Questions about N0, Standard Dev. and SCORE from a Newbie

[Kyzz]

Hey guys, just wanted to start off by saying thanks to those who have contributed to my threads. Your replies help tremendously for someone starting out.

I have CVCX, and have never really paid attention to std dev, N0 or SCORE. I play a pretty good game imo, (6D, 3:2, H17, DAS, RSA, LS, 80% PEN), so I only really paid attention to the EV & RoR.

So I guess my question is, what are N0, std dev, and score in simplified terms? is N0 the amount of hands played before I reach the "long run"? Is standard deviation the short term of how far i'll actually be from expected results? I'm not necessarily a math wizard, so It's somewhat difficult to understand when looking on the web.

Thanks in advance!

[Three]

So I guess my question is, what are N0, std dev, and score in simplified terms?

n0 and SCORE are pretty much the same stat but in inverse form with a constant. n0 = 1,000,000/SCORE. n0 is the number of rounds played when EV equals SD. It is a sort of gauge as to how long you have play to get a certain degree of certainty for results. After n0 (in rounds) there is an 84% chance you will not be behind for your play. If you skip to the bell curve section in this article it gives you the percentage of the time you will be in the various ranges of SD from the EV at n0.
https://n8henrie.com/2012/01/sigmoid...ttractiveness/

Now how certainty is increased as you play EV increases literally as rounds accumulate (total EV = (number of rounds played)*(EV/round)) and SD increases as the square root of the number of rounds played ((SD for any number of rounds) = (SD/round)*sqrt(number of rounds played). So as you accumulate rounds the EV increases linearly, multiple of rounds played, while SD increases much slower, the square root of rounds played. So as EV moves further into positive territory as rounds accumulate more and more, the EV bell curve moves into positive territory because EV is its center or peak. SD increases slower at the sqrt function so more and more of the EV bell curve moves into positive territory as rounds accumulate as indicated by the multiple of SD that is at 0 EV in negative territory. At n0 rounds played, -1SD is at 0 EV (indicating 84.1% of the time you will not be in negative territory). Since SD increases by square root, at -2SD is at 0 EV at 4*n0 rounds played (indicating 97.7% of the time you will not be in negative territory). Continuing at -3SD equals 0 EV is 9 times n0 (indicating 99.9% of the time you will not be in negative territory).

Obviously when counting you want to reach a high degree of certainty fast, low n0 which means high SCORE, and you want to have a high EV. Now SCORE is the relationship between EV and the square of SD, which is variance. And n0 is the has the inverse relationship. SD is the square root of the average sum of the squares of the distance of each point from the mean of the distribution (SD = sqrt(sum of (x-mean)^2/n)) for each of the n points in a distribution where x is the value of each individual point in the distribution.

Minimizing n0 or maximizing CE may be the best things to do. If you have the same EV with a high n0, say 42,000 rounds and you average 100 rounds per hour, you will need to play 420 hours to be 84% certain that you will not be showing a loss at that point. With a n0 of 10,500 you will be 84% certain you are ahead after 105 hours of play. And 97.7% certain you will be ahead after 42,000 rounds or 420 hours. The higher n0 would need to play 168,000 rounds or 1,680 hours to get 97.7% certainty that you will not be in negative territory. Now both would have the same EV after the same number of rounds but with the higher n0 having a much higher SD results are more random (lower degree of certainty) for a much longer time as is sown in this example.

[Kyzz]

Thanks T3, I can always count on you on helping clear things up.

[Kyzz]

No and SCORE are pretty much the same stat but in inverse form with a constant. n0 = 1,000,000/SCORE. n0 is the number of rounds played when EV equals SD. It is a sort of gauge as to how long you have play to get a certain degree of certainty for results. After n0 (in rounds) there is an 84% chance you will not be behind for your play. If you skip to the bell curve section in this article it gives you the percentage of the time you will be in the various ranges of SD from the EV at n0.
https://n8henrie.com/2012/01/sigmoid...ttractiveness/

Now how certainty is increased as you play EV increases literally as rounds accumulate (total EV = (number of rounds played)*(EV/round)) and SD increases as the square root of the number of rounds played ((SD for any number of rounds) = (SD/round)*sqrt(number of rounds played). So as you accumulate rounds the EV increases linearly, multiple of rounds played, while SD increases much slower, the square root of rounds played. So as EV moves further into positive territory as rounds accumulate more and more, the EV bell curve moves into positive territory because EV is its center or peak. SD increases slower at the sqrt function so more and more of the EV bell curve moves into positive territory as rounds accumulate as indicated by the multiple of SD that is at 0 EV in negative territory. At n0 rounds played, -1SD is at 0 EV (indicating 84.1% of the time you will not be in negative territory). Since SD increases by square root, at -2SD is at 0 EV at 4*n0 rounds played (indicating 97.7% of the time you will not be in negative territory). Continuing at -3SD equals 0 EV is 9 times n0 (indicating 99.9% of the time you will not be in negative territory).

Obviously when counting you want to reach a high degree of certainty fast, low n0 which means high SCORE, and you want to have a high EV. Now SCORE is the relationship between EV and the square of SD, which is variance. And n0 is the has the inverse relationship. SD is the square root of the average sum of the squares of the distance of each point from the mean of the distribution (SD = sqrt(sum of (x-mean)^2/n)) for each of the n points in a distribution where x is the value of each individual point in the distribution.

Minimizing n0 or maximizing CE may be the best things to do. If you have the same EV with a high n0, say 42,000 rounds and you average 100 rounds per hour, you will need to play 420 hours to be 84% certain that you will not be showing a loss at that point. With a n0 of 10,500 you will be 84% certain you are ahead after 105 hours of play. And 97.7% certain you will be ahead after 42,000 rounds or 420 hours. The higher n0 would need to play 168,000 rounds or 1,680 hours to get 97.7% certainty that you will not be in negative territory. Now both would have the same EV after the same number of rounds but with the higher n0 having a much higher SD results are more random (lower degree of certainty) for a much longer time as is sown in this example.

Although I sort of understand N0 and Score, I just don't completely understand SD. So to get this completely straight, negative SD is a good thing? -1SD = my simmed N0 played, giving me a 84% chance of not being in the red after reaching those numbers? -2SD 4 times my N0, = 97.7%?

Also for example, my N0 is 14,033, 100 hands/hour. This means I have an 84% chance of being within my EV at 140 hours or 14k hands? Or does it just mean I'll have an 84% chance of being positive?

[Monty]

SD has to do with your results on the bell curve. At the top of the bell is your EV. -1sd is one standard deviation below your EV. Negative is not a good thing.
Having played enough to fulfill 1 n0 the -1 SD on the bell curve would = 0. It is true, as you stated, after playing enough to fulfill 1 n0 you would have an 84% chance of being at or above $0.

[Kyzz]

SD has to do with your results on the bell curve. At the top of the bell is your EV. -1sd is one standard deviation below your EV. Negative is not a good thing.
Having played enough to fulfill 1 n0 the -1 SD on the bell curve would = 0. It is true, as you stated, after playing enough to fulfill 1 n0 you would have an 84% chance of being at or above $0.

Oh I see. So it's 1SD = 84% of being above $0, and 2SD = 97.7% of being above $0? So I guess my question is, at what point do you reach the infamous "long run"? When does your hours played start to equate to your EV?

[Kyzz]

SD has to do with your results on the bell curve. At the top of the bell is your EV. -1sd is one standard deviation below your EV. Negative is not a good thing.
Having played enough to fulfill 1 n0 the -1 SD on the bell curve would = 0. It is true, as you stated, after playing enough to fulfill 1 n0 you would have an 84% chance of being at or above $0.

Also, how does one even get to a negative SD? By playing a losing game?

[RS]

SD (Standard Deviation) is how far away from expected results (EV) you can be within different degrees of certainty. Check out https://en.wikipedia.org/wiki/Standard_deviation for more information. Notice the picture/graph at the right, where it shows you'll be within 1 SD 68.2% of the time (34.1% between -1 SD and 0, and 34.1% between 0 and +1 SD). Look at 2 SD's and 3 SD's as well.

If your EV for 1 hour is $100 and your hourly SD is $1400, then you'll be within 1 SD 68.2% of the time. In other words, 68.2% of the time you'll land somewhere between -$1300 and +$1500 (that's $100 EV +/- $1400).

NOTE: Standard Deviation is NOT additive. EV on the other hand, IS additive. This means that if you have an hourly SD of $1400, you cannot say, "Ah, then my SD for 2 hours is $2800...and my SD for 10 hours is thus $14,000!"


Variance is additive. Variance is standard deviation squared. So for one hour, your variance would be $1400^2 or 1,960,000 (which, doesn't really mean anything at face value). If you want to figure out what your SD is for 10 hours, then you multiply your hourly variance by 10 to come up with 19,600,000. To find the SD, you have to find the square root of that, which is $4,427.




N0 is the number of rounds such that 1 SD = EV (IOW: Once you've played this many hands, if you're DOWN 1 SD, then you're breaking even). If your EV is $100 and your SD is $1400, then your N0 in hours would be found by doing something like this:

EV = Hours * Hourly_EV
SD = (SD^2 * Hours) ^ 0.5

EV = SD:
H * E = SD^2 * H
H * 100 = (1400^2 * H) ^ 0.5
H * 100 = (1,960,000 * H) ^ 0.5

https://www.wolframalpha.com/input/?...2+*+H)+%5E+0.5

Hours = 0
Hours = 196

It'd make sense that 0 is a solution, because if you play 0 hours, then your EV is 0 and your SD is 0 (ie: they're equals).

EV = $100 * 196 = $19,600.
SD = ((1400^2)*196)^0.5 = $19,600.

$19,600 == $19,600



4 N0's is the number of rounds needed to play such that EV = 2*SD, IOW: After this many rounds, if you're down 2 SD's from expectation, then you're breaking even.



Knowing this stuff helps with figuring out certainty and risk. You can look at the sim and think, "Okay, my ROR is 2%" -- but that doesn't really tell you anything in the 'tangible' sense. But knowing what your SD and N0 are, as well as how likely you are to be up or down some certain amount after some certain amount of time, is much more tangible -- ie: You plan a trip and expect to play 20 hours, you can figure out X% of the time you'll be somewhere between -Y (loss) or +Z (win) for the trip, and you can plan accordingly (how much $$$ to bring) as well.

It can also help you determine how strong (or weak) your game is and what kind of tweaks you may want to incorporate -- game rules, penetration, game speed, bet ramp/spread, and number of hours played. Obviously, it works for other games. IE: Maybe all you do is grind out full pay deuces and you want/need $X amount of income every year.....you can figure out how many hours you need to play in that year such that you'll have a 90% chance (or however certain you want to be) of hitting that amount in profit for the year. On the other hand, if you only look at EV, then you'll be cutting yourself short if you run bad and profit less than expected.


Perhaps you have 2 games to choose from, both have the same EV/hour, but one has a higher SD than the other but you can play more hours at it. You might be better off playing the game with lower SD and fewer hours, even if you're making less EV for the trip than playing the more volatile game. Or you might be better off playing the more volatile game for more hours. It's a matter of your opinion and what kind of risk (or lack of risk) you want to take.

[Three]

If you look at the SD formula you see you can't have a negative SD because you add the sum of squares where each addend would be positive since it is a number squared. Also the last step is taking the square root of the average of the addends. Square root either produces positive or imaginary numbers. RS explained things well enough. -1SD is 1 SD below the mean EV which for a bell curve is the peak most of the time.

https://www.mathsisfun.com/data/imag...tion-large.gif

I chose this graph because it has useful extra info at the bottom. You first have cumulative percentage going from the negative to positive. The first graduations are at the 1 SD increments. To see the likelihood you are ahead you need the go in from the positive but the info is symmetrical so you can use the info and flip the sign of the SD from negative to positive. Your expected EV will be the peak of the bell curve. When you have played n0 rounds -1SD is at 0 EV in the bell curve. At 4*n0 rounds -2SD is at 0 EV in the bell curve. At 9*n0 -3SD is at 0 EV in the bell curve. This allows you to see how card counting works in the long run. The chances you will be in any sector of the graph is given in the graph and is the area of the graph in that sector or SD range. The chances you will be at any point is given by the height of the bell curve at that point. So as you play more rounds the percentage of the bell curve that is in positive territory becomes larger (0 EV keeps becoming a more negative multiple of SD). This is because EV is a straight linear multiple of rounds played by EV/round and SD is multiplied by the square root of the number of rounds played by SD/round.

Now the higher the ratio of EV to SD the quicker you make the EV bell curve move into positive territory. This stat is desirability index (DI) but it is a logarithmic function not a linear function. Don came up with the idea (or at least made it popular) to make a stat that has a linear relationship for comparing games. By taking the ratio of EV to variance (SD^2) the resulting stat, SCORE, is a linear function. The movement of the EV bell curve in SD units into positive territory is a logarithmic function as you see by the 1, 4, 9 progression of rounds played for n0 to get 0 EV at SD increments. If it was expressed as a function of variance the increments would be 1, 2, 3 etc multiples of variance but the rounds played to get there would not change. So SCORE being the latter (a function of SD^2 or variance) allows us to quickly and intuitively compare things on a linear basis. Most people have trouble thinking in logarithmic terms. They may understand the skewing but quantifying it is not intuitive since the slide rule was replaced by the calculator.

[ericfarmer]

Variance is additive. Variance is standard deviation squared. So for one hour, your variance would be $1400^2 or 1,960,000 (which, doesn't really mean anything at face value). If you want to figure out what your SD is for 10 hours, then you multiply your hourly variance by 10 to come up with 19,600,000. To find the SD, you have to find the square root of that, which is $4,427.

Nitpicky, but we have to be careful here: variance is only "additive" (i.e., linear) when the summands are independent (or more precisely, uncorrelated). In your example, they are, but consider, for example, the variance of the net outcomes of the two different halves of a pair split. As you point out, the *expected value* of the overall round is the sum of the expected values of the two halves of the split. But the *variance* of the outcome isn't the sum of the variances, since the two halves are correlated.

[Three]

If you are the type that likes to understand why SD is not additive:

When you add the sqrt(a) + sqrt(b) it is not the sqrt(a+b). When you add more rounds you want the sqrt(a+b) where a and b are the average sum of squares of the difference between data points and the mean. Assuming the mean doesn't change:

If you square both sides of:
Givens:
SD1= sqrt(a)
SD2 = sqrt(b)
X= SD of both samples combined into 1 sample.

X = sqrt(a) + sqrt(b) you get:
X^2 = a + 2*sqrt(a)*sqrt(b) +b
Then take the square root of both sides:
X = sqrt(a + 2*sqrt(a)*sqrt(b) +b)
X does not equal sqrt(a + b) but you want sqrt(a + b) because you are just adding more data points to the sample. So you can't add SD's to get the SD of the combined sample set.

[DSchles]

Nitpicky, but we have to be careful here: variance is only "additive" (i.e., linear) when the summands are independent (or more precisely, uncorrelated). In your example, they are, but consider, for example, the variance of the net outcomes of the two different halves of a pair split. As you point out, the *expected value* of the overall round is the sum of the expected values of the two halves of the split. But the *variance* of the outcome isn't the sum of the variances, since the two halves are correlated.

And, of course, we have a formula for that, in blackjack, which includes the covariance (roughly 0.50), for when we calculate the total variance of two (or more) simultaneous hands against the same dealer upcard. See Griffin, bottom of page 142, or BJA3, page 20.

Don

[ericfarmer]

And, of course, we have a formula for that, in blackjack, which includes the covariance (roughly 0.50), for when we calculate the total variance of two (or more) simultaneous hands against the same dealer upcard. See Griffin, bottom of page 142, or BJA3, page 20.

Don

The Table 2.1 in BJA3 is a good example of my point here (although, to be fair, the error in this case is slightly different than the non-linearity described in my initial comment). The overall variance calculation neglects the impact of changing expectation vs. true count, as described here-- and as pointed out in recent email exchange as well. The table does not list the per-count expected values (as was pointed out to me by Gronbog, that's all we need, not the entire distribution of outcomes), so I can't comment on the magnitude of the correction.

E