I am really sorry but that statement is not answering the problem.Quote:AxiomOfChoiceNo, 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.

It would be if the cards dealt were dealt at random. This, I agree. Consequently, I also agree that, anticipating the moment when one more deck is dealt, the TC would be expected to be stationary. The argument is based on the fact that the cards are ordered randomly. OK.

But my point is that the number of cards dealt during a round depends on the value of the cards dealt. For example, if the high cards happen to be clustered in the beginning of your remaining decks, the players will tend to ask for fewer hits. Can you confidently say that the TC after these few hands is stationary?

Here we are not anticipating TC at a future deck-size stage, but at the next round.

In your numerical example, there is an expected 5 high cards dealt from one deck. OK. But that moment may come in the middle of a playing hand.

Is there an expected 5/52 of the total cards dealt after the completing of a number of hands?

N.B. I don't know the answer. I'm just saying it is NOT obvious and requires some calculation.

Quote:kubikulannI am really sorry but that statement is not answering the problem.

It would be if the cards dealt were dealt at random. This, I agree. Consequently, I also agree that, anticipating the moment when one more deck is dealt, the TC would be expected to be stationary. The argument is based on the fact that the cards are ordered randomly. OK.

But my point is that the number of cards dealt during a round depends on the value of the cards dealt. For example, if the high cards happen to be clustered in the beginning of your remaining decks, the players will tend to ask for fewer hits. Can you confidently say that the TC after these few hands is stationary?

Here we are not anticipating TC at a future deck-size stage, but at the next round.

In your numerical example, there is an expected 5 high cards dealt from one deck. OK. But that moment may come in the middle of a playing hand.

Is there an expected 5/52 of the total cards dealt after the completing of a number of hands?

N.B. I don't know the answer. I'm just saying it is NOT obvious and requires some calculation.

When each card dealt does not change the expected count, the number of cards dealt is irrelevant. The fact that the number of cards dealt depends on the value of the previous card is irrelevant. You are still adding 0s together; all that you are changing is the number of 0's that you add together.

Do you really teach math? At what level? I'm sorry but this should be extremely obvious to anyone with a university-level math education or higher. Saying "the expectation of each card is 0, and expectation is additive, regardless of the dependence between the variables" should be enough to explain it. E(X) + E(Y) = E(X+Y) is a hammer; 95% of the gambling questions asked on this forum (including this one) are nails.

Here's a puzzle: n people arrive at a party and leave their hats at the door (each person is wearing 1 hat). As they leave, they each grab a hat at random (uniformly distributed) from the hats that are left. After everyone has left, what is the expected number of people who got their own hat back (as opposed to someone else's hat)? If it takes you more than two seconds to blurt out that

And, why are you still on this? I proved it in a previous post. Was that not sufficient?

No, I'm teaching statistics. Maybe that's why you can't grasp. It's more complicated than math.Quote:AxiomOfChoiceDo you really teach math?

Repeating a statement with much force has never been a proof of validity, in my book.

Your choice of the (famous) hats example shows you are treating another problem.

Look, I have set up a simplified problem, and the result clearly shows the expected relative composition of the deck changes after a player's hand.

Three types of cards : values are 10, 6 and 3.

A player receives one card, then chooses to hit (several times if wanted) or stand.

You bust if you go past 10.

Say that the player's strategy is :

" Hit on 3-6, stand on 7-10 "

Develop the decision tree, with associated probabilities. With each possible play, you end up with a new composition.

Compute the expected posterior composition.

(Absolute composition is RC; relative composition is TC.)

According to your statement, this expected posterior should be equal to the prior ("adding zeros").

Well, the figures show otherwise: there is an expectation of a slightly lower proportion of 10's and a slightly higher proportion of 3's.

Do the math ! Otherwise, I'm afraid you will stubbornly hold to your mantra E(X)+E(Y), as if that exempted you from calculation.

*Note: your mantra is correct. It's just not applicable in this instance.

Quote:kubikulannNo, I'm teaching statistics. Maybe that's why you can't grasp. It's more complicated than math.

Repeating a statement with much force has never been a proof of validity, in my book.

Your choice of the (famous) hats example shows you are treating another problem.

Look, I have set up a simplified problem, and the result clearly shows the expected relative composition of the deck changes after a player's hand.Start with an arbitrary deck composition (for instance 10 of each, or more generally p, q & r ). That is the prior composition.Three types of cards : values are 10, 6 and 3.

A player receives one card, then chooses to hit or stand.

You bust if you go past 10.

Say that the (basic) strategy is :

" Hit on 3-6, stand on 7-10 "

Develop the decision tree, with associated probabilities. With each possible play, you end up with a new composition.

Compute the expected posterior distribution.

According to your statement, this expected posterior should be equal to the prior ("adding zeros").

Well, the figures show otherwise: there is an expectation of a slightly lower proportion of 10's and a slightly higher proportion of 3's.

Do the math ! Otherwise, I'm afraid you will stubbornly hold to your mantra E(X)+E(Y), as if that exempted you from calculation.

You keep talking about how the number of cards is variable. I keep pointing out that this is true, but also irrelevant, because each of them have an expected value of 0. It's like going to a roulette wheel with no zeros and saying that you are going to keep betting on red until you win twice, and then stop, and asking if that changes your expectation. Of course it doesn't; you can add 0's as much or as little as you want, conditionally, or not, and you still end up with 0. If you have a 47.248% chance of adding another 0, that's still 0.

If you think that statistics is more complicated than math then you obviously haven't done any real math. That is the difference here; you are insisting on using brute force calculation because you are a statistician (or, at least, a statistics teacher) and you don't understand the simple theory (ie, the math). I am a mathematician, so I understand the theory and therefore refuse to waste my time with long calculations, because I already know what the result will be when I add up a bunch of 0's. Feel free to add them up, though, and tell me what you come up with. Although, be a little more careful when you add them up, since you made a mistake with your example. That tends to happen when you insist on a brute force approach instead of taking the time to understand the math.

I don't think that. I was saying it was more complicated for you, because I'm fed up with your insulting innuendos.Quote:AxiomOfChoiceIf you think that statistics is more complicated than math then you obviously haven't done any real math.

I'm fed up of people chanting their mantra just saying "it's obvious, if you don't see it you're an idiot."

Nothing is obvious. You'll never convince me unless you offer a mathematical proof. It's the one who is convinced without a proof who is an idiot.

It is also "obvious" that the Earth does not move, isn't it? Well, it's wrong !

It is YOU who are similar to all those system proponents who claim it works but never provide the details.

Quote:kubikulannI don't think that. I was saying it was more complicated for you, because I'm fed up with your insulting innuendos.

I'm fed up of people chanting their mantra just saying "it's obvious, if you don't see it you're an idiot."

Nothing is obvious. You'll never convince me unless you offer a mathematical proof. It's the one who is convinced without a proof who is an idiot.

It is also "obvious" that the Earth does not move, isn't it? Well, it's wrong !

It is YOU who are similar to all those system proponents who claim it works but never provide the details.

My time is valuable to me, so how about this:

I will bet you 5000 US dollars (or more, if you'll agree to it) that I can provide a rigorous mathematical proof that playing a hand does not change the expected value of the true count. I will take the time to write up the proof and post it here. We will agree beforehand on a trusted party who will settle any dispute as to whether the proof that I provide is rigorous and correct. I'd nominate teliot, if he'd agree (since he is a real mathematician, with experience in academia, and is therefore used to reading and writing proofs. I have been critical of him in the past so you can trust that he will give me no "special treatment"). If he wishes to charge a fee for his validation (which I would, if I were him -- his time is valuable too) the loser of the bet must pay for it, in addition to paying the winner of the bet (so the fee does not come out of the amount of the bet; it is extra). Of course the fee can be avoided if the loser simply accepts that the proof is correct/incorrect without involving the judge, and pays off the bet.

I do not suggest that you take this bet. You will lose, for sure.

If you wish to get my work for free, I'm not going to take the time to write up a formal proof of what amounts to the statement that 0x + 0f(x) = 0 for all functions f and all values of x, so long as f is defined at x. However, I'd be happy to refute the example that you gave; perhaps this will convince you. for your simplified games, simply tell me which initial distribution you feel leads to a different expected distribution after the hand is played, and I will show the calculation that proves you wrong. I am letting you choose the initial distribution simply so you don't think that I'm cherry-picking one.

Don't bother, I can make the proof myself at no charge. Being a teacher, I have a large provision of free time, have I not?

Moreover, I said I do not know what the answer is, so I would certainly not bet on one result against the other.

In the sex-biased society problem, I have produced two cases. One agrees with your feeling (no change in expected value); the other leads to a different answer. My next step is to determine which is applicable to the blackjack situation.

I'll check my back-of-the-envelope equations and post them here.

Three card values, 10, 6 and 3. The deck composition is p 10's, q 6's and r 3's. Define n as deck size (p+q+r).

The table shows all possible cases according to the defined strategy.

1st card | Prob | 2d card | Probab | 3d card | Probab | New p | New q | New r |
---|---|---|---|---|---|---|---|---|

10 | p/n | - | 1 | - | 1 | p-1 | q | r |

6 | q/n | 10 | p/n-1 | - | 1 | p-1 | q-1 | r |

6 | q/n | 6 | q-1/n-1 | - | 1 | p | q-2 | r |

6 | q/n | 3 | r/n-1 | - | 1 | p | q-1 | r-1 |

3 | r/n | 10 | p/n-1 | - | 1 | p-1 | q | r-1 |

3 | r/n | 6 | q/n-1 | - | 1 | p | q-1 | r-1 |

3 | r/n | 3 | r-1/n-1 | 10 | p/n-2 | p-1 | q | r-2 |

3 | r/n | 3 | r-1/n-1 | 6 | q/n-2 | p | q-1 | r-2 |

3 | r/n | 3 | r-1/n-1 | 3 | r-2/n-2 | p | q | r-3 |

(For the expectations, I have extracted the common term p (resp. q or r) and manipulated the "-1", "-2" etc.)

E(New p)= p - [p(n-1) + pq + pr + pr(r-1)/(n-2)] / n(n-1)

E(New q)= q - [pq + 2q (q-1) + qr + rq + qr(r-1)/(n-2)] / n(n-1)

E(New r)= r - [qr + rp + rq + r(r-1)( 2p+2q+3(r-2) )/(n-2)] / n(n-1)

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.

The expectation of the relative composition of the deck after one hand is played, is that there are relatively more 3's and relatively less 10's.

This is counter-intuitive, like many probabilistic results.

Never trust what you think obvious when dealing with probabilities.

Yet... AxiomOfChoice is correct in one respect. I'll leave it to him to point what. ;-)