I am trying to use the CVBJ calculator to perform various calculations such as NO, ROR, etc. The problem I keep bumping into is that regardless which calculation I want to do I don't know what the SD is per 100 hands nor the Win Rate per 100 hands. Where can I get this information?
I realize CVCX gives both of these values however they are Per/Hr and NOT per 100 hands so I'm not sure what to make of that. I'm using KO Preferred. Thanks in advance for any help.
In most cases you can set a simulation to play 100 hands per hour. If this is done your SD per 100 hands is your SD per hr. If lets say you plan on playing 150 hands an hour then you need to do some multiplication to find your SD per 100 hands. Below is an example.
Simulation based on 150 hands/hr.
SD 27 units/hr
Win Rate 2 units/hr
Must calculate SD per 100 hands and WR per 100 hands.
27 units/hr * 1 hr / 150 hand * 100
This results in 18 units per 100 hands.
Similarly the Win Rate is 1.33 units per 100 hands.
"In most cases you can set a simulation to play 100 hands per hour. If this is done your SD per 100 hands is your SD per hr. If let's say you plan on playing 150 hands an hour then you need to do some multiplication to find your SD per 100 hands. Below is an example.
"Simulation based on 150 hands/hr.
SD 27 units/hr
Win Rate 2 units/hr
"Must calculate SD per 100 hands and WR per 100 hands.
27 units/hr * 1 hr / 150 hand * 100
"This results in 18 units per 100 hands."
No, that's not correct. While EV is a linear function, SD is not; it's a square-root function. So, instead of dividing the 27 units by 1.5, to get 18, you need to divide the 27 by the square roo of 1.5, which is 1.225. The correct 100-hand SD would then be 22.0 units per hour (or per 100 hands).
"Similarly the Win Rate is 1.33 units per 100 hands."
This part is fine.
Don
Note that the general CVCX setup is 100 hands per hour. So hourly SD is the same as 100 hands... I think Norm used that for all his sims that are canned in CVCX, at least the ones I have looked at have been that way...
Excellent, thank you Don. Fortunately I have run most of my sims at 100 hands an hour so my data is correct. But thanks again for the correction.
were of great help over the past 20 years or so, as were your books, of course.
AB
Rather busy over at advantageplayer.com, as you know.
Don
allows you to set hands per hour to anything you want.
I had not noticed you can change that after a sim had been run, but I tried it and it worked perfectly. I had noticed that everytime I open up an archived sim, or when I run one of my own, the default hands per hour seems to be 100... After thinking about it, it makes sense as the number of hands per hour has nothing to do with the sim itself, other than dictating how many hours it spans, since you fix the number of rounds up front when you run the thing...
thanks...
Thanks BP I didnt realize that. But how does one know HOW MANY hands they will play an hour while back counting? Do you just have to GUESS the number?
Lastly, what does a SD of 5.140/Hand and SD of 18.820/Hr mean? Can somebody explain this without getting too complex?
Thanks,
MJ
1. hands per hour varies. Most "humans" relate to hours, but sims relate to number of hands. So we want a convenient way of converting from "sim terms" (hands/rounds) to human terms, namely hours. How much will I win or lose per hour? how much will things vary over an hour? Etc... How fast you play depends on lots of things. 6D shoe heads-up with a fast dealer can go well over 100 to 150 hands per hour. If the dealer is willing. Hand-held games, or games with side bets can crawl along since for hand-held games, the dealer has to turn your cards over, add them up, and pay you off. The side-bets add another delay. 100 per hour is not a bad number. Every now and then you might get a dealer that is up to a challenge and that will play "full-speed" for a while, and you can double that if he is willing. Just remember he has to deal the cards, you have to indicate hit/stand/double/split, and then he has to play his hands, pay you off, take your chips, put them in the rack, etc. And he has to shuffle from time to time.
2. standard deviation is the statistical variation you can expect. A simple explanation is to add up all your hourly win/lose amounts, and divide by N. This is your "mean" win/lose amount. Now for each individual hourly win rate, subtract the mean from it, square the result, and divide by the number of total observations you have. Add these up. This is called the variance or "sigma^2". The standard deviation is simply the square root of that, or "sigma". What this is is basically an estimate of how far off your average outcome is from the observed outcomes. It is not the maximum your observed data can differ from the mean data, but it is an estimate of a "sane deviation". You can have observed data that is 2x or 3x the std dev, easily, but not frequently.
If you express your hourly win rate in "units won/lost per hour" then the standard deviation could be interpreted as this:
my expected range of win/losses is the hourly win / lose rate (aka the mean hourly win/lose rate) but it can be off by one std deviation in either direction without being unusual. It could be off by two std deviations in either direction but this is less common. And yes, you could even play a session where your expected win/loss rate is 10 std deviations away from the mean, on very rare occasions.
You will see terms like "one sigma or two sigma deviations" and that is exactly what they are talking about. You are way more likely to see a one sigma deviation than a two sigma deviation. Ditto for two sigma vs three sigma, etc...
The problem with BJ is that those std deviation numbers are pretty big. And our advantage is pretty small. That's why we see those +300 unit sessions and those -300 unit sessions on occasion.
Hope that helps. You can also search for "standard deviation and variance" on the internet and you'll probably find some good simple info..
And note that if you are interested in dollars per hour, rather than units per hour, you multiply your hourly win rate and standard deviation by the betting unit amount (CVCX does this under hourly win rate already to give it to us in dollars and cents).
sigma/sigma^2 are really independent of dollar amounts. They are related to the absolute win/lose/push results for a game and its specific rules. But since when counting, the rules and count influence our betting ramp, things tend to get a bit intermixed
the term "one standard deviation" generally means that if you look at all of your observed data, and compare it to the mean of that data, about 68% of the observations will be within one standard deviation of the mean. If you widen this to two standard deviations, about 95% of all observed data will fall within that range.
This dates back to the central limit theorem, and the so-called "normal distribution" or "normal bell-shaped curve" of distributions.
For BJ, take your hourly win/lose rate, +/- the std deviation, and you now have a "bracket" that says about 68% of your playing results are going to lie within that range. Note that the exact +mean+ of your observed data is not 0.00, because you do play with an advantage when counting. So the exact axis of the bell is just to the right of the y-axis. But notice how much of the damned thing is to the left of the y-axis, showing that we can lose significantly no matter what, just due to the luck factor associated with a shuffled deck of cards.
Not addressed specifically to you, but any aspiring counter ought to study the std dev stuff just a bit to see just how wild the roller-coaster can get. If you are a "easy puker" it might be the wrong ride to get on in the first plafce. :) Of course, over time, our average hourly win rate approaches the mean hourly win rate our advantage predicts, so that we eventually have a good chance of coming out ahead (the so-called N0 number of trials).
The 100 rounds becomes 100 rounds observed not 100 rounds actually played.
Thanks SSR. I sort of get it. Lets clarify. On CVCX my Win Rate is $24.00/Hr. Does that mean the 7.832 SD/Hand represents +/- 1 one sigma? In other words 68% of my hands played will fall within the range of $24/Hr +/- $7.83? Is this correct? If so then 95% of my hands will earn $24/Hr +/- $15.86(2 sigmas)?
If that is correct I can apply the same idea to SD per hour I suppose. But something tells me I am adding apples and oranges in the above example because I was never given the Mean earning per Hand.
Lastly I don't understand how 1 SD is MORE likely to occur then 2 SD. If 1 SD falls within a range that occurs 68% of the time, and 2 SD is falls within a range occuring 95% of the time, then 2 SD clearly occurs MORE OFTEN then 1 SD. Oddly enough it seems as though the further we veer away from the mean the more likely an event will occur. Where am I going wrong? Thanks for any answers.
-MJ
"SD" _is_ "one sigma". That is the definition. :)
The point of probable confusion is that the SD is not dollars, at least as presented by CVCX. It is in terms of "units".
So if your hourly win rate is X, and the StdDev is Y, then your hourly win rate is X +/- Y*unit
You have to look at the data to see what the betting unit is at TC < 0, to figure out what the unit is, unless you do as I usually do and set the min bet unit to $5, which is sane for most games unless you play at a $10 or $25 min table. I also set the min chip size to $5 as well so I get bets that are multiples of $5.
Your last question comes from the math involved. Remember that the base calculation is to compute the mean of all observations, then sum up the squares of the differences between the mean and each individual observation, and divide by N, the number of observations. So we work with the square of the differences, divided by N. Std Dev is then the square root of this term. That's where the term "standard deviation" and "one sigma" comes from. Roughly 68% of observations will be within 1 SD of the mean of all observations. 95% within 2SD, etc...
Now for the next problem. Remember that 68% of your observations will fall within 1SD. 95% of all observations will fall within 2SD, but you have to remember that of that 95%, many will _also_ be within 1SD, only 27% will be between 1SD and 2SD. Does that help? Remember this is the classic bell-shaped "normal curve"
If you remember your calculus, and can construct an equation for the normal curve, integrating it and solving with limits -inf, +inf to get the total area under the curve. If you substitute -SD, +SD for the limits and solve, you get about 68% of the amount you get using +/- infinity. If you then plug in -2SD and +2SD, you get about 95% of the total area under the curve. But much of that area is also inside the -SD,+SD area as well.
What this all means is that about 2/3 of the time, you can expect your results to fall within 1SD of the expected value. 95% of the time you can expect your results to fall within 2SD of the expected value. If you play for the rest of the life of the universe, you could expect your results to fall within +/- infinity. :) Of course, 68% of your results will fall within +/- 1SD still. But with an infinite number of hands, you could certainly expect to play a long session where you lose every last hand.
Thanks SSR. Ok I think I finally got it. Now let me if I can finish the calculation:
Unit = $25.00
Win Rate/Hr = $52.15
SD/Hand = 4.375 units = $110.00
SD/Hr = 43.75 units = $1100.00
100 Hands/Hr
So the occurences are broken down as follows:
68% fall within -$1047.85 to $1152.15
95% fall within -$2147.85 to $2252.15
99.7% fall within -$3247.85 to $3352.15
Ok so let me see if I can make sense of the overall picture. I guess what this does is provide the counter with an EXTREMELY likely range where he can expect his earnings/losses to fall after playing a certain # of hands. It would be NEARLY impossible for earnings/losses to be outside of 3SD right?
Now lets analyze the bell curve a bit. There is a 34% chance( SD) earnings will be in the range of $52.15(Mu) to $1152.15. Now lets say I win $2000 at the tables. Would that occur 2SD - 1SD = 48% - 34% = 14% of the time?
$2000 is between 1 SD and 2 SD but I'm not certain if this logic is precise. Is it EXACTLY 14%? If not is there anyway I can calculate exactly how often I would be ahead $2000?
One last question. If my results are 1.5 SD to the right of the mean what range would my results fall into? How would compute this?
Thanks for any answers you can provide.
-MJ
1. "nearly impossible to fall outside 3SD". nearly, but _not_ impossible. You really ought not see numbers outside 2SD but they will happen. Good or bad...
2. For your second question, let me make sure I understand it. You have a win rate that is on the "edge" of the 1SD spread. That is, 68% of the time, your win/lose will be +/- 1SD. So if you are within 1SD, that is what you should see 68% of the time. Technically then, since you ended up positive, you will see that about 34% of the time (positive means on the + side of the mean, where the mean should be positive to start with because of your +EV when playing by counting). So when you end up over +1SD, but less than 2SD, then yes, you are in the 50% of that particular probability, which is about 14% of the time as you said...
3. Now things get more fuzzy. Your $2k winnings will not happen 14% of the time however. 14% of the time you will end up somewhere between 1SD and 2SD, but that is integrated over that entire range. For example, you are _more_ likely to end up +1.1SD than you are to end up at +1.9SD. the normal curve and SD stuff is about the area underneath the curve between two points on the X axis. Trying to predict the exact probability of a specific sample is a different issue. There is a formula given a mean and SD, to compute the probability of a specific outcome, but I'm not at my office with my stat books handy. You might have luck searching with google or whatever. But what you are wanting to do is to pick a particular point on the normal curve, then find the x-coordinate value on the X-axis (the standard deviation axis) and find the corresponding y-coordinate value which will be the probability that you hit that exact win rate. Note also that there will be an associated error term with that value since it is an estimate based on a finite number of observations used to compute the original SD.
4. Your last question is also one that I'd have to look up, but yes, you can solve for any of the variables. You can find the probability of a specific outcome, which is what you are asking for... One simple way is to buy some graph paper, and plot the normal curve with your specific std dev. Now you can look at the intersectin of the X and Y coordinate values for a point on the curve to find the probability of hitting that specific x or y value, within the error bound...
Sorry I couldn't help more with the last couple of questions without a book handy. I use this kind of stat frequently, as a computer scientist, but I don't really care about the inverse operations you are asking about. For example, when I am testing a parallel algorithm, I care about the speedup, which can be variable for the same application and data, and the standard deviation of that speedup to get an idea of how consistent (or inconsistent) the parallel performance is.
