Statistically Significant

I'm sure it's been asked a million times before, but I keep hearing varying answers. I just posed this question to the group, but we got sidetracked on another topic. When computing blackjack strategy, be it card counting or BS, how many hands need to be dealt before you finally reach the statistically significant threshold, where it's no longer short term but now long term? Is it 10 million, 100 million, more? Just wondering for my own computations.

Thanks!

There is no such definite answer when you deal with this kind of situation. Well, let's say it's infinity... as it comes closer to reality. The world of numbers is one that one must decide how much is adequate to get a fair sampling. This problem in the real world in commonly encountered in engineering. What is statistically significant, so to speak, for a given situation when it deals with public safety and wellbeing. Most of the time, the precision that is being looked for is not that significant to make much of an impact.

Your question is one of interest I assume. Because whether it's 10M or 100M or more the answers will not be not too different as far as to their significance.

I don't know what exactly constitutes statistically significant, but I'll take a stab at it. For whatever game you play, you have a certain win rate and standard deviation. After you play n number of hands, your expected win is EV/hand * n. Your standard deviation is SD/hand * sqrt(n).

If you are a basic strategy player, your EV/hand is about -.004 units(.4% house advantage), and SD/hand is 1.15 units. After you have played 10,000 hands at say, $10/hand, your EV is $-400, and SD is $1150, so your expected win/loss = $-400 +- $1150*SD. Now, after you have played 1,000,000 hands, your EV is $-40,000, and SD is $11,500, so your expected win/loss = $-40,000 +- $11,500*SD. You can see that as the number of hands go up, the SD/EV goes down. In the first case, the SD is 287.5% of the mean. In the second case, it is only 28.75%.

As a counter, your EV/hand may be something like .015 units, and SD/hand something like 2.80 units. When counting, many people like to refer to N0, which is a measure of how many hands it takes to overcome a negative fluctuation of one standard deviation. In other words, it is the number of hands that must be played so that EV = SD. To determine N0:

EV * N0 = SD * sqrt(N0)
sqrt(N0) = SD/EV
N0 = (SD/EV)^2

So in the example case, N0 = (2.80/.015)^2 = 34,850 hands. This says that after 34,850 hands, you will have earned money, unless you have suffered from greater than -1 SD's (16% chance). You may also determine how many hands it takes to overcome 2 and 3 negative standard deviations. These values are 4*N0 (139,400) and 9*N0 (313,650). This says that after having played 139,400 hands, you will have made a profit unless you have suffered from more than 2 negative SD's (about a 2.5% chance), and that after having played 313,650 hands, you will be on the plus side unless you have suffered more than 3SD's on the sad side (.15% chance).

To put these numbers into perspective, if you played a game with this EV and SD, 8 hrs a day and 5 days a week (100 hands/hr), for 35 weeks (about 139,400 hands), you would have a 1/40 chance of winning nothing or losing money. Then again, you would have the same chance of winning 4182 units (104K with a $25 unit). You can see that even though playing this number of hands gives you a sound chance of making money, you could win $5 or you could win $100K. This isn't something that I would want to have to rely upon to make serious money (aka professionally).

Hopefully this answered something...I'm sure there are others who could supply better answers. Good luck.

--Homer J.