I am due for luck

Cardcounter

Well-Known Member
#41
I'm due for bad luck!

Recently I have won like $1,200 playing blackjack in my last 3 sessions. Plus I sold Cit 2 days before the price collapsed saving me about $4,000 or 80% of my investment! I should step carefully and not feel to bad when I have bad luck at the blackjack tables in a couple of days.
 

iCountNTrack

Well-Known Member
#42
StandardDeviant said:
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 :confused:, 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. :confused:

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. :eek::grin:
Glad you finally get it :), the only thing i would like to point out is the number of players, it has no relevance to the discussion, we could have a million players playing the same game (same ev/hand, sd/hand) independently all over the world, each player will have a different accumulated ev and SD depending on the number of hands that each of them has played.
 

johndoe

Well-Known Member
#43
StandardDeviant said:
I think, but I am not sure now :confused:, 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).
You're still technically right, but remember that for large (non-infinite) N, the standard deviation gets really, really small, so for practical reasons it's not worth worrying about which side you're on. The distribution is always described by EV and SD, it's just that SD gets small, and eventually to zero as N approaches infinity.

This is often what the "long run" refers to - when the SD is small enough that you're "pretty certain" to be "really close" to EV.

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. :eek::grin:
Apology not at all necessary, but happily accepted. I understand how frustrating it can be sometimes when trying to work through this stuff.
 

psyduck

Well-Known Member
#44
Assuming SD for one hand is A. Then SD for N hands is sqrt(N)*A. As N increases, sqrt(N)*A should also increase, not decrease. Am I missing something? How can sqrt(N)*A approach zero with N?
 

iCountNTrack

Well-Known Member
#45
psyduck said:
Assuming SD for one hand is A. Then SD for N hands is sqrt(N)*A. As N increases, sqrt(N)*A should also increase, not decrease. Am I missing something? How can sqrt(N)*A approach zero with N?
sqrt(N)*A does not approach zero, it becomes negligible when compared to the much larger accumulated expectation. So basically it is like saying i have $100 give or take $5000 vs saying i have 1 billion dollars give or take $10000. For all practical purposes you can safely say person number 2 has 1 billion dollars, but you cant say that person number one has $100
 

Dopple

Well-Known Member
#46
So SD = EV @ 1.1%?

I am glad to see this post result is such a lively.... dialog. I do not totally grasp all the concepts real well but will in time.

Some strong player wrote that they seek to find EV and SD as equal in value.

From the basic SQRT(hands played) x 1.1 = SD does it not follow:

....................SQRT(100) x 1.1 = SD

.....................100 x 1.1 = 11 hands

now I am lost again, I was thinking 1.1/100 as SD per hand to give 1.1%

Could somebody help this junior rocket scientist back on his tricycle?
 

Kasi

Well-Known Member
#47
Dopple said:
From the basic SQRT(hands played) x 1.1 = SD does it not follow:
....................SQRT(100) x 1.1 = SD
.....................100 x 1.1 = 11 hands
now I am lost again, I was thinking 1.1/100 as SD per hand to give 1.1%
Hi Dopple - don't worry about it too much - I couldn't make heads or tails of it for a long time either. And still have troubles.

To make it simple - say you have a BJ game with such favorable rules the player has a 5% House advantage (on average). So a flat-betting BS player, after initially betting 1 unit for 100 hands will have an EV of 5% of 100 units or 5 units. If his unit was $10, his EV is $50.

After 100 hands it's, like you say for SD, 1.1*sqrt(100)=10*1.1=except 11 UNITS, not hands.

It means one Standard dev after 100 hands is 11 units. It means 2 stan dev after 100 hands would be 22 units, etc.

And, sort of importantlly, which I think you understand, it always means "how far actual results are from expected".

So, in this example, with an EV of +5 units after 100 hands, one's results will fall between =5+11=16 and +5-11=-6 units and be called within one stan dev.

Which also means, such results will happen about 68% of the time. (Don't ask me why ~68% lol. Just accept it as by "definition" lol.)

And don't worry too much about when EV=1 Stan Dev, although that is usually what "N0" means. Basically the point at which if one's results are 1 SD to the bad, one will not have lost money but be at zero. Often used to describe "long-run" the assumption apparently being it becomes likely that it is indeed your "skill" that explains such results rather than random chance.

Like, in this unrealistic example, it would occur after 484 hands becasuse one's EV would be 5% * 484=24.2 units. And one SD would be sqrt(484)*1.1=24.2 units. So after 484 hands, one's actual results will be between 24.2+24.2 or 24.2-24.2=0 and be within one SD of expected.

Hope that helps a little. In real AP BJ, SD will never be anywhere close to 1.1 anyway. I think. And an "avg bet" will almost never be just 1 unit on average like it only is for a flat-betting guy.

And, basically, it's only a sim that will tell your EV and SD in the first place which is why I consider it to be pretty much indispensable if betting any serious money over any extended period of time.

But I guess that's just a personal choice so don't worry about that either.
 

Dopple

Well-Known Member
#48
Sorry to be so late with the thanks. I really the time you take to help me understand the math better. I find nothing more enjoyable than watching the numbers work for you and see the money before your eyes as a result of ones study of the game.
 
#49
NO, NO & Oh NO

Two things to keep in mind when considering long run and SD issues.

For fixed bets once one play's 9NO (3sd) one should be around expectation.

If one were to continuously resize bets one needs to play 36NO (3sd), if one resizes less frequently it will be close to but less then 36NO.

So the goal is to bet properly and play to one's 3sd NO.:joker::whip:

If one's playing career only lasted about 1 NO then yes they could be substanitally behind the curve, so to speak!:joker::whip: Only earning a few $ over many hours is not much better then losing.:joker::whip:
 

StandardDeviant

Well-Known Member
#50
blackjack avenger said:
Two things to keep in mind when considering long run and SD issues.

For fixed bets once one play's 9NO (3sd) one should be around expectation.
Well, not exactly. 9N0 would be the number of hands one would need play to play in order for 99.5% of all outcomes to be > 0. This does not guarantee that one's outcome will be around expectation, unless we define the word "around" pretty broadly. It does however say that after 9N0 hands, it would be pretty unlikely for a player to still be losing money.
 
#51
Semantics?

StandardDeviant said:
Well, not exactly. 9N0 would be the number of hands one would need play to play in order for 99.5% of all outcomes to be > 0. This does not guarantee that one's outcome will be around expectation, unless we define the word "around" pretty broadly. It does however say that after 9N0 hands, it would be pretty unlikely for a player to still be losing money.
I agree
However, I did not say "guarantee" and yes the word "around" is pretty broad. Where will most players outcomes be at 9NO? Most players will have results clustered around EV. We agree that most will be winning.

So if you go back to the original post and someone wondering about their luck or when they are due for a win. My response is to play to the long run NOs.:joker::whip:

So
at 9NO fixed
36NO resizing
most will be winning
many will be around expectation

Nothing wrong with clarification!:joker::whip:
 

StandardDeviant

Well-Known Member
#52
blackjack avenger said:
Nothing wrong with clarification!:joker::whip:
I think we're in violent agreement.

I confess that I am on a bit of a one-woman campaign to keep people from assuming that counting guarantees that they will realize EV in the near term. :rolleyes:
 

Jack_Black

Well-Known Member
#55
what the F? you really are woman? I thought that was kind of strange when I saw a post of yours that said. "what's a woman to do?" I thought to myself, I think a woman could care less about what we were talking about. what the hell is this guy smoking? Most other females on here make it known right away, maybe that's why it is so shocking.
 
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