It is ok, I have my way to understand how and why the EV is the same in different round.Originally Posted by JohnGalt007
But there is still some confusion:
Since more big cards are used in the first round on average, so it should be the same, more big cards are used in the second round, so where is the end of "more big cards are used"?
Or it is a different meaning of "on average"?
Can you be more specific about your question 'where is the end of "more big cards are used"?'? I'm not sure where exactly your confusion lies in this case.
As for the meaning of "on average", it should be the same as the interpretation of the expectation operator defined in probability.
In the first round, more big cards are used on average than small cards(this is confirmed at another theme posted)
By one-to-one mapping, it is true for the second round, and for the third round, and so on.
This is beyond my common sense.
Imagine a game of Red-and-Black, so that 1 red and 1 black card are randomly shuffled and dealt one at a time to the player without replacement (two deals total). Clearly, the possible orderings of the deck are RB or BR. Before each deal, he guesses the color of the next card according to a strategy S.
Let the player employ the completely random myopic strategy, so that he guesses Red with probability 1/2 and Black with probability 1/2 regardless of which card comes out on each deal.
Let X denote the random variable modeling the number of correct guesses given by the player for a round of this game, where a "round" consists of two guesses: what color is the first card, and what color is the second card. The support of X is {0, 1, or 2}. We are interested in E(X).
If we guess completely uniformly at random both deals without regard to what came out first, then E(X) = (0)(1/2)^2 + (1)(1/2) + (2)(1/2)^2 = 1 by the binomial distribution. Your current confusion seems to be how to reconcile this result with the claim that the expectation for the second guess remains the same as for that the first guess, that is, 1 in this case.
Importantly, E(X | red first card) = E(X) and E(X | black first card) = E(X) by assumption because the player's strategy is "myopic". That is, even though the deck's relative proportion of Red and Black cards are in fact dependent trials so that the conditional expectation of the number of correct guesses given the cards that have been dealt do change, the strategy by which either player makes their decision for the current round does not take this information into account. The player will always guess uniformly at random regardless of which cards come out (red first black second, or black first red second). Thus, he treats the game like two coin flips which will result in an expected number of successes of 1.
Furthermore (and this is where the "mapping" argument helps), the myopic strategy yields the same expectation even though the ordering of the deck on the second deal is in fact conditioned on the ordering of the pack on the first deal, as follows: Let the player's guesses be "red, red". Then we have two cases.
(1) If the deck's ordering is BR, then his expectation is 1.
(2) If the deck's ordering is RB, then his expectation is 1.
The deck is assumed to have each of these orderings occur with probability (1/(2!)) = 1/2. Thus, by the law of conditional expectation we have (1/2)(1)+(1/2)(1) = 1, which is the same as the expectation calculated earlier. In other words, it doesn't matter whether we calculated the expectation of this strategy by using the orderings of the deck as our sample space or by ignoring the orderings entirely. The orderings which yielded a given outcome (i.e. the ordering BR yielded an incorrect guess on the first card and a correct guess on the second card) were exactly balanced by an equal number of orderings which yielded the opposite outcome (i.e. the ordering RB yielded a correct guess on the first card and an incorrect guess on the second card). This is, again, because the strategy in both cases is assumed to be myopic (no card counting going on!) and the deck is assumed to have all 2! = 2 orderings equally likely. Only one ordering is used per game, but over a large number of games both orderings will be used roughly equally often by the Law of Large Numbers.
The mapping argument holds true for n cards; that is, any ordering which has a higher proportion of red cards used up on the first round is counterbalanced by just as many equally likely orderings which have a higher proportion of black cards used up on the second round, so that the conditional expectation remains the same between rounds as long as you don't use the previous cards as extra information. Replace "red cards" with "high cards" and "black cards" with "low cards", and you have the same argument applied to blackjack in the paper. It's not exactly the same game, but the same underlying mathematical ideas are used to justify the theorem.
Hope this helps!
Thanks for explaining in such detail.
I think I understand the concept of "The same expectation for each round".
Now I would like to move forward with something that was not mentioned in the paper.
In blackjack, on the first round, on average, more big cards(ten's) are used.(you have to agree of this before continue). When asked why the expectation for the second round is the same even though there are more small cards in the remaining cards, it seems that not all of the experts here know how to explain it.
And suprisingly, Edward Thorp even participated in the discussion. (Has this ever happened before?)"
https://www.blackjacktheforum.com/sh...-Players/page6
The two questions are solved, "The same expectation for each round", and "The same expectation for each round even if the running count is not the same"
Now my third question: If the running count drops on average after the first round, would it drop even more after the second round, and the third round, and so on?
+++
[Randomly shuffled deck of 2 red and two black cards. There are six equally probable sequences.
Draw until first red card appears.]
In Thorp's example, more red cards are drawn both the first round and the second round.
Last edited by peterlee; 08-05-2023 at 11:05 PM.
Happy to help! I also recommend looking up the notion of exchangeable random variables on Wikipedia to clarify further why the mapping argument works, and especially the section on exchangeability and its application to card games in Chapter 11 of "The Doctrine of Chances" by Stewart Ethier (2010), specifically Chapter 11.2, pages 400 to 408.
Let's make your third question more precise. Assume a balanced count, say Hi-Lo, where the count tags assigned to all ten-valued cards and aces are -1 and the count tags assigned to all 2-6s are 1. Let a ten come out on the first round, so the running count drops from 0 to -1. What is the conditional expectation of the running count in the next round?
With probability 19/51, the next card to be dealt has count tag -1.
With probability 12/51, the next card to be dealt has count tag 0.
With probability 20/51, the next card to be dealt has count tag 1.
We are currently at running count -1.
Thus, the expected running count after the second card is dealt is (19/51)(-2)+(12/51)(-1)+(20/51)(0) = -50/51.
Note that this is very slightly larger than -1, so that we expect the running count to increase very slightly. This should make intuitive sense because after a ten-card is dealt the deck now has one less card with tag -1, so that it is slightly less probable that the running count will decrease further because of the preponderance of cards with tag +1.
We take away two lessons from this observation:
(1) We bet less when the count is low not simply because the count is low; we bet less when the count is low because we expect (given only the current information) the count to increase on subsequent rounds. However, the amount of expected increase is quite small, so we Wong out of negative counts because we expect to wait a while before we get a favorable count.
(2) A distinction must be made between the paper's conclusion, which assumed myopic playing and betting strategies on the part of the players, and your question about expectations of change in the running count, which of course assumes a composition-dependent strategy based on cards already seen. The latter was not mentioned in the paper simply because the implicit assumption that players count cards using a myopic strategy is a contradiction in terms.
Hope this helps!
Last edited by JohnGalt007; 08-06-2023 at 11:42 AM.
My third question is about the fact that Big cards are dealt more than small card ON AVERAGE, every round!
This description is beyond our commond sense.
If we randomly pick some cards from a shoe, or place a cut card somewhere in the shoe and choose one side, those cards we picked, should be running count zero on average.
However, after a BS player plays the first round, running count negative on average. So as the second round, and so on, until there are not enough cards to complete a game(Let's call it "end of the shoe")
Base on this logic, on average, the remaining cards in the "end of the shoe" should be the sum of the "running count negative" from all the rounds played.
The Thorp's example:
[Randomly shuffled deck of two red and two black cards. There are six equally probable sequences. Draw until first red card appears.]
(1 is red, two rounds played)
1 1 0 0
1 0 0 1
1 0 1 0
0 1 1 0
0 1 0 1
0 0 1 1
In this example, more reds are dealt, same expectation for two rounds.
But there are four blacks will never be used in the game.
Are these four blacks as the same as [the sum of the "running count negative" from all the rounds played] memtioned above?
There are 24 elements in all the combinations listed, but only 20 elements are involved by the rule of the game. Should we describe this game as having 24 or 20 elements?
There are 52!*52 elements listing for one deck game, but some of the elements will never be used by the conditions of "enough cards to finish a hand". Then, the elements in the game, big cards are more then small cards, distribute to all the rounds play, on average. Maybe this is the new concept that I need...?
Understood, friend. I want the same thing you want: to make sure the issue is clarified for all parties involved. I know you're frustrated, but keep in mind that probability theory has the dubious honor of being the one branch of mathematics which defies common sense even for experts. Thus, it's okay to feel frustrated and impatient with the process, this is supposed to be non-intuitive. I'm a reasonable guy, you're an intelligent guy, let's meet halfway and communicate sensibly, without all caps. That way, we're all incentivized to help you and you're more likely to receive a positive response next time.
In a word, yes. That's exactly right.
If this is the case, are there any games or math problems related to this kind of problem?Originally Posted by peterlee
There are 52!*52 elements listing for one deck game, but some of the elements will never be used by the conditions of "enough cards to finish a hand". Then, the elements in the game, big cards are more then small cards, distribute to all the rounds play, on average. Maybe this is the new concept that I need...?
In a word, yes. That's exactly right.
Several. Check out the section on exchangeability and its application to card games in Chapter 11 of "The Doctrine of Chances" by Stewart Ethier (2010), specifically Chapter 11.2, pages 400 to 408.
https://www.amazon.com/Doctrine-Chances-Probabilistic-Probability-Applications/dp/3540787828/
For a seven-spot BJ table, dealer's hole card is the 16th card dealt. Chance of getting a picture is 4/13.
For a no-hole-card game, dealer's hole card would be the 16th or later, due to the players hitting or taking other actions.
Since it is more often end up with a big card on hitting, the chance of the dealer's hole card being a picture is not exactly 4/13.
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