Theory EV/SD hours to double:

[doublesnapper]

I currently play 10-20 online or 2 tables of 5-10 simultaneously. I'm trying to guess at my "Hours to Double" my poker bankroll of currently ~6K. My "guess" is at 4.5 BB per hour (playing 2 tables of 5-10 simult., i.e.: $45 per hour).
I read somewhere that EV and SD for a good solid player is 1.5 BB and 12.5 BB at a brick and mortar, so I'm using 1.5 times the speed for online and times 2 because the 2 tables for the EV, but I don't know what to do with the SD figure for 2 tables simultaneously, the only thing I know is that it is LESS than double the SD for 1 table.

My actual figures are higher for the EV and lower for the SD, but I started playing at the micro-limits and many times short handed. That must be the reason, worse players and many more hands per hour.

I would appreciate some advice on calculating my "Hours to Double" figure. Also would appreciate comments on higher limit players' actual EV and SD in bigger games.

One last question: How high is the limit holdem sky?. I.e.: What are the highest games available that still contain "live ones"?. I have no experience of playing other than online.

Thanks in advance.

DoubleSnapper.

Thanks in advance.

[BruceZ]

I read somewhere that EV and SD for a good solid player is 1.5 BB and 12.5 BB at a brick and mortar

That's quite good. A rule of thumb is that SD = 10*EV for B&M. Lower is better. This is meant to apply for higher limits also.


so I'm using 1.5 times the speed for online and times 2 because the 2 tables for the EV, but I don't know what to do with the SD figure for 2 tables simultaneously, the only thing I know is that it is LESS than double the SD for 1 table.

SD for 2 tables is sqrt(2) times what it is for 1 table. Since you are playing 1.5 times as fast as B&M, you are playing 3 times more hands per hour than B&M, so your SD will be sqrt(3) times higher than B&M. For you that is sqrt(3)*12.5 = 21.65 bb.


would appreciate some advice on calculating my "Hours to Double" figure.

If you just want average hours to double, you don't need SD of course. Just divide your bankroll by your win rate per hour. (600 bb) / (4.5 bb/hr) = 133 hours. For the time to double your bankroll with a particular probability, solve this for n:

n*EV - sqrt(n)*SD*x = BR

where EV and SD are for 1 hour
n is the number of hours
BR is bankroll in same units as EV and SD
x is the number of SDs for the probability of doubling

x is from the normal distribution as follows:

84%, x = 1
90%, x = 1.28
95%, x = 1.64

Get other x values from Excel function NORMSINV(p) or a table of the standard normal distribution.

This equation says that even if you are x SDs below the average value of EV*n, you will still have won BR dollars.

Your BR is 600 bb, EV = 4.5 bb, and SD = 21.65 BB, so for a 90% chance of doubling, solve

n*4.5 - sqrt(n)*21.65*1.28 = 600.

That comes out to n = 226. You can solve this by trial and error, or use the goal seek feature in Excel.

The number of hours to be ahead is (x*SD/EV)^2. This comes from solving the above equation for n with BR = 0. You have a 90% probability of being ahead after 38 hours.

[doublesnapper]

Thanks a lot for a very clear explanation Bruce.

What level is the highest limit game, that you know of, that would still be reasonably loose for the 1.5 BB EV to hold for a solid player?. In other words, how high can I aim at with the proper bankroll?.

Best Regards.

DoubleSnapper

[TheLoser]

I have played as high as 100/200, people there are still crazy, and in the really high limits you then get rich people who don't care about the money so the sky really is the limit.

[AliasMrJones]

[ QUOTE ]
x is from the normal distribution as follows:

84%, x = 1
90%, x = 1.28
95%, x = 1.64


[/ QUOTE ]

I understand all of the above, but this. For a standard curve, 1 SD = 68% of outcomes, 2 SD = 95% of outcomes and 3 SD = 99% of outcomes. To get 95% confidence, why don't you use a multiple of 2 rather than the 1.64 you reference above? (and a multiple of 1 for 68% confidence, 3 for 99% confidence based on the % of outcomes each SD multiple represents?) TIA as this has perplexed me since reading your previous excellent post about SD.

[Big Dave D]

I think its unlikely that you will be able to continue to win 4.5 BB/hr for two table play. My feel, based on others empirical evidence, is that a number around 3 BB/hr is more likely.

gl

dd

[doublesnapper]

Thanks for your reply. That's my strong suspicion too. It'd be too easy if one could double on average every 70 hours.

I wonder about the results of solid players in 100-200 games and above. Do they still make 1 to 1.5 BB per hour?.

Best

[Big Dave D]

Not playing that high, Im not sure. I suspect that its very game dependent. For example, online the 50-100 and sometimes the 100-200 is notoriously weak on Ladbrokes. And sometimes you get "fish out of water" scenarios in the Stars game, such as short handed players ending up in a full game and visa-versa.

gl

dd

[BruceZ]

For a standard curve, 1 SD = 68% of outcomes, 2 SD = 95% of outcomes and 3 SD = 99% of outcomes. To get 95% confidence, why don't you use a multiple of 2 rather than the 1.64 you reference above? (and a multiple of 1 for 68% confidence, 3 for 99% confidence based on the % of outcomes each SD multiple represents?)

We want the 1-sided confidence interval here, that is, the probability of being no more than x SDs below the mean. The figures you suggest are for the 2-sided confidence interval, for the proability of being +/- x SDs from the mean. So instead of 68% for +/- 1 SD, for > -1 SD we want 1 - 16% = 84%. 16% is 1 tail. > - 2 SDs would be 1 - 2.5% = 97.5%. 1.64 SDs is +/- 90%, but > -1.64 SDs is 95%.

[AliasMrJones]

[ QUOTE ]
We want the 1-sided confidence interval here, that is, the probability of being no more than x SDs below the mean. The figures you suggest are for the 2-sided confidence interval, for the proability of being +/- x SDs from the mean. So instead of 68% for +/- 1 SD, for > -1 SD we want 1 - 16% = 84%. 16% is 1 tail. > - 2 SDs would be 1 - 2.5% = 97.5%. 1.64 SDs is +/- 90%, but > -1.64 SDs is 95%.

[/ QUOTE ]

OK, thanks. I thought it must have something to do with the no less than (one-sided) aspect. One more question. I have created a spreadsheet that computes # of hours to break even and computes worst case scenario at 10,50,100,200 hours, etc. And, with a bit of calculus, when the max loss occurs and what the max loss will be. But, for fairly typical numbers (SD of 10 BB and 1 BB/hr. win rate) it shows a max loss of about 100 BB. Standard rule of thumb is a bankroll of 300 BB is needed. Why is that?

Also, one of the reasons I began investigating SD, etc. is a recent string of bad cards that took me down about 80 BB. The number of hours and max loss amounts I experienced were almost exactly what my spreadsheet shows I should have seen given my actual numbers as shown in PokerTracker. Perhaps some validation of the theory?

Also, FWIW, in another thread someone mentioned that since SD, win rate, etc. doesn't change your play it isn't worth much. I very much beg to differ. It is crucial to determining what limits to play, bankroll management and, as was the case for me, how to determine when to become concerned (or as was the case for me when not to be concerned) about a losing streak.

[AliasMrJones]

Sorry to keep firing more questions, but this is very interesting to me and I think I'm starting to get it. I found a table in a statistics book that I believe corresponds to what we're talking about that is labeled critical values of t. It shows a one-sided tail has rows with lables degrees of freedom (1-infinity) and columns of t.100 t.050 t.025 t.010 t.005. If I look at infinite degrees of freedom row and go to the t.050 colume (which I assume means excluding the tail with 1-.05 = 95% confidence) I find the number 1.645, which just happens to be very very close to the number you reference above. If I think this one step further, looking down the .010 column I find the number 2.326 SD for 99% confidence and 2.576 SD for 99.5% confidence. Is this all correct? (I am sort of assuming degrees of freedom is somehow related to sample size?)

[AliasMrJones]

OK, more research, got it now. T test is for small sample sizes. Normsdist Excel function returns probability of X being less than number input for standard curve with standard deviation 1. Normsinv returns the number for which X is the number input % likely to be less than. These two functions are already one-tailed so they can be used as-is to determine the number of standard deviations to use in our functions. So for our purposes, 95% confidence is 1.645 SD and 99% is 2.33 SD and 99.9% is 3.09 (or approx. my original 3!). The numbers look good, I think. For 99.9% confidence you might lose about 200 big bets, which given that you still need some bankroll to play with makes the 300 BB number for bankroll look reasonable. However, you are 99% likely to only lose 100 BB and 95% likely to only lose 58 BB. (Given SD of 13 BB and win rate of 2 BB.)

Don't know if anyone has mentioned it before, so I'll mention it now. The formula for determining number of hours of max loss is (normsinv(conf. %) * SD / 2 / WR)^2. To get that you take the derivitive of the original function for determining worst case profit/loss and solve for where it equals 0. (i.e. where the slope of the line through point = 0 or flat line, which would be a max/min of the curve.) You plug that number of hours into the original equation for worst case profit/loss, (WR * h) - (normsinv(conf. %) * SD * h^1/2), and you get the max loss amount.

This is very cool stuff, particularly since PokerTracker gives you your Win Rate and Standard Deviation, making it very easy to plug these numbers into simple excel functions and find out the answer to some very useful questions like "How confident can I be that I am a winning player?" "Should I be concerned about this losing streak?" and "How much money do I need to play at X limit?"

[doublesnapper]

Great stuff. Thanks a lot.

Best

DoubleSnapper

[BruceZ]

OK, more research, got it now. T test is for small sample sizes. Normsdist Excel function returns probability of X being less than number input for standard curve with standard deviation 1. Normsinv returns the number for which X is the number input % likely to be less than. These two functions are already one-tailed so they can be used as-is to determine the number of standard deviations to use in our functions. So for our purposes, 95% confidence is 1.645 SD and 99% is 2.33 SD and 99.9% is 3.09 (or approx. my original 3!).

That's correct, it is the integral from -infinity to x of the standard normal distribution. This area under the curve is called the cumulative distribution function, or cdf, while the normal distribution itself (bell curve) is the probability density function, or pdf. The last line of the t-distribution table for infinite degrees of freedom is the same as the cumulative standard normal distribution. You can use the t-distribution with lower values of N to take into account the uncertainty in your SD. Use N-1 degrees of freedom for N sessions. You also need to scale your SD by a factor of sqrt(N/N-1) if you use the maximum likelihood estimate for the SD, that is, the one that divides by N, or you can use the formula for SD that divides by N-1 directly without any additional scaling. I have given some examples of this in recent posts. The effect of this is small once you have a reasonable number of sessions, like 20-30.


For 99.9% confidence you might lose about 200 big bets, which given that you still need some bankroll to play with makes the 300 BB number for bankroll look reasonable. However, you are 99% likely to only lose 100 BB and 95% likely to only lose 58 BB. (Given SD of 13 BB and win rate of 2 BB.)

You are the latest in a long line of people to make a very common and subtle error. You cannot calculate risk of ruin and bankroll requirements this way. The problem is that you can go broke before you reach the point where your loss is a maximum. You're not allowed to go negative in a risk of ruin calculation. See my posts in this thread for the proper bankroll and risk of ruin formulas (http://forumserver.twoplustwo.com/favlinker.php?Cat=&Entry=1883&F_Board=genpok&Threa d=207100&partnumber=&postmarker=). These are derived in a manner similar to the classic gambler's ruin problem or random walk. To see how large a difference this makes, for an SD of 13 BB and a win rate of 2 BB/hr, for 0.1% risk of ruin you need -(13^2/2/2)ln(.001) = 292 BB, for 1% you need 195 BB, and for 5% you need 127 BB. Your bankrolls of 200, 100, 58 will give risks of ruin of 0.88%, 9.4%, and 25.3%.

[AliasMrJones]

Boy, just when I think I've got it and make a nice spreadsheet to calculate everything...

OK. So this seems to imply that the equation given for maximum loss at h number of hours is not (WR * h) - (SD * 2.33 * h^1/2). (For 99% confidence.) Is that correct? Because all the other formulas seem to be derived from this.

Or, is it just that the max loss formula smooths out the curve and the extremely short-term fluctuations can cause ruin?

I read the thread you reference, which says that the formula I used is incorrect and gives the correct formula and says that the incorrect formula is wrong because it doesn't take into account the possibility of ruin before the max loss hours. But, if it is really the minimum of the curve, I don't get how you could go broke before that point unless either of the two possibilities above are true.

Can you give a simple, but slightly more detailed explanation of why the one formula is correct and the other isn't?

[BruceZ]

So this seems to imply that the equation given for maximum loss at h number of hours is not (WR * h) - (SD * 2.33 * h^1/2). (For 99% confidence.) Is that correct?

This formula tells you that if you play for h hours, your total win (which is a negative number for a loss) will be greater than this value with probability 99%. You can use this to determine your confidence interval for your WR after h hours. For example, if this comes out to -$10,000, it means that 99% of the time, you will be no worse than $10,000 behind if you play to h hours. You CANNOT use this to say that $10,000 is the bankroll required to play these h hours in order to have a 99% probability that you will not go broke, or a 1% risk of ruin. This is because out of that 99% of the time that you reach h hours, some of those times you will have lost more than $10,000 before you reach h hours. In these cases, if you start with $10,000, your bankroll will go negative, and you will have to find some more money in order to continue to play to the h hours. This means that your risk of going broke before h hours is greater than 1%. In fact, it can be proven that the risk of going broke before h hours with this bankroll must be greater than 2%. More generally, for any h, the probability of going broke before reaching h hours is always greater than twice the probability of being broke at h hours, even if h represents the number of hours for which this loss is a maximum, which you correctly computed to be h = (2.33*SD/2WR)^2.

The question of how much bankroll you need for a given risk of ruin is a different one, and is not answered by this elementary equation. For a nice discussion of this fallacy, and the correct risk of ruin equations I have given (without derivation), you can refer to Don Schlessinger's book Blackjack Attack. I'm sure you would enjoy this book for the math content even if you aren't interested in blackjack; however, I think you can understand the fallacy from what I've written above. The bankroll formula I have given tells you how much bankroll is required to play forever with a given risk of ruin. There is an additional formula given in Blackjack Attack for computing the amount you need to play for a given length of time, and this has also been given before on this site. This latter formula can be used to determine how much money to bring on a trip. It isn't used to determine your total bankroll, because after a few hundred hours, the bankroll requirements become essentially the same as what you need to play forever, as I demonstrated in this post (http://forumserver.twoplustwo.com/showthreaded.php?Cat=&Board=mediumholdem&Number=34 0951&Forum=All_Forums&Words=play%20forever&Match=E ntire%20Phrase&Searchpage=0&Limit=25&Old=6months&M ain=339841&Search=true#Post340951). That is, if you don't go broke relatively early, it is unlikely that you ever will, assuming the conditions don't change.

[doublesnapper]

Thanks for all your input Bruce. You have clarified several lagoons I had in my gambling/poker risk management knowledge.

A rule-of-thumb question: When one has a bad run, at what level do you suggest stepping down in size?. I use a bank of 300 BB and when/if (it has happened once only so far) I lose 80 BB, I come down to half the previous playing limit.

When I played BJ many years ago, I use to adjust my bet size every 10% fluctuation in bankroll, but there the advantage was smaller and hence the adjustment more critical, and easier to do than in poker where the limits are fixed whereas in BJ one basically can size the max bet at will (pit scrutiny not withstanding... /images/graemlins/grin.gif).

Your rule-of-thumb for poker?.

Thanks in advance.

DoubleSnapper

[BruceZ]

The risk of ruin given by the bankroll formulas assumes that you will play at the same level forever, or until you lose your last dollar. If you are willing to step down in level when you lose a certain amount, then your bankroll requirement will be greatly reduced. Of course if you start out losing, your risk of ruin will immediately increase. The level you decide to step down is a personal decision based on the maximum risk of ruin under which you are willing to play. Keep in mind that when bankroll is reduced by a factor x, your risk of ruin will be raised to the x power. For example, if you start with a 1% risk of ruin, and then lose half your bankroll, then your risk of ruin will be .01^.5 = 10%, or 10 times greater. If you lose 25% of your bankroll, your risk of ruin will be .01^.75 = 3.2%, and so on.