is an application:
Your bankroll should be at least c*SD^2/ROI, where SD is standard deviation and c is a number that depends on your risk tolerance and ability/willingness to move down if you hit a bad streak. (based on the "kelly criterion" in the post below.) Most people seem happy with a value of c between 2 and 4. (1/2 to 1/4 kelly)
If your ROI is 40%=0.4 buy-ins with a SD of 5 buy-ins, then your bankroll should be at least c*5^2/0.4 = c*62.5. For c=2, that is 125 buy-ins. For c=4, that is 250 buy-ins.
For comparison, the SD for 1-table SNGs is about 1.7 buy-ins. In LHE, it is about 15 BB/100 for online full games, or 17 BB/100 for 6-max. In NLHE, it is about 20-60 BB/100.
If you know your win rates in each game, you can use these to determine what gives you the best hourly rate for your bankroll.
This is from an excellent post by Tom Weideman on RGP.
Suppose you are playing a game in which you have some sort of edge. This edge can consist of any combination of probability and odds such that each bet has a positive expectation. Now you ask yourself, "What fraction of my bankroll should I wager on this game?" If your answer is, "The amount that if I keep repeating this same strategy over a long period of time, my bankroll increases at its maximum rate.", then your answer is to bet "Kelly".
To see how we can find such an answer, we should look at it from the beginning. Suppose you bet a fraction f (0 Bet size = f*B, odds = v-to-1, amount won = v*f*B, amount lost = f*B
If you win, your new bankroll B' is going to equal your old bankroll plus your winnings:
Win: B' = B + v*f*B = (1+v*f)*B
If you lose, your new bankroll will equal your old bankroll minus your losses:
Lose: B' = B - f*B = (1-f)*B
Let's take a short timeout to make sure these funny looking equations make sense by looking at an example (I'm a huge fan of concrete examples).
Say you have $100, and you bet $20, getting 2-to-1. For the sake of future considerations, we'll assume the game is a fair coin flip (but the need for the probabilities does not come into play for awhile yet). If you win this bet, you win $40, raising your bankroll to $140. The equation above shows this correctly:
B = $100, f = 1/5 = 0.2, v = 2: B' = (1+v*f)*B = (1+2*0.2)*($100) = $140
The "Lose" equation works in a similar manner.
Now, suppose you play the game multiple times. Assuming no "pushes" (which I will be assuming throughout), then the "new" bankroll (which we have called B') after a game is played becomes the "old" bankroll for the next game to be played. We are assuming the game and the odds offered don't change, and that we do not choose to change our strategy (the fraction of bankroll to be wagered), so the equations given remain the same for each game, with the previous game's B' becoming the next game's B. Your result after winning twice in a row would be:
B after two wins = (1+v*f)*[B after one win] = (1+v*f)*[(1+v*f)*(starting B)]
B after two wins = [(1+v*f)^2]*(starting B)
A check: Suppose in the example above you won again, employing the same strategy. Your new bankroll was $140, so you wagered 1/5 of it ($28) and won at 2-to-1, for a total win of $56. Now your new bankroll is: $140+$56 = $196. Plugging into the equation gives the same result:
B after two wins = [(1+v*f)^2]*(starting B) = [(1+2*0.2)^2]*($100)
B after two wins = [(1.4)^2]*($100) = [1.96]*($100) = $196
The two losses results work out the same way. What about a win and a loss, you ask? You just multiply the starting bankroll by the "lose factor" (1-f), and the "win factor" (1+v*f), and the result is the new bankroll. Note that it doesn't even matter if you lost first or won first. Back to our example:
You won the first game and lost the second: $100 + $40 = $140 after first game, then lost $28 (you wagered 1/5 of it) in the second game for a total of $140-$28 = $112. If you lost first, you have $100 - $20 = $80 after the first game, and then you wager 1/5th of $80 (=$16) at 2-to-1 in the second game and win, for a total of $80 + 2*($16) = $112. Same amount as if you win first. [BTW, this less than obvious fact may affect retirement planning for those of you thinking about Roth IRA's vs. traditional ones. I won't digress any further on this topic.]
Using the equation gives the same result:
B after 1 win and 1 loss = (1+v*f)*(1-f)*(starting B) B after 1 win and 1 loss = (1+2*0.2)*(1-0.2)*($100) = (1.4)*(0.8)*$100 = $112
Okay, so lets say we have won "w" times out of n total games. This means we have lost n-w times (since we assumed no pushes). To find the new bankroll, we need to multiply the starting bankroll by (1+v*f) a total of w times, and multiply it by (1-f) a total of n-w times. In other words, our "new bankroll equation" has gotten much more complicated:
B' = [(1+v*f)^w]*[(1-f)^(n-w)]*B
The first factor in brackets is just the "win factor" multiplied by itself w times, and the second is the "lose factor" multiplied by itself n-w times. The two "B's" in this equation are not important, so we will instead look at just the factor multiplying B, as this is what determines the bankroll's growth (or lack thereof):
B'/B = [(1+v*f)^w]*[(1-f)^(n-w)]
This gives us the factor that the bankroll has changed from the beginning (after n games). We want to look at the bankroll's growth rate PER GAME, so if we call the average-per-game-factor "y", then after n games, the bankroll has grown by a factor of y^n. The average-per-game-factor is then found to be:
B'/B = y^n
y = (B'/B)^(1/n) = {[(1+v*f)^w]*[(1-f)^(n-w)]}^(1/n)
Again we return to our example. After 2 games where we won 1 and lost 1, we have gone from $100 to $112. This is an increase of a factor of 1.12 for 2 games. On average, this is an increase PER GAME of a factor of sqrt(1.12) = 1.058 . In other words, if in the game described, we win the same number that we lose (i.e. we assumed way back when that the probability of winning is 1/2), on average we will increase our bankroll each game by a factor of 1.058. Now here's the important part:
*** If we employ a different strategy (risk a different fraction "f" of the bankroll), then this factor will also change. We seek to find the f for which this factor is a MAXIMUM. ***
Now finding the value of f for which this function peaks is no small matter. It involves calculus. If this intimidates you, I invite you to jump down to below the second set of "*'s" to see the answer. I include the calculus for the math.weenies that may find it interesting...
********
The value of f for which y(f) is a maximum is the same value for which ln[y(f)] is a maximum, so we can equivalently seek to maximize:
z(f) = ln[y(f)] = (1/n)*[w*ln(1+v*f) + (n-w)*ln(1-f)]
The derivative is:
dz/df = (w/n)*v/(1+v*f) - [1-w/n]/(1-f)
Setting equal to zero and finding f gives:
********
f = [p*(v+1)-1]/v, where p = w/n.
Note that in the long run, the fractional number of times you will win a game (w/n) equals the probability of winning a single game, so p = probability of winning.
Okay, this is our answer. Given that your chance of winning is p, and that you are receiving v-to-1 odds on your bet, then the fraction of your bankroll that you should wager to maximize your rate of bankroll growth is the f given above. This value can also be plugged back in above to find out what the maximum growth rate actually comes out to be.
Let's try it for the game we've been using as an example. We have a probability of p=0.5 of winning, and odds of v=2, so the fraction of our bankroll we should risk each time is:
f = [p*(v+1)-1]/v = [0.5*(2+1)-1]/2 = 1/4
y = [(1+v*f)^p]*[(1-f)^(1-p)] = 1.061 (recall p=w/n)
The bankroll increases an average of 6.1% over its preceding amount every game. [Using the "rule of 72" familiar to bankers and investors, this means the bankroll will double roughly every 12 games.]
Most people like to remember the Kelly criterion using a mnemonic that goes something like: "Bet the fraction of your bankroll that equals your percentage advantage." It should be understood that this ONLY applies to bets with even money odds (v=1). Note that with v=1, f comes out to be equal to 2p-1, which is exactly your percentage advantage. The origin of this mnemonic is probably from blackjack, which would explain why the even money assumption is made. A better, more general mnemonic would be:
"Bet the fraction of your bankroll equal your percentage advantage divided by the 'to-1' odds."
For example, if you have a 10% advantage and you are getting 5-to-2 odds, then the fraction of your bankroll to bet is (0.10)/2.5 = 0.04 = 4%.
If you plowed through this whole post, my compliments. It's possible that the only people willing (able?) to follow it all the way through are the people who already understand all this, which means I was drawing dead as I wrote it. I hate when that happens.
How does the above post on the kelly criterion apply to poker?
If we want to maximize our expectation, we would maintain a "kelly" bankroll of [sigma^2/u]. If you never adjusted this bankroll, the risk of ruin formula shows that it would produce a risk of ruin of 13.5%. Most everyone agrees that maintaining only the Kelly bankroll is much too risky, even if you step up and down as required, because the fluctuations will be too great. Most people play some fraction of Kelly, like 1/2 or 1/4 Kelly, meaning that you maintain 2 or 4 times the Kelly bankroll. Just remember that in poker, unlike blackjack, stepping up usually means that the game gets tougher, so you need to wait longer before stepping up.
I often see the question "Which has more variance, limit or No limit?"
Hopefully this post should clear things up.
The swings in NL for a given blind structure are much greater. This is why NL requires a much larger bankroll for a given blind structure.
e.g. About $3000 for 200NL compared to about $1200 for 2/4 limit, both with blinds of 1/2.
By the mathematical definition of variance
"The average of the squares of the differences from the mean"
NL has much higher variance when looking at it from a pure $$$ perspective.
However, a skilled player's edge is much greater in NL.
A "good" win rate in NL is 10BB (20x the bb) per 100 hands.
A "good' win rate in limit is 3BB/100 hands.
So looking back at our 1/2 blind structure, A skilled player would be averaging $40/100 hands in NL, and $12/100 hands in limit.
Standard Deviation is the square root of variance.
A good standard deviation for limit is 15BB/100 hands
For NL, it is about 30BB/100 hands
So, while standard deviation is higher in NL, when you compare it to your winrate it is actually lower.
Looking at it from another perspective:
To get approximately the same winrate as our 2/4 limit game you could play 50NL.
Your winrate would be around $10/100 hands with a standard deviation of $30/100 hands,
compared to $12/100 at 2/4 with a standard deviation of $60/100 hands.
This is why the bankroll requirements for the 50NL game are only about $750, compared to our $1200 for 2/4 limit.
Another cool table:
Code:a 0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09
0.0 0.0000 0.0040 0.0080 0.0120 0.0160 0.0199 0.0239 0.0279 0.0319 0.0359
0.1 0.0398 0.0438 0.0478 0.0517 0.0557 0.0596 0.0636 0.0675 0.0714 0.0753
0.2 0.0793 0.0832 0.0871 0.0910 0.0948 0.0987 0.1026 0.1064 0.1103 0.1141
0.3 0.1179 0.1217 0.1255 0.1293 0.1331 0.1368 0.1406 0.1443 0.1480 0.1517
0.4 0.1554 0.1591 0.1628 0.1664 0.1700 0.1736 0.1772 0.1808 0.1844 0.1879
0.5 0.1915 0.1950 0.1985 0.2019 0.2054 0.2088 0.2123 0.2157 0.2190 0.2224
0.6 0.2257 0.2291 0.2324 0.2357 0.2389 0.2422 0.2454 0.2486 0.2517 0.2549
0.7 0.2580 0.2611 0.2642 0.2673 0.2704 0.2734 0.2764 0.2794 0.2823 0.2852
0.8 0.2881 0.2910 0.2939 0.2967 0.2995 0.3023 0.3051 0.3078 0.3106 0.3133
0.9 0.3159 0.3186 0.3212 0.3238 0.3264 0.3289 0.3315 0.3340 0.3365 0.3389
1.0 0.3413 0.3438 0.3461 0.3485 0.3508 0.3531 0.3554 0.3577 0.3599 0.3621
1.1 0.3643 0.3665 0.3686 0.3708 0.3729 0.3749 0.3770 0.3790 0.3810 0.3830
1.2 0.3849 0.3869 0.3888 0.3907 0.3925 0.3944 0.3962 0.3980 0.3997 0.4015
1.3 0.4032 0.4049 0.4066 0.4082 0.4099 0.4115 0.4131 0.4147 0.4162 0.4177
1.4 0.4192 0.4207 0.4222 0.4236 0.4251 0.4265 0.4279 0.4292 0.4306 0.4319
1.5 0.4332 0.4345 0.4357 0.4370 0.4382 0.4394 0.4406 0.4418 0.4429 0.4441
1.6 0.4452 0.4463 0.4474 0.4484 0.4495 0.4505 0.4515 0.4525 0.4535 0.4545
1.7 0.4554 0.4564 0.4573 0.4582 0.4591 0.4599 0.4608 0.4616 0.4625 0.4633
1.8 0.4641 0.4649 0.4656 0.4664 0.4671 0.4678 0.4686 0.4693 0.4699 0.4706
1.9 0.4713 0.4719 0.4726 0.4732 0.4738 0.4744 0.4750 0.4756 0.4761 0.4767
2.0 0.4772 0.4778 0.4783 0.4788 0.4793 0.4798 0.4803 0.4808 0.4812 0.4817
2.1 0.4821 0.4826 0.4830 0.4834 0.4838 0.4842 0.4846 0.4850 0.4854 0.4857
2.2 0.4861 0.4864 0.4868 0.4871 0.4875 0.4878 0.4881 0.4884 0.4887 0.4890
2.3 0.4893 0.4896 0.4898 0.4901 0.4904 0.4906 0.4909 0.4911 0.4913 0.4916
2.4 0.4918 0.4920 0.4922 0.4925 0.4927 0.4929 0.4931 0.4932 0.4934 0.4936
2.5 0.4938 0.4940 0.4941 0.4943 0.4945 0.4946 0.4948 0.4949 0.4951 0.4952
2.6 0.4953 0.4955 0.4956 0.4957 0.4959 0.4960 0.4961 0.4962 0.4963 0.4964
2.7 0.4965 0.4966 0.4967 0.4968 0.4969 0.4970 0.4971 0.4972 0.4973 0.4974
2.8 0.4974 0.4975 0.4976 0.4977 0.4977 0.4978 0.4979 0.4979 0.4980 0.4981
2.9 0.4981 0.4982 0.4982 0.4983 0.4984 0.4984 0.4985 0.4985 0.4986 0.4986
3.0 0.4987 0.4987 0.4987 0.4988 0.4988 0.4989 0.4989 0.4989 0.4990 0.4990
hm... never finished this thread. Oh well deserves a bump. I just posted this somewhere else, and figured it fit in here:
How many hands must be played in order to accurately know one's winrate within +/- .1BB/100, assuming respective standard variations of 10BB/100 and 20BB/100?
It depends on what level of confidence you want. I will compute this for 95% confidence and 99% confidence, but you probably won’t achieve 0.1 bb/100 accuracy in your lifetime at these levels.
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 0.1 bb to be about 1.96 standard errors. For 99% confidence, we need 0.1 bb to be about 2.58 standard errors. These are from a table of the standard normal distribution. (like the one above)
Call this number s, for the number of standard errors. Then we have:
0.1 = s*SD/sqrt(N/100)
N = 100*(s*SD/0.1)^2
We can use this equation to get the following results:
95% confidence:
For SD = 20 bb/100, N =~ 15.4 million hands.
For SD = 10 bb/100, N =~ 3.8 million hands.
99% confidence:
For SD = 20 bb/100, N =~ 26.5 million hands.
For SD = 10 bb/100, N =~ 6.6 million hands.
So "true win rate" is effectively impossible to determine, and therefore should not be relied too heavily upon in determining what your future profits may be.