kubikulann
kubikulann
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May 27th, 2014 at 3:21:54 PM permalink
(Unkind comment deleted)
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AxiomOfChoice
AxiomOfChoice
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May 27th, 2014 at 3:26:07 PM permalink
Quote: kubikulann

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 I'm not sure that we are answering the same question.

The question I am answering is this:

Supposing I back-count a blackjack game, and observe that the true count is n. What is the expectation of the true count after the next hand is played? What is the expectation of the true count after the next two hands are played? The next 3?

The answer to all those questions is n.

The only caveat is that the number of additional hands to be dealt will depend on how the count moves, so it may be that the 3rd hand is played only if the count goes down, but not if it goes up (so if we restrict the question only to the cases where it is dealt, we are introducing bias and the expectation is no longer n. This is precisely the cut card effect. It's still true that for all hands from now until the end of the shoe, the expected true count is n, but you may play more hands if the count goes down than if it goes up, so your expectation if you play until the end of the shoe may not be so trivial to calculate)
Sonuvabish
Sonuvabish
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May 27th, 2014 at 3:35:56 PM permalink
Quote: kubikulann



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.



Expectations of what? If the count changes, then it changed. Your new expectation is the same as is was before--the count will not change. Are you trying to determine the likelihood that the count will change? It constantly fluctuates. It has already been said that the count tends to remain static. Changes tend to be incremental. The running count tends toward zero. I give up. Good luck with your analysis.
AxiomOfChoice
AxiomOfChoice
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May 27th, 2014 at 3:58:03 PM permalink
Hopefully we can put this issue to rest with a simple example.

Recall that TC = RC * 52 / (number of cards left)

Suppose we are left with 25 high cards, 15 low cards, and 12 neutral cards. TC = 10 * 52 / 52 = 10.

What is the expectation of the true count after drawing 1 card?

TC_h = TC after drawing a high card is 9*52/51
TC_n = TC after drawing a neutral card is 10*52/51
TC_l = TC after drawing a low card is 11*52/51

E(TC after drawing a card)
= (25/52 * TC_h) + (15/52 * TC_l) + (12/52 * TC_n)
= (25/52 * 9*52/51) + (15/52 * 11*52/51) + (12/52 * 10*52/51)
= 510*52 / 52*51
= 510 / 51
= 10

By replacing numbers with variables, it is not hard to show that for any deck with more than 1 card, E(TC after drawing a card) = current TC. I leave this as an exercise to the reader.

Ok, now suppose we draw one card, and, if it's a high card, we draw another. If it's neural or low, we draw no more cards.

To get the expected true count after this sequence of 1 or 2 draws (where the 2nd draw is conditional on the first), in the bolded line above, we reply TC_h with the expected TC after drawing a card from the deck with 24 high cards, 15 low cards, and 12 neutral cards. But this is the same as TC_h, by what we established above. So, the answer is still 10. The answer will continue to be 10 no matter how we decide to draw or not draw cards (unless we deplete the whole deck, since the TC on an empty deck is undefined) because, by conditionally drawing more cards, all we are doing is conditionally replacing some true count with the expected true count after drawing one more card, ie, conditionally replacing a value with the same value. We are replacing x with p*x + (1-p)*x.

That is as close to a formal proof as I'm going to get. Replacing numbers with variables and doing some annoying algebra will give you your formal proof.
MangoJ
MangoJ
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May 27th, 2014 at 3:59:36 PM permalink
Quote: Boney526

I disagree with the couple of people who said the count wouldn't tend to go towards 0 f you stopped counting, because obviously at the end of the deck the count will be 0 again. I don't really think that's a useful piece of information, though, since if you missed a part of the deck it'd be more accurate to count those as burned or undealt.



Oh no... not again. The running count tend towards zero, true. The remaining decks N also. After the last card dealt (if it would), RC=0, but N=0 also. TC = RC / N is not defined here ... claiming it is zero is simply a false statement.


The essence is, the shoe is assumed to be shuffled well, and any undealt card is likely to be anywhere in the remaining shoe with equal probability. The true count is a precisely defined property of the shoes remaining composition - which statistically speaking is the same for any portion of the undealt shoe.
Not counting any part of the shoe is equivalent as taking that part out of the shoe and placeing it in the discard unseen. Hence the true count - as a property of the remaining cards - stay the same when you miss cards. The only adjustment is to take as N all unseen cards, not the cards left in the shoe.
kubikulann
kubikulann
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May 27th, 2014 at 4:00:52 PM permalink
Thanks, Axiom.

( Note: is that what you called "obvious"? )
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AxiomOfChoice
AxiomOfChoice
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May 27th, 2014 at 4:07:58 PM permalink
Quote: kubikulann

Thanks, Axiom.

( Note: is that what you called "obvious"? )



Yes, IMO it is intuitively obvious. The proof is long but it's all simple high school algebra. I understood why it was true before I wrote down the proof.

There may be a simpler way to state it in terms of expectations of random variables.
Sonuvabish
Sonuvabish
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May 27th, 2014 at 4:10:42 PM permalink
Quote: kubikulann

Thanks, Axiom.

( Note: is that what you called "obvious"? )



I used to let AOC hang out with me if he did my math homework and bought me cigarettes. Then I crashed his car and abandoned it as a practical joke, and we haven't hung out since.
Boney526
Boney526
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May 27th, 2014 at 7:54:03 PM permalink
Quote: AxiomOfChoice

You suspect it would what? True count tends to remain the same as more cards are dealt -- ie, it is equally likely to go up as it is to go down, regardless of what the current TC is. This is obvious from the definition of true count.

The TC at the end of the shoe is not 0, it is undefined.



I suspect that on average, if you started at a True Count of 5 say 5 decks in, then the average true count, say 2 decks, in would be less than 5 - since the closer you get until the end of the deck, the closer you should get to (an average count of) 0 - before it becomes undefined as you pointed out.

I mean even if I'm right it's not useful information because you're never going to come across a scenario to utilize it.
AxiomOfChoice
AxiomOfChoice
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May 27th, 2014 at 8:00:01 PM permalink
Quote: Boney526

I suspect that on average, if you started at a True Count of 5 say 5 decks in, then the average true count, say 2 decks, in would be less than 5 - since the closer you get until the end of the deck, the closer you should get to (an average count of) 0 - before it becomes undefined as you pointed out.



No, that is incorrect. The expected value of the true count remains at 5. It does not get closer to 0. I'm not sure why you think that it should.

If the true count is 5, that means that there are 5 extra high cards per deck left. If there are 3 decks left, that means that the running count is 15.

If you deal one out of those 3 decks, you would expect an extra 5 high cards to be dealt. Of course there may be more or less, but the expected value of the number of extra cards to come out is 5. That leaves the expected running count at 10, which leaves the expected true count at 5, since there are now 2 decks left.

If you deal another deck, you would expect another 5 extra high cards to come out, leaving the expected running count at 5 with 1 deck left, which would leave the expected true count at 5.

No matter how many cards you deal, the expected value of the expected true count remains 5.

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