I am due for luck

[johndoe]

Originally Posted by StandardDeviant

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?

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]

Originally Posted by johndoe

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]

Originally Posted by StandardDeviant

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]

Originally Posted by StandardDeviant

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]

Originally Posted by Sonny

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]

Originally Posted by StandardDeviant

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]

Originally Posted by 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.

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!... (Although I was thinking "No, it's more like empirical)

[StandardDeviant]

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.

[psyduck]

Very interesting thread!

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