What is the "long run" in blackjack?
https://wizardofvegas.com/forum/questions-and-answers/gambling/18140-what-is-the-long-run/#post357678
Here we are 3 years later. I read the entire thread once, and could not find an answer I could understand. I read the thread a 2nd time, and either got lost in tangential discussions, or the math went way over my head.
There did not appear to be a finite answer, even from the Wiz.
I am just hoping there might be some new opinions from the WoV forum mathletes? Please express in non-technical terms, if possible, or add explanations to terms used.
My question is "what is generally considered as the long run in Blackjack?"
75,000 hands? More? Less? Do you run a simulation calculating the number of hands after which you will experience getting positive returns? How do you account for betting, and different house rules?
I don't believe I'll be able to play "the long run". I have quite a ways to go, since I've only made 2 trips to Las Vegas to play BJ, totaling maybe 6 hours.
I just wanted to enjoy reading a clearer explanation to the "long run" compared to the 2014 post.
The long run is different for every one. It's when nothing surprises you, anymore, at which point you've seen enough, or just grown old. Same goes for the universe.Quote: LostWagesI just wanted to enjoy reading a clearer explanation to the "long run" compared to the 2014 post.
EDIT: In other words, not a lot to look forward to.
Thanks for your insights! While I've seen and experienced many things in only 67 years, I am not yet satisfied that I've "seen enough". Here's to another 10-20 years walking on reasonably steady feet! :-)Quote: InTimeForSpace1The long run is different for every one. It's when nothing surprises you, anymore, at which point you've seen enough, or just grown old. Same goes for the universe.
EDIT: In other words, not a lot to look forward to.
Some say the long run is when EV = actual. Of course, you can't calculate a number of rounds to play such that both will be exactly the same since variance exists. Instead, you need to determine how "off" your results can be from EV, and use that figure, as well as how likely you are to be within that range.
Say your Blackjack game is worth $100/hour and a 1% advantage. You can calculate how many rounds you'll need such that you'll have a 99.95% (or whatever degree of certainty you want) chance to be within 2% of your EV (0.98% to 1.02% edge).
RS - thanks for a MUCH clearer explanation than what I've been able to gather so far. I am math-challenged and not ready to ask more in equations or formula.Quote: RSIt's different for different games. Some will be the same or similar just by mere coincidence.
Some say the long run is when EV = actual. Of course, you can't calculate a number of rounds to play such that both will be exactly the same since variance exists. Instead, you need to determine how "off" your results can be from EV, and use that figure, as well as how likely you are to be within that range.
Say your Blackjack game is worth $100/hour and a 1% advantage. You can calculate how many rounds you'll need such that you'll have a 99.95% (or whatever degree of certainty you want) chance to be within 2% of your EV (0.98% to 1.02% edge).
My understanding of your explanation is that while there may not be a FINITE answer to the "long run", I should be better off determining my comfort level and acceptability RANGE of how "off" my results can be from EV. That is, how much variance can my bankroll support.
If I still didn't get it right, that's ok, I'll be patient and read some more. Thanks again, though! Appreciate your input.
It must be how well you described your insight about "long run". OK, I'm now on my looooooooong journey. Have a great week ahead!Quote: RSThat's right -- you got it.
Say you have the following:
AvgBet = $100
AvgAdv = 1%
OriginalSD = 1.1 * AvgBet = $110
EV(50,000 hands) = (50,000 * 100) * (.01) = $50,000
SD(50,000 hands) = Sqrt(50,000) * 110 = ~$24,600... 2SD = ~$49,000... 3SD = ~$73,800 ...so you could still be down!
EV(100,000 hands) = (100,000 * 100) * (.01) = $100,000
SD(100,000 hands) = Sqrt(100,000) * 110 = ~$34,800... 2SD = ~$69,600... 3SD = ~$104,000
Thus, to me 100,000 hands would basically be your N0, because with 3SD of confidence it's impossible for you to mathematically be down. The reason it's different for everyone as it refers to blackjack (in my opinion) is because everyone has different bankrolls, spreads (.5xkelly vs 1x kelly vs 1.5xkelly), average bets, and skill level / Average Advantage. An average counter might have a 1% AvgAdv, but a great counter with lots of deviations also selecting good situations could have a 2% AvgAdv. Then again if he wonged a ton more then perhaps he would have a 2% AvgAdv but it would take him twice as long to get the same amount of hands as the first guy. It's all relative to the choices you want to make with your game... which often times aren't even our choices but our available options around us. If someone only has 1 casino within 5 hours of them (and it's 5 min away) but it's a H17 8D game, well then they're going to have to play a crappy game AND a camp out strategy if they want to play and get hours/hands. However, if someone lived near 10 casinos and 3-4 of them had good rules, then this person could afford to make more picky choices/decisions to play only the better games.
Thus, it's different for everyone because we all live in different areas and have different situations as it comes to not only casinos/games, but also to our personal finances/bankrolls and the risk tolerance we want to play with our spreads/etc.
Thanks for a very eloquent and easy-to-follow (not necessarily grasp and understand!) mathematical description. At last, I now know what N0 means! It's been bothering me. I understand a little bit about SD and the bell curve for normal distributions.Quote: RomesThus, it's different for everyone because we all live in different areas and have different situations as it comes to not only casinos/games, but also to our personal finances/bankrolls and the risk tolerance we want to play with our spreads/etc.
I understand now why in other posts you mentioned that the long run would at least be 50,000 hands; then in another post, it was 75,000 hands. I forgot to take into consideration the situations you were describing, and they were both different - the bankrolls, the spreads, the betting strategy, and the types of games. I don't think I'll do much ABOUT the long run, but I just wanted to know the animal a little better and respect it for what it is.
Thanks to you & RS for the math-spirin!
1. SD is going to be substantially higher than 1.1 * AvgBet if you are not flat betting.
2. There is no number of hands where it is “mathematically impossible” to be down.
3. N0 is defined as 1SD, not 3SD. The exact definition is: “the number of rounds that must be played, with a fixed betting spread, such that the accumulated expectation equals the accumulated standard deviation. As such, it is a measure of how many rounds must be played to overcome a negative fluctuation of one standard deviation with such a fixed spread.”
4. At N0, the probability of being down is about 16%, not 0%.
5. Incidentally, the theoretic chance that you are ahead at N0 hands by 50% of your EV or better is about 69%, and the chance you are ahead by double your EV or better is about 16%. Many players consider this the “long run”.
Quote: RomesMy interpretation of the long run has always been the point at which it's mathematically impossible for you to be "down". In the mathematics of blackjack this has often been referred to as NO.
That is not what NO is, I've seen you state it several times without correcting it but have to at some point, it is a measure of the "long run", however. I think you are referring to 3x-4x NO as being NO. At 1 NO, you have I believe an 84% chance of being ahead, 2x NO about a 95% chance of being ahead; and 3x NO you will be ahead all but 3 times out of 1,000. That assumes perfect play as you simmed and NL mid shoe backoffs etc combination of which is impossible so may as well bump the numbers for long run up some.QFIT or EJ can correct that if it is wrong.
Number of rounds to NO will vary widely by the game very good DD or SD game played aggressively may be as low as 6,000 for 1 NO, marginal 6d 75% h17 DS is going to be about 30k NO with a 1-32 or so spread, not counting rounds below -1.
QFIT - appreciate your additional comments. I am unable to argue any issues, as the level of math conversation here is a bit over my head, but thanks anyway!Quote: QFITA coupla points:
1. SD is going to be substantially higher than 1.1 * AvgBet if you are not flat betting.
2. There is no number of hands where it is “mathematically impossible” to be down.
3. N0 is defined as 1SD, not 3SD. The exact definition is: “the number of rounds that must be played, with a fixed betting spread, such that the accumulated expectation equals the accumulated standard deviation. As such, it is a measure of how many rounds must be played to overcome a negative fluctuation of one standard deviation with such a fixed spread.”
4. At N0, the probability of being down is about 16%.
5. Incidentally, the theoretic chance that you are ahead at N0 hands by 50% of your EV or better is about 69%, and the chance you are ahead by double your EV or better is about 16%. Many players consider this the “long run”.
Quote: mcallister3200That is not what NO is, I've seen you state it several times without correcting it but have to at some point, it is a measure of the "long run", however. I think you are referring to 3x-4x NO as being NO. At 1 NO, you have I believe an 84% chance of being ahead, 2x NO about a 95% chance of being ahead; and 3x NO you will be ahead all but 3 times out of 1,000. That assumes perfect play as you simmed and NL mid shoe backoffs etc combination of which is impossible so may as well bump the numbers for long run up some.QFIT or EJ can correct that if it is wrong.
Number of rounds to NO will vary widely by the game very good DD or SD game played aggressively may be as low as 6,000 for 1 NO, marginal 6d 75% h17 DS is going to be about 30k NO with a 1-32 or so spread, not counting rounds below -1.
You’re on the right track. 84% (100-16) is correct. 2SD would be 97.5% (midway between 95% and 100%) and 3SD would be 1.5 times out of 1,000, again halfway. This is because the SDs straddle the standard normal curve (bell curve). Half ahead, half behind. But, 2SD would be 4xN0 and 3SD would be 9xN0 as this is squared, not linear. Takes rather a lot of hands to get down to that tiny a difference.
(Just came back from a meeting which involved fermented grapes; so EJ or the Wiz are welcome to correct. “Alcohol may be man's worst enemy, but the bible says love your enemy.”)
N0 and SD only behave this way for univariate normal distributions (68-97-99.5% rule).Quote: QFITYou’re on the right track. 84% (100-16) is correct. 2SD would be 97.5% (midway between 95% and 100%) and 3SD would be 1.5 times out of 1,000, again halfway. This is because the SDs straddle the standard normal curve (bell curve).
For distributions and games which have "long tails", you can still compute 2SD and 3SD, but the probabilities of being >2 SD and >3 SD will be much higher.
Conversely, some non-normal distributions will have "skinnier tails".
Quite right and I apologize for my ambiguity. Similar to SD when I say "SD" for something I generally refer to 2SD or 3SD because 68% confidence is a lousy assumption in my opinion and 1SD means practically nothing =P.Quote: mcallister3200I think you are referring to 3x-4x NO as being NO.
I think your #2 and #3 disagree with each other if you think of 3x-4x N0 similar to 3SD.Quote: QFIT2. There is no number of hands where it is “mathematically impossible” to be down.
3. N0 is defined as 1SD, not 3SD. The exact definition is: “the number of rounds that must be played, with a fixed betting spread, such that the accumulated expectation equals the accumulated standard deviation. As such, it is a measure of how many rounds must be played to overcome a negative fluctuation of one standard deviation with such a fixed spread.”
If your EV for a set number of hands is $1,000,000 and your 3SD is +/- $400,000... yes, I consider that a mathematical certainty to "not be down" at that point. The only way you would be down is if you're playing incorrectly or being cheated.
Quote: RomesI think your #2 and #3 disagree with each other if you think of 3x-4x N0 similar to 3SD.
If your EV for a set number of hands is $1,000,000 and your 3SD is +/- $400,000... yes, I consider that a mathematical certainty to "not be down" at that point. The only way you would be down is if you're playing incorrectly or being cheated.
At N0, it is not that unlikely to be down. 3SD equates to 9xN0. That can be a great deal of hands. In any case, I believe that precision in language is a must when discussing mathematics as whatever is stated may be used as a part of a further proof. Certainly at 9xN0 the odds of being behind are tiny, assuming perfect play/betting and no forced variations (table closings or limit changes, backoffs, chip confiscations, cover play, cover betting, low funds on hand with a hi count, wongers entering and eating cards at hi counts, etc.). But, there is no number of hands where it is “mathematically impossible” to be down.
Quote: FleaStiffI think long run is an illusory concept that we resort to when we've lost our money making decisions and tell ourselves that 'hit' and 'stand' were the correct decisions just made at the wrong time. If we had had a sufficient bankroll to attain this illusory 'long run' that is constantly over the horizon we would be leaving the table flush.
What?
Quite correct - but (genuine complement) you're so advanced in them that you often seem to forget your audience on different topics. A good teacher must curve their verbiage to the understanding of their pupils. Less you can be technically correct all day and no one would ever know it.Quote: QFITAt N0, it is not that unlikely to be down. 3SD equates to 9xN0. That can be a great deal of hands. In any case, I believe that precision in language is a must when discussing mathematics as whatever is stated may be used as a part of a further proof.
Depends on what you determine "mathematically impossible" to be. Many in the math community agree after 4 decimal places (for most calculations, certainly not molecular/atom level for physics calculations) the rest of the decimals are erroneous. Thus, if after 1,000,000 hands I have a .00001 chance of being down, I take that as a mathematical certainty that I won't be. I definitely believe there is a "mathematical certainty" with which each counter (counting/playing correctly and not getting bar'd over the world) will not be down anymore.Quote: QFITBut, there is no number of hands where it is “mathematically impossible” to be down.
The difference in N0 from rule and penetration changes is striking. Merely changing from S17 to H17 can increase N0 by two-thirds. Which is to say, the long run is pushed substantially farther away by seemingly minor changes.
Quote: RomesYes, so in the simple case of discussion on an online forums of counting cards in blackjack and the expected value +/- standard deviations I'm quite comfortable using the word "never" or the phrase "mathematical certainty" with regards to a player not being down after X hands. I'm not undervaluing the precision, as I'd still expect it out of any paper/book/etc, but in an online forum with an understood target audience I think a physics level of precision is complete overkill =P.
Understood. Problem is, if you say something that is highly unlikely never happens; I can prove that Martingale works.:)
rgds,
norm
My point is when the numbers are so minute... as in "history of the universe" status, then yes, mathematically speaking that's the same thing as 0. After 4 decimal places, no one usually worries anymore.Quote: DiscreteMaths2I just wouldn't use the word *impossible*. Because at some point in the history of our universe those extremely unlucky/lucky events still have to occur.
Hmmm, we're all "gambling folk" here. When people disagree, usually they put their money to the test and that straightens things up quickly. I'll propose the following:
I'll play 100,000 hands of blackjack, spreading 1-12, at a mixture of H17 and S17 games.
I'll bet anyone willing to take it, $25,000 that after 100,000 hands I WILL BE POSITIVE from my starting bankroll.
Anyone willing to book that action? Obv not... making a point (though let me know if someone wants to book that lol).
My other point was that newbies have enough to figure out rather than worrying about the 5th and beyond decimal places of N0 and SD's. Get to "the long run" is a widely preached lesson and it leaves many newbies asking "well, what is the long run???" Eventually, since there are so many different answers (look at my first post) we have to give them an "average" or "rounded" answer to give them an answer they can understand. That was literally the point of this thread, which is why my answer was very specific for the OP asking the question... and yes, still correct. Maybe not 5 decimal places correct, but 4 decimal places correct... lol.
Oh, I figured I was covered 10 ways from tuesday, though if someone was seriously interested more details would be drawn up and I was planning on actually counting a real shoe game.Quote: QFITAhh, one must be very careful when framing a bet. I could lead you to a table with a CSM. OTOH, you could still easily win that bet. Just bet $1 until the last few hands, and then use a progression. Progressions are not long-term winners. But, with a $25,000 bonus, they can certainly be positive EV.
(The blue line is theoretical, green yellow and red are deviations in 1,2,3 SD )
It seems that you need to play about 15 thousand hands, this is the confidence distance
Much more likely if Roams is in the room (-;Quote: QFITTheoretically, it is possible for all the oxygen in a room to move to one side of the room for an instant. It's just absurdly unlikely.
For the eradication of smallpox and measles (possibly in the future), 0 is very different from .0000001.Quote: RomesMy point is when the numbers are so minute... as in "history of the universe" status, then yes, mathematically speaking that's the same thing as 0. After 4 decimal places, no one usually worries anymore.
Mathematically speaking, small numbers are highly investigated. Consider "infinitesimals" and their use in the Newton-Leibnitz calculus.
As a gambler, I enjoy seeing some rare events; e.g. winning 15-17 hands in a row in Blackjack.
...and not others. Losing 29 of 30 hands (and pushing the other). Why couldn't I win 29 of 30 (and push the other)?
-----
For newbies, you can use (1) less than 1% (2) impossible.
It is helpful for newbies to know HOW RARE something is when you say "man I had really bad luck today".
----
In the 1990s, I had to decide if I wanted Lasix on my eyes.
My optometrist said
(1) he had Lasix, and considered it fairly safe
(2) But if i didn't want a 0.5% chance of screwed up eyes (bad night vision, etc...), don't do it
(3) But if I did want to do it, his doctor never had a problem, and fixed other people's problems.
I chose not to take the 0.5% chance on Lasix, and stuck with contacts & glasses.
Sometimes people want the "sure thing" (not the "very likely").
This is kind of exactly the idea behind my previous post.Quote: mamatFor the eradication of smallpox and measles (possibly in the future), 0 is very different from .0000001.
OMG you misspelled my name (like you did yours when signing up for this forum). ;-)Quote: AxelWolfMuch more likely if Roams is in the room (-;
I read that any given real number, say, the square root of two, has zero chance of being selected randomly among all reals, but that it could still happen. More likely, there is no way to thus select a number of all the reals.Quote: mamatMathematically speaking, small numbers are highly investigated. Consider "infinitesimals" and their use in the Newton-Leibnitz calculus.
Quote: QFITProblem is, if you say something that is highly unlikely never happens; I can prove that Martingale works.:)
Yup! I can confirm that Martingale works.
.
.
.
It's just a matter of how you define 'works'
$:o)
By my thinking, if time is infinite, and I will live for 100 years, then the odds of me being alive now are 100 divided by infinity, or zero. So, either I'm not alive, time isn't infinite, reincarnation exists, or all time instances exist at once (the most probable). But then, I'm not a mathematician.
Time is not infinite from your perspective (which you're looking at for your individual life span).Quote: QFIT...By my thinking, if time is infinite, and I will live for 100 years, then the odds of me being alive now are 100 divided by infinity, or zero. So, either I'm not alive, time isn't infinite, reincarnation exists, or all time instances exist at once (the most probable). But then, I'm not a mathematician.
https://en.wikipedia.org/wiki/Eternalism_(philosophy_of_time)
Quote: QFITBut, if time is infinite, and my life span 100 years, then there is no chance that I am currently experiencing any perspective. Unless you take the Eternalist approach to the concept of time. But then, I'm not a metaphysicist.
https://en.wikipedia.org/wiki/Eternalism_(philosophy_of_time)
basic probability is number of events you are looking for / total possible events. The assumption that time is infinite does not make the numerator or denominator infinity unless you believe in you are simultaneously living in many universes or something. There is a very large finite number of ways you might live and die and you would just divide them appropriately.
is no long run, the wheel resets after
every spin. Every pocket has an equal
chance of getting the ball every time.
The wheel has no memory. Nothing
changes. You are the variable in the
game, the game itself is a rock.
I think that if you found a way to combine all of the above, then you would have the answer. If the infinite becomes countable. It seems to me that at any given time, that mass and motion, etc, "all" converge into "one" centered stream "at a time" within all possible streams, which are analogous to how we view time, itself, as a deterministic dimension in which all events of our time may be spread out. The universe just seems to know which stream is centered (within the "infinity" of time) when. It keeps the when on the what, ie, (physical and mental) arrangement on its (physical and mental) stuff, and versa.Quote: QFITBy my thinking, if time is infinite, and I will live for 100 years, then the odds of me being alive now are 100 divided by infinity, or zero. So, either I'm not alive, time isn't infinite, reincarnation exists, or all time instances exist at once (the most probable). But then, I'm not a mathematician.
Quote: EvenBobDepends on the game. In roulette there
is no long run, the wheel resets after
every spin. Every pocket has an equal
chance of getting the ball every time.
The wheel has no memory. Nothing
changes. You are the variable in the
game, the game itself is a rock.
Well, there's a long run. It's just that it's not good. Unless you've mastered Laurence Scott's aural techniques. Or, you find one of those British machines that shot balls mechanically. That was a mistake.:)
Quote: QFITOr, you find one of those British machines that shot balls mechanically. That was a mistake.:)
All the machines that shoot balls are a
mistake, they're rigged to play like a
slot machine. They are not roulette, that
are not random.
Quote: QFITThere was a window where the UK machines were beatable. Alas, windows don't stay upon long. Must be the rain in London.
They run an algorithm. A few years ago some
machine repairmen in Chicago figured out
how to beat it and they took machines at
several casinos for a few hundred K. It's
been fixed now, mostly by speeding up
the spins and cutting the max bets.
Quote: RomesDepends on what you determine "mathematically impossible" to be. Many in the math community agree after 4 decimal places (for most calculations, certainly not molecular/atom level for physics calculations) the rest of the decimals are erroneous.
Name one person in the math community who thinks this? A Powerball ticket has something like a one in 300,000,000 chance of winning, yet they do hit regularly. Does this really mean lotteries somehow violate the rules of mathematical possibilities.
Quote: RomesI'll play 100,000 hands of blackjack, spreading 1-12, at a mixture of H17 and S17 games.
I'll bet anyone willing to take it, $25,000 that after 100,000 hands I WILL BE POSITIVE from my starting bankroll.
I'll take that bet to win $25,000 provided you give me something close to true odds. My 25-cents against your $25,000. You should have a mathematical edge, as you claim it is mathematically impossible to be behind
Quote: RomesMy interpretation of the long run has always been the point at which it's mathematically impossible for you to be "down".
This is very wrong, even if we were to change the wording a bit to be "near certain" or something similar that doesn't allow for picking nits out of shit. This means a basic strategy player can never reach the long run no matter how many millions of hands they play ... someone with an extremely slim edge can never reach the long run in a life time of playing ... while someone who only uses match-play, free-play, and fun chips can reach the long run after a few dozen hands.
I think RS nailed it in the very beginning. Rather than think about a specific number of hands, think about what x% chance you have of being within y% of your expected profits after z amount of time.
Instead of saying you can reach the long run with one year of full time (40 hour weeks) play, it would be better to say after one year you have a 95% chance of being within 25% of expected profits.
Quote: QFITBy my thinking, if time is infinite, and I will live for 100 years, then the odds of me being alive now are 100 divided by infinity, or zero. So, either I'm not alive, time isn't infinite, reincarnation exists, or all time instances exist at once (the most probable). But then, I'm not a mathematician.
We could also entertain the argument that 100 divided by infinity does not equal zero
You mean see a profit?Quote: TomGwhile someone who only uses match-play, free-play, and fun chips can reach the long run after a few dozen hands.
Even if you played thousands of match plays you probably wouldn't be at exactly the EV.
Year's ago, I played a nice little some of $25 match plays. I ran horrible, especially on my double downs. I ended up breaking even during that period.
I ran teriffic on the $50 match plays before and after that.
Quote: AxelWolfQuote: TomGwhile someone who only uses match-play, free-play, and fun chips can reach the long run after a few dozen hands.
You mean see a profit?
Certainty of seeing a profit is Romes' definition of long-run.
