AxiomOfChoice
AxiomOfChoice
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May 28th, 2014 at 7:11:38 PM permalink
Quote: kubikulann

Now if the quotient of change E(New x)/x is different for each value, then evidently the relative proportions are not in the p/q/r ratio anymore. You can easily verify this by giving specific values to p, q and r.



That does not follow. You are adding numerators of fractions with different denominators. All that is happening is that more cards are used up when small cards come out (obviously) but the ratios remain the same.

Try computing E(p'/n'), E(q'/n'), and E(r'/n'), where "prime" denotes the new values. You will find that they equal p/n, q/n, and r/n, respectively. They have to. It's obvious. I know this without doing any computation of my own. Just like I know that when someone proposes a betting system to beat roulette, their system does not work, even though I do not take the time to do all the calculations. The number crunching is not necessary if you understand the math.
kubikulann
kubikulann
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May 29th, 2014 at 3:56:19 AM permalink
Quote: AxiomOfChoice

That does not follow. You are adding numerators of fractions with different denominators.

I'm not adding, I'm comparing. Anyways, that is not the problem. You don't need to use the quotients I proposed. You can simply show that the ratio of E(p') to E(r') is different from that of p to r.
Quote: AxiomOfChoice

Try computing E(p'/n'), E(q'/n'), and E(r'/n'), where "prime" denotes the new values. You will find that they equal p/n, q/n, and r/n, respectively.

EXACTLY.
The culprit here is the difference between E(p')/E(n') and E(p'/n'). They are not the same.
That (among other things) is what makes probabilities so counter-intuitive to so many people. (Cfr the Wizard's concept of Element of Risk, which is E(loss)/E(bet) instead of E(loss/bet). )

Now the question is: which of the two is the one that should be used in the present context? Both have their pro's and contra's.

Quote: AxiomOfChoice

They have to. It's obvious.

:-D Do you have a special intimate relationship with that sentence? That is almost a religious belief.
Like I said, nothing is obvious. Or rather, some so-called "obvious" things are in fact false. Especially when handling conditional probabilities. For example, the birthday problem, or the two coins problem, or Monty Hall. The intuitive solution is wrong.
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kubikulann
kubikulann
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May 29th, 2014 at 4:03:57 AM permalink
OBVIOUSNESS

1. Let's agree that it is a subjective concept, not an objective one. Something may be "obvious to this individual", but it cannot be "obvious" per se.

2. Let's agree that obviousness and truth are, if corrrelated, not perfectly correlated. It has long been obvious (to everybody) that the Sun was moving around the Earth, or that two parallels never crossed, or that gravity could not affect trajectories, or that gods exist. All showed eventually wrong.

3. The scientific method is
(a) to check. Even what seems obvious to one's eyes.
(b) to convince others. You can't transmit "obviousness", anymore than you can convince someone a thing is "beautiful" or "bad". What you can transmit is objective truth, and for this you need experimental or theoretical proof.
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AxiomOfChoice
AxiomOfChoice
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May 29th, 2014 at 11:42:21 AM permalink
Quote: kubikulann

EXACTLY



So, you admit that you are wrong?

Quote:

Now the question is: which of the two is the one that should be used in the present context? Both have their pro's and contra's.



Wow, it's rare that I'm overly-optimistic.

Do you understand what true count is? Or how it is defined? The original question was asking about the true count. Please don't go changing the question now. Obviously if you ask a different question, the answer is different. All I have been saying ALL ALONG is that the expected true count after a hand has been played is the true count before the hand is played.

Quote:

:-D Do you have a special intimate relationship with that sentence? That is almost a religious belief.
Like I said, nothing is obvious. Or rather, some so-called "obvious" things are in fact false. Especially when handling conditional probabilities. For example, the birthday problem, or the two coins problem, or Monty Hall. The intuitive solution is wrong.



My intuition in the two coins problem and the Monte Hall problem is right. I asked a lawyer-friend of mine the Monte Hall problem and she got it (intuitively) right away. She had never heard it before and has no math background. She is very smart. So, no, the intuitive solution is not wrong. Some peoples' intuition is wrong.

As for the birthday problem, it's intuitive to me that the answer is not 365. I won't pretend that the exact solution is intuitive, but intuitively I do understand that it's not going to be linear in the number of people. That's just silly.
98Clubs
98Clubs
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May 29th, 2014 at 12:25:13 PM permalink
I have followed this OP a bit, and here are my thoughts;
1.) When exposing the last card of the shoe, and when using a balanced count like hi/lo, the running count will be zero
2.) Because of the cut-card (stop-card) the final running count is uncertain until such final card is exposed. It may be above or below the average running count of the shoe at that point.
...a.) If there is a mathematical average running count at the cut-card, it will depend upon the fraction of penetration.
...b.) There is a corollary in that whatever the running count is, the closer one is to the cut-card, the less likely the
remaining cards IN PLAY will balance to an average value.
3.) Basic Strategy presumes that because no card is given a value other than zero, that the decisions made are based
upon an equal (uniform) distribution of rank and suit. Giving ranks a value other than zero allows changes to Basic
Strategy based upon non-uniform distribution. This leads to an advantage or disadvantage to the Player, rather than
a constant House advantage when playing Basic Strategy.
Some people need to reimagine their thinking.
MangoJ
MangoJ
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May 29th, 2014 at 12:53:44 PM permalink
Quote: 98Clubs


...a.) If there is a mathematical average running count at the cut-card, it will depend upon the fraction of penetration.
...b.) There is a corollary in that whatever the running count is, the closer one is to the cut-card, the less likely the
remaining cards IN PLAY will balance to an average value.



Could you explain in more detail what you mean ? The average running count at any penetration is always zero, if you are using a balanced count and have a fair deck. Especially the running count at the cut-card is zero on average. What changes is the fluctuation of the running count, it will be zero before the first and after the last card, and be maximal at the middle of the shoe.

to b) conditionally a given running count RC1 at penetration N1, the expected running count RC2 at a later penetration N2 is easy to calculate: E(RC2) = E(RC1)*(N-N2)/(N-N1), because the constant expected true count E(RC2/(N-N2)) = RC1 / (N-N1) (and N2 not a random variable).

So, for an observed running count, the further you advance into the game the more closer it will get to 0 on average. This is trivial, as the count is balanced and the shoe is fair, i.e. RC=0 at the very end of the shoe.
AxiomOfChoice
AxiomOfChoice
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May 29th, 2014 at 1:04:20 PM permalink
Yes, absolutely, starting in the middle of a shoe with running count RC != 0, that running count tends towards 0 as the shoe progresses.

However, the point is, the true count does NOT tend towards 0. For a partially-depleted shoe with a specific true count TC != 0, then expected true count after the next hand is played is TC -- it does not tend towards 0.

The fact that you are more likely to draw cards when the count goes up than you are when the count goes down is not relevant, because the expected change from drawing a card is 0. Multiplying 0 by some probability still yields 0. The correlation between the count going up and drawing another card does not matter for this (it does change the number of hands dealt before the cut card is reached, and therefore overall edge for the shoe, though -- this is known as the cut card effect. But that is a different question).

The original question is, does playing later (3rd base vs 1st base) confer an advantage in high counts, because the expected TC is higher by the time you get to play your hand. The answer is no, because the expected TC is not higher. The expected RC is higher, but that is irrelevant.
MangoJ
MangoJ
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May 29th, 2014 at 1:39:33 PM permalink
Quote: AxiomOfChoice

The original question is, does playing later (3rd base vs 1st base) confer an advantage in high counts, because the expected TC is higher by the time you get to play your hand. The answer is no, because the expected TC is not higher. The expected RC is higher, but that is irrelevant.



Still, 3rd base is better than 1st base. Both player have the same betting efficiency for the reason stated above (i.e. the expected TC when 3rd base makes his play is the same as the TC for first base when the bets are made). But 3rd base has a better playing efficiency when using indices as he sees more cards when he makes his decision on the updated TC.
AxiomOfChoice
AxiomOfChoice
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May 29th, 2014 at 1:45:45 PM permalink
Quote: MangoJ

Still, 3rd base is better than 1st base. Both player have the same betting efficiency for the reason stated above (i.e. the expected TC when 3rd base makes his play is the same as the TC for first base when the bets are made). But 3rd base has a better playing efficiency when using indices as he sees more cards when he makes his decision on the updated TC.



Of course it's better for that reason, but that's not really what this thread is about.
kubikulann
kubikulann
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May 29th, 2014 at 3:23:17 PM permalink
Quote: AxiomOfChoice

So, you admit that you are wrong?

There we are...

You are not searching the truth collaboratively. You are in a contest of egos and you need to be the winner.

For the record, a person cannot be "wrong". It is a statement that is wrong. (Or I am missing some specificity of English?)
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