Theory EV/SD hours to double:

[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. 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. 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... [img]/images/graemlins/grin.gif[/img]).

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.