Let's consider the red/black game, since that is definitely going to be easier to analyze. Suppose, as a player, you are allowed to sit out any particular round of play, so that you can wait out the cold spells. If the house has a 1-to-1 payoff for red, then, obviously, you can make money by betting when the deck is hot, i.e., when the proportion of reds remaining in the deck exceeds 50%, and sitting out when the deck is cold. Obviously, this scheme makes money even when there is a shill playing, since you will be sitting out when the deck gets cold. The question is whether the shill can hamper your ability to make money. One side-point to note is that the game is zero-sum between each individual player and the house: what the players win the house loses, and vice versa, but it is not zero-sum among the players. Obviously, if each player plays the smart strategy of only betting when the deck is hot, then every single player has positive expected return. In the long run, they will all win. Now, let's assume there are three players: S (standing for shill), A, and B. S works for the house and his job is to hamper A's and B's ability to make money. Further, suppose A is smart and plays only when the deck is hot, while B is dumb and plays all the time. When the deck is cold, S sits out, because he wants A and B to get as many cold cards as possible. Of course A sits out as well when the deck is cold, so the only person playing is B. When the deck is hot, S plays, because he wants to rob A and B of as many hot cards as possible. A plays, because the deck is hot, and B plays because B always plays. Note that A and S play exactly the same way. Now, let's suppose that the current state of the deck is (k,n), which means that out of n cards left, k of them are red. Further, let's suppose the deck is hot, so k/n > 1/2, and hence S and A are both playing. S goes first, taking a card from the deck before A gets a chance to get his card. We'll calculate the expected return to A for a unit bet, given that S takes a card first. But before we calculate it formally, note that this is equivalent to saying that A gets the *second* card from the top of the deck (because S has taken the first card). Since all the cards remaining in the deck are equally probable to be red, the fact of S taking the first card does not affect A's expected return on this round. The expected return must be k/n. Now, more formally: E[A|S] = E[A|S gets black] * p(black) + E[A|S gets red] * p(red) = k/(n-1) * (n-k)/n + (k-1)/(n-1) * (k/n) = (nk - k^2 + k^2 - k)/[n(n-1)] = k(n-1)/[n(n-1)] = k/n This is the only point I was making in my initial response. When S draws a card, it certainly doesn't affect A's expected return on the round in question. The argument that S by his play makes a hot deck end sooner seems valid, but I'd need to see a careful argument, and we should formalize exactly how we'll measure A's return rate. I'm guessing it should be the expected return per round. * * * Here's an interesting related problem: Consider the following version of red/black: you post one dollar up front, and then the dealer will start rolling the deck one card at a time until you tell him to stop. At that point, the house pays 1-to-1 if the *next* card is red; otherwise, you lose. Surprisingly, this is a fair game. That is, you cannot make money at it. JSS Bernie Cosell wrote:
On 1 Dec 2003 at 18:51, Joshua Singer wrote:
Regarding the method outlined below, in which a shill burns through a blackjack deck when the state favors the player, I'm not sure this would accomplish anything. It seems to me that this is as likely to make the deck even more favorable to the player as to make it less favorable.
Well, I've gotten fooled by probability args in the past, and surely will be in the future, but I think this one is OK.
The question turns on whether you believe that knowing something about the distribution of the cards remaining in the deck affects the player's odds or not. If you *DO*, then the shill _can_ affect things to favor the house:
As a thought experiment: suppose that the shill had the option, at his turn, to take the *bottom* four cards off the deck. Does it seem like this would affect the other players' odds? Since the next four cards and the bottom four cards are equivalent, from a probabilistic standpoint, the answers to the thought experiment and to the actual proposed method must be the same.
Indeed, and it works for me: if the deck is 'hot', that means that the probability of getting "good cards" is higher than expected, which is disadvantageous to the house. Taking the bottom four cards, or ANY four cards, out of the deck in such a situation is *more* likely to remove "good cards" from the deck than would be expected, and so should, on average, 'cool' the deck [and in a way that deprives the player from winning money from the house with the 'good' cards]. Yes, taking the bottom cards won't affect the *next* hand, but even taking the TOP card might not help the house: even with the deck being 'hot', the shill might get a deuce next and make the deck "hotter"for the player, but I think that *on*average* the shill will help the house.
Consider a simpler game: red wins and black loses. The dealer just turns over a single card per player and it is 50/50 [no house edge!] whether you win or lose. It seems obvious to me that in this game [which, IMO, has the same basic hotdeck/colddeck properties as in BJ] that the player that ONLY plays when there are more reds than black left in the deck will end up ahead [which means that the other players will have ended up behind, since the complete game is zero-sum].. Am I missing something here?
/Bernie\
-- Bernie Cosell Fantasy Farm Fibers mailto:bernie@fantasyfarm.com Pearisburg, VA --> Too many people, too few sheep <--
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