I am due for luck

[Kasi]

..but what about for those on this forum that constantly post about their 6 month losing streak, and that it is natural in this game?

Ever notice how absolutely none of them ever say "I'm 3sd's down over the last 6 months" (even better a specific number of (guessed) hands played) or somesuch? That would be a player I highly regard.

Seemingly treating almost any amount of losses over almost any period of time as "natural". Often, in apparent disbelief of how "bad" their "luck" has been.

A player who would actually know, had a good guess,of how likely his 6 month losing streak was wouldn't likely even bother mentioning it in the first place if was less than 3 sd's.

 

[Jack_Black]

ok well, 6 months was a random number I threw out there. After re reading some of the posts on here about people's losing streaks, it has run the gamut, but I could have sworn I read somewhere that you must be mentally prepared to live through a losing session, and still play accurately because as long as you are playing accurately, your losing sessions will only be temporary. But ypou must know that losing sessions can last days weeks and even months. Months is where I should have stopped at. But then there was was the post about black swans, and being that you could do your best to reduce your risk, even down to 1%. 1% is still not 0% and thus there have been those occasional unlucky good players that have hit that 1% losing chance.

http://bj21.com/bj_reference/pages/playingwithborrowedmoney.shtml

Here was one of those horror stories. Icountntrack, I will keep that in mind that shitty games can contribute to long losing streaks. But good thing I analyzed my games to note an extremely low N0, so I SHOULD be safe.

 

[StandardDeviant]

Would you care to elaborate on how EV is not guaranteed?

You can demonstrate this quite easily. Take a quarter. Label 1 side "1" and the other side "-1". What is the expected value of any number of coin tosses? Zero , right? So, if expected value was "guaranteed," then we would always realize the expected value after any number of coin tosses.

Now, toss the coin 1 time. Did it come up zero? Why not? I thought this was a guarantee!

Oh the law of large numbers comes into play? Toss the coin 999,999 times, or 999,999,999 times. I'll wait. Did you get zero? I thought this was a guarantee!!

Now assume you toss that coin another 999,999,999 times and have a running count of -1. You have one more toss to go and you have a 50% chance of getting 1 which would give you an actual value of 0. I thought EV was a guarantee in the long run, not a 50:50 proposition?

EV is the value of the mean of a probability distribution. It is not a guarantee. All else held equal, in the long run, half of all players will do better than EV and half will do worse.

 

[StandardDeviant]

Anyone who has a 6-month losing streak would not be a "highly regarded" player in my book. A 6-month losing streak indicates a very weak overall game, possibly negative.

The player could just be unlucky. Given enough players, someone has to occupy the far left side of the standard distribution of returns from playing BJ.

But I agree with you that the unlucky player shouldn't be highly regarded. After all, if she's constantly losing, she won't have extra money to go shopping. And if she can't go shopping ... well, what fun is that? :laugh:

 

[iCountNTrack]

The player could just be unlucky. Given enough players, someone has to occupy the far left side of the standard distribution of returns from playing BJ.

I am sorry StandardDeviant but i cannot agree with your explanation, you are talking about the ev for one hand, that is not what players aspire for, they aspire for accumulated expectation and to the point where no matter how unlucky they are they will still be in a positive regime. I have attached a table with 4 columns for a strong counting game (N0=~5000 hands). The titles of each column are pretty indicative. You can see that for instance after 100000 hands (which could be achieved in 6 months) 4 times the accumulated SD is ±3863 which not enough to overcome the accumulated expectation (4537), meaning that there is a 99.994% chance you will be up after six months.

 

[StandardDeviant]

I am sorry StandardDeviant but i cannot agree with your explanation, you are talking about the ev for one hand...

Actually I am talking about playing a large number of hands. EV describes the mean of a probability distribution. This means that, after N hands, as N approaches infinity (i.e., "the long run"), half of all players are expected to have earned more than EV and half of them are expected to have earned less. EV is no magic guarantee that all players, even all APs will make money. The truth is that, given enough APs, some number of them will lose money no matter how long or how well they play. Don't blame me; blame Gauss.

This is what the math tells us, and the math holds true for all games and all casinos...except for Garrison Keillor's casino in Lake Wobegon. I'm told that there all players are above average.

 

[johndoe]

Actually I am talking about playing a large number of hands. EV describes the mean of a probability distribution. This means that, after N hands, as N approaches infinity (i.e., "the long run"), half of all players are expected to have earned more than EV and half of them are expected to have earned less. EV is no magic guarantee that all players, even all APs will make money. The truth is that, given enough APs, some number of them will lose money no matter how long or how well they play. Don't blame me; blame Gauss.

This is what the math tells us, and the math holds true for all games and all casinos...except for Garrison Keillor's casino in Lake Wobegon. I'm told that there all players are above average.

The beginning of this is technically correct, but you're reaching the wrong conclusion. Remember by the central limit theorem, as N increases, the variance of the distribution is scaled by 1/N. This narrows the 'width' of the Gaussian distribution as the number of samples increase.

http://en.wikipedia.org/wiki/Central_limit_theorem

So as the number of hands approaches infinity, variance approaches zero, and the real result approaches theoretical EV for everyone.

In your example above as N approaches infinity, with half being above and below EV, it's still technically true, but they're only infinitesimally above and below EV. Rounding to the penny and for all practical purposes, they all get exactly to EV eventually!

So all (sufficiently bankrolled) +EV players will eventually make money, guaranteed. Even if they're on the trailing end of the bell curve, as long as the bell is narrow enough (and EV is high enough), that trailing edge is still in positive territory.

 

[iCountNTrack]

Actually I am talking about playing a large number of hands. EV describes the mean of a probability distribution. This means that, after N hands, as N approaches infinity (i.e., "the long run"), half of all players are expected to have earned more than EV and half of them are expected to have earned less. EV is no magic guarantee that all players, even all APs will make money. The truth is that, given enough APs, some number of them will lose money no matter how long or how well they play. Don't blame me; blame Gauss.

This is what the math tells us, and the math holds true for all games and all casinos...except for Garrison Keillor's casino in Lake Wobegon. I'm told that there all players are above average.

Again you still dont get it When you play A LOT of hands you get to a point where accumulated becomes negligible when compared to accumulated expectation, at that point it is an excellent approximation to say that everybody will have the same result for instance after 1 billion hands everybody will have about 45 million units, but you cant say the same thing when you have only played 10 hands

 

[StandardDeviant]

Again you still dont get it When you play A LOT of hands you get to a point where accumulated becomes negligible when compared to accumulated expectation, at that point it is an excellent approximation to say that everybody will have the same result for instance after 1 billion hands everybody will have about 45 million units, but you cant say the same thing when you have only played 10 hands

So...if an unlucky player loses a bunch of money in her first few outings, the probabilities of the game somehow magically change so that she somehow earns more than her expected value in subsequent hands to arrive back at the expected value she had when she first started. Very funny. Is the Tooth Fairy involved in this somewhere? :laugh:

After 1 billion hands everyone who plays the will have earned 45 million units. Even funnier.

 

[johndoe]

Actually iCount prompted this comment, maybe it will clarify:

How would sims possibly work if each result (from near-infinite number of hands) could be anywhere on a (wide) bell curve?


(Uh oh, I have a feeling that JSTAT will use this as proof that sim's don't work..! )

 

[johndoe]

So...if an unlucky player loses a bunch of money in her first few outings, the probabilities of the game somehow magically change so that she somehow earns more than her expected value in subsequent hands to arrive back at the expected value she had when she first started. Very funny. Is the Tooth Fairy involved in this somewhere? :laugh:

After 1 billion hands everyone who plays the will have earned 45 million units. Even funnier.

That's exactly what it predicts! But not through the mechanism you're stating. The probabilities don't have to change.

The "magic" is that as the number of hands increase, the contribution of that brief case of bad luck (or any short event) shrinks, and eventually becomes buried in the noise. Nothing has to change, it's just that these events become less significant with more play time.

As for the 1 billion hands example, everyone who plays will have earned *very close* to 45 million units (assuming the EV predicts it). And how close that "very close" means is also quite predictable based on the statistics (variance).

 

[StandardDeviant]

That's exactly what it predicts! But not through the mechanism you're stating. The probabilities don't have to change.

If a player is expected to earn EV after N hands, but the player loses $10K after the first 1000 hands, then the player will be expected to earn EV-10K after playing the remaining N-1000 hands.

The probabilities of the game will not magically change so that the player is guaranteed to win EV+10K playing the remaining N-1000 hands.

 

[johndoe]

If a player is expected to earn EV after N hands, but the player loses $10K after the first 1000 hands, then the player will be expected to earn EV-10K after playing the remaining N-1000 hands.

The probabilities of the game will not magically change so that the player is guaranteed to win EV+10K playing the remaining N-1000 hands.

You still don't get it.

Yes, if you start off with -$10k then your "new" ev goes from there. But the more hands that are played, the less that initial $10k matters, because it will be swamped by wins and become less and less significant as the number of hands increases.

EV is a "slope", or a "trend", and is unrelated to where you "start".

 

[Sonny]

If a player is expected to earn EV after N hands, but the player loses $10K after the first 1000 hands, then the player will be expected to earn EV-10K after playing the remaining N-1000 hands.

Yes, but after N hands that $10k will be insignificant compared to the EV that he has earned. He will expect to be $10k below EV for the rest of his life, but he will also expect to earn many times that by the time he is done. The dollar amounts do not need to "even out" in order for the percentages to converge on the theoretical EV.

http://www.blackjackinfo.com/bb/showthread.php?p=67906

The game doesn’t need to magically change. As the number of hours increases, not only does the curve become narrower but the peak of the curve moves towards the right. Even a big loss at the beginning will not absorb all of the EV for the entire career. It is true that about half of the players will be below their exact EV and about half will be above it. It is also true that most of those players will not care about being a few dollars away.

-Sonny-

 

[StandardDeviant]

As the number of hours increases, not only does the curve become narrower but the peak of the curve moves towards the right.

Nice curve. What is it supposed to represent? Standard deviation of outcomes becomes wider as more hands are played, not narrower.

 

[johndoe]

Nice curve. What is it supposed to represent? Standard deviation of outcomes becomes wider as more hands are played, not narrower.

The standard deviation of the result is inversely proportional to the number of hands played. 1/N. It gets narrower.

 

[iCountNTrack]

At this point i am convinced that you are just trying to flame. After all the different phrasing, wording, table, graph, explanation, you still dont get it.

Ok for the last time:

Given that the expectation of one hand is A
The standard Deviation of one hand is B

After N hands, the accumulated expectation is N*A, while the accumulated standard deviation is sqrt(N)*B. Which means that after MANY HANDS, N*A>>>sqrt(N)*B because the term in N will win over the term with sqrt(N).

So in other words the more hands you play the normal distribution will get narrower, which in other words brings back to the example that after 1 billion hands everyone will have an expectation value very very close to 45 million units.

 

[daddybo]

That's Theoretical!

At this point i am convinced that you are just trying to flame. After all the different phrasing, wording, table, graph, explanation, you still dont get it.

Ok for the last time:

Given that the expectation of one hand is A
The standard Deviation of one hand is B

After N hands, the accumulated expectation is N*A, while the accumulated standard deviation is sqrt(N)*B. Which means that after MANY HANDS, N*A>>>sqrt(N)*B because the term in N will win over the term with sqrt(N).

So in other words the more hands you play the normal distribution will get narrower, which in other words brings back to the example that after 1 billion hands everyone will have an expectation value very very close to 45 million units.

This thread reminds me of a boss I once had.

I worked at a place that used lots of industrial gases. I once calculated the flow rates used to plug in a bid. Mine was considerably lower than the one used by my boss, who was quite experienced. I actually operated much of the equipment used in the process from time to time and knew my numbers were very accurate. When he questioned my numbers I explained how I arrived at them...his reply "Oh, that's just theoretical!"

I could not think of a reply!... :laugh: (Although I was thinking "No, it's more like empirical)

 

[StandardDeviant]

On further reflection...

My head hurts!

I'm trying to think thru this, now that the day is done and I can concentrate on this.

ICount is right in saying that, if cumulative EV is given by A*N, where A is the per hand EV and N is the number of hands, and if cumulative SD is given by sqrt(N)*B, where B is the per hand SD, then in the limit, as N approaches infinity, the ratio of cumulative SD over cumulative EV goes to zero. At large values of N, A and B are insignificant and can be dropped out of the equation. So we are essentially left with the limit as N->infinity of sqrt(N)/N = 0.

johndoe and Sonny must also be right when they say that "everyone" always gets to the predicted EV in the limit. After all, if in the limit, cumulative SD/EV is zero, SD can be ignored and we are only left with EV.

I think, but I am not sure now , that I am correct in saying that at any large N, the expected result will still be described by an EV and a SD, and that therefore half of all players will have realized more than EV and half less (for N < infinity).

And then there is the question of what happens as the number of players (X) approaches infinity. Since the normal distribution curve asymptotically approaches 0, there must still be some area under the curve for very small values of the Z score. Would this not mean that, as X approaches infinity, and as Z also approaches negative infinity, that the product of X times the area under the normal distribution curve from Z = very small to negative infinity is not zero. In other words, that despite EV being very large, and SD being very small in proportion, that there is still some probability that some poor player will have a negative result, if there are an infinite number of players.

I need a drink.

But first...I want to apologize. I was pretty flip earlier today. And you guys were right, I didn't get it (or at least parts of it anyway ). And I was particularly rude to johndoe, so a double down on apologies to you. Now, suitably humbled, I'm going shopping. :grin:

 

[psyduck]

Very interesting thread!

Isn't it true that one will need nine lives to reach the "long run"?