AxiomOfChoice
AxiomOfChoice
Joined: Sep 12, 2012
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May 27th, 2014 at 3:03:02 PM permalink
Quote: kubikulann

Can you prove it mathematically?



I can, but I'm not about to take the time to write a formal proof.

Stop and think about it for a second. It's really, really obvious.

As for the large vs small cards, just remember that expectation of random variables is additive, regardless of dependence between the variables (in this case, the expectation of the true count). I feel like a broken record repeating this every day, but E(X) + E(Y) = E(X+Y), always.
kubikulann
kubikulann
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May 27th, 2014 at 3:05:44 PM permalink
If it were obvious, I'd seen it. Allow me some intelligence.

It is obvious if cards come out at random. But they don't ! They come out faster before a low hand than before a high hand. They come out faster when they are low themselves.

So definitely NO, it is NOT obvious that there should be an equal probability of going up or down.
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Sonuvabish
Sonuvabish
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May 27th, 2014 at 3:06:12 PM permalink
Quote: kubikulann

Irrelevant...

If it helps you to understand, consider players using BS only.



Sorry, I don't know what else to tell you. It has no effect. I am an AP counter, as is AOC. If you are not convinced, then you probably want a different answer.
Sonuvabish
Sonuvabish
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May 27th, 2014 at 3:07:32 PM permalink
Quote: kubikulann

If it were obvious, I'd seen it. Allow me some intelligence.

It is obvious if cards come out at random. But they don't ! They come out faster before a low hand than before a high hand. They come out faster when they are low themselves.

So definitely NO, it is NOT obvious that there should be an equal probability of going up or down.



I think you are confusing TC and RC. You mentioned you are playing at a CSM; if so, I don't really understand the context.
kubikulann
kubikulann
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May 27th, 2014 at 3:08:11 PM permalink
Quote: AxiomOfChoice

I feel like a broken record repeating this every day, but E(X) + E(Y) = E(X+Y), always.

I do repeat it day after day to my students:
E(X|A) + E(Y|B) does not equal E(X+Y)
You need the priors of A and B.
Furthermore, your "obvious" assertion seems to say that E(X)=E(Y) which is untrue.
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AxiomOfChoice
AxiomOfChoice
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May 27th, 2014 at 3:11:04 PM permalink
Quote: kubikulann

If it were obvious, I'd seen it. Allow me some intelligence.



Smart people miss obvious stuff all the time. That is why I asked you to stop and think about it.

Quote:

It is obvious if cards come out at random. But they don't ! They come out faster before a low hand than before a high hand. They come out faster when they are low themselves.

So definitely NO, it is NOT obvious that there should be an equal probability of going up or down.



The speed that the cards come out is irrelevant. Next you will be telling me that you have a betting system to make money in a 0EV game.

Do you agree that, if the current remaining shoe composition has a true count of n, and I draw exactly one card, the expectation of the true count after drawing that card is still n? So the expectation of the effect of that one card on the true count is 0.

Now suppose I have some system where I will decide whether to draw or not draw more cards based on what cards were previously drawn. How will you compute the expected change in true count? 0, plus 0, plus 0... you see where this is going.
Sonuvabish
Sonuvabish
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May 27th, 2014 at 3:13:11 PM permalink
Quote: kubikulann

I do repeat it day after day to my students:
E(X|A) + E(Y|B) does not equal E(X+Y)
You need the priors of A and B.
Furthermore, your "obvious" assertion seems to say that E(X)=E(Y) which is untrue.



Although I do not understand his math talk, I am fluent enough to know that is clearly not what the equation means. Lets make the X = 1 and Y = 2. E(1) + E(2) = E(3)

We are just trying to help you. Don't get mad because you are not getting the answer you want.
kubikulann
kubikulann
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May 27th, 2014 at 3:14:10 PM permalink
OK, let's recapitulate.

If, in a round, there are many "good" hands (with 9's, 10's, A's) there will be few Hits. At the end of the round, there are relatively less good cards in the deck (running count) but also many cards left, so that your true count is less badly affected.

If there were many "bad" hands (small cards) there have been many Hits. If those Hits were themselves low cards, at the end, there are reatively better cards in the deck, but also less cards overall, which increases your TC in two ways.

So in case one your negative evolution is tempered by the division, in the second your positive evolution is enhanced by division. Overall, TC should be better on average.

IF... the prior probabilities of cases One and Two confirm it.

Do they?
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AxiomOfChoice
AxiomOfChoice
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May 27th, 2014 at 3:15:44 PM permalink
Quote: kubikulann

I do repeat it day after day to my students:
E(X|A) + E(Y|B) does not equal E(X+Y)
You need the priors of A and B.



Yes, this is a good point. Be sure to remember to multiply 0 by the conditional probability that you end up drawing the card before adding it.

so instead of 0 + 0 + 0 + ....

we have 0 * p(A) + 0 * p(B) + 0 * p(C) + ...

Oh, look, they are the same...
kubikulann
kubikulann
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May 27th, 2014 at 3:18:28 PM permalink
Quote: AxiomOfChoice

Next you will be telling me that you have a betting system to make money in a 0EV game.

Please don't be insulting. Why such hatred when you can't understand something? Ego problems?

Quote: AxiomOfChoice

Do you agree that, if the current remaining shoe composition has a true count of n, and I draw exactly one card, the expectation of the true count after drawing that card is still n?

May I respectfully point out that you do not seem to take into account the difference between prior and conditional probabilities. Before I saw the card, expectations are unchanged. AFTER I saw the card, it has changed.

Now, if it is not ONE card, but the number of cards depends on what people have seen, Then definitely the independence is gone.
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