Blackjack Paradox

[Mickey Brausch]

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My thinking appears to be producing a paradox. If the deck on average becomes unfavorable after the first hand then it seems to imply that the first hand must be enjoying an average edge.

[/ QUOTE ]We'll call it the Central First-Round Theorem.

[img]/images/graemlins/laugh.gif[/img]

[Mickey Brausch]

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Over the long run, since player 2 tends to get to see more total cards before making his decisions, he should have an edge over player 1.

[/ QUOTE ]That is not what the problem says.

[VK_Rick]

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Over the long run, since player 2 tends to get to see more total cards before making his decisions, he should have an edge over player 1.

[/ QUOTE ]That is not what the problem says.

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I think this misses the point of the analysis. The problem says we must not take into account which player acts first on any given hand (i.e., that the player acting second on a particular hand has the added advantage of seeing additional cards -- those cards exposed in the first player's hand, either by hitting or busting -- before making a decision).

That said, in the third hand dealt (the second hand for player 2), player 2 still has an advantage because before that hand is dealt, he will have seen more cards than player 1 saw before his second has was dealt. This is what should create an advantage for player 2.

[Daliman]

I'm pretty sure the "floating advantage", proven by Don Schlesinger, would denote that the answer to #1 is that you have a slightly better edge in the actual hands played than if you started from hand one, but the edge you DIDN'T gain from hand one makes it slightly less so. In either case, the edge difference is likely so small as to hardly matter.

For #2, I'm guessing the FA comes into play again, as better counts are worth more the deeper you play, and your edge comparitive would be somewhat significant.

Ugh, I just re-realized we are flat betting here. I'd say what I say still holds true, but to a much reduced effect in each case.

[Lama]

Question One: Its not as good cuze if the pit boss counted the card he may have trick you otherwize it almost as good.

on the first 3 hand is edge is greater then yours on the 4 hand he is not playing you get the edge he had on is 4th hand so on average your edge is smaller but if you compare with the 4 hand he played the edge is the same.

Question Two:no,cuze u get diferent count.

[imsobad]

Question One:

The number of unseen cards is still 100% accurate. But, the number of cards in the deck is now less then 100% accurate and is off by at least four cards and possibly more. The accuracy of our algorithm may have been reduced by the inaccurate count of cards still in the deck. This could adversely affect our play if we wanted to take a hit, but there were no more cards in the deck. But, this cannot happen in this scenario as there are more than enough cards in the deck given the limited number of players and hands before the shuffle. Even if the pro's were not playing optimally and if the dealer was hitting on 20, there would be enough cards to take a hit that would ultimately break or make 21 for the second pro on their forth hand. Therefore, the inaccurate count of cards in the deck should not affect the accuracy of our solution which depends heavily on the accuracy of cards seen and unseen.

Question 2:

The first pro could have seen as many as 13 cards in the first hand if they took 22223333 for 20 and the dealer took 44445 for 21. (Note, the first pro would likely stand on 17 and would have seen 12 cards.) With those cards gone, the second pro would not see any hand that had more than 4 cards per player (and wouldn't know why this was happening). The first pro would have a decided advantage given the amount of additional information that the second pro would never overcome in this sitution.

[bobman0330]

I don't know a lot about blackjack, but I can imagine two scenarios where the second player might have an advantage:
1. If you can sight-count the deck to a fairly accurate extent, you might be able to infer something about the remaining cards. For example, if 8 cards are out, a lot of low cards are likely gone.
2. In certain situations, you might be able to pick up information about the current count from the other player's action. I don't know enough to create an example, but I'm sure there are certain unusual actions he might take that would reveal information about the count.

[Buccaneer]

Slansky it must be past your bed time. Easy question. Easier answer.

If you sit down to a fresh deck then you have perfect information. You know you have 52 cards seperated into 4 suits and with a numeric value 1 - 11. You know that Aces are 1 or 11, that the numbered cards are valued by thier number and the face cards are valued at 10. Perfect information.

If you sit at a table with a deck with unknown cards missing then you have imperfect information.

Now we know that a card counter must count each and every card with no mistakes or his performance suffers greatly. This is simply because he/she is now working with imperfect information making it impossible to make perfect decisions.

Imperfect information makes it impossible to play the game 100% properly. You see why don't you!

Card counters do not enter a game mid deck start counting. You see why don't you?

Card counters wait for the shuffle to begin thier count. You see why don't you?

Question One: No.

Your next question requires a more detailed answer.
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Question Two: Are both of you playing with the exact same edge?

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Hell no. The other player has perfect information you have imperfect information.
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Even though you will often be playing hands two and three assuming different counts.

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No. If the other player has more and perfect information then how can you out play him if you do not have the same advantage. Maybe a reread of TOP is in order. Different game but same ideas.
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In fact you might sometimes have the exact same hands and play them differently? If so how could this be?

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The same hands may be played differently because the other player has more knowledge about the condition of the deck. Perfect information.

[VK_Rick]

Unless it can be shown that the deck will have a bias after a single hand -- i.e., that based on the way blackjack is played, one cannot assume a "random" distribution of cards -- the fact that player 2 does not have "perfect" information when he sits down for hand 2 should not matter. Player 2 has exactly the same information in hand 2 as Player 1 had in hand 1 (none). In the next hand, however, player 2 gains an advantage by having seen more cards in hand 2 than player 1 saw in hand 1 (because now we have three players). Thus, player 2's advantage in hand 3 has to be greater than player 1's advantage was in hand 2. And player 2's advantage only grows with each successive hand.

That said (and has someone already pointed out), it could be that the course of play biases the deck slightly. If the first hand contains large cards, neither the player nor dealer will hit, and the deck will have a negative count. On the other hand, if low cards come out, it is likely the player and/or dealer will hit and, with low cards already out, it is then more likely that the "hit" card will be a high card, balancing the count, rather than another low card.

Looking at it somewhwhat differently, far fewer high cards can be dealt in a single hand of blackjack than low cards, meaning the possibility for a strong plus count exists after one hand, while there is no possibility for a strong minus count. This could also create a bias.

All that said, without a computer simulation showing a fairly strong, predictable bias (and I'd love to see someone run one), I still say player 2's advantage in having seen more cards at the outset of each of his hands trumps.

[PairTheBoard]

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PairTheBoard -
**** At this point I have to say that my thinking appears to be producing a paradox. If the deck on average becomes unfavorable after the first hand then it seems to imply that the first hand must be enjoying an average edge. But that doesn't make sense. Might have to think some more about this. ****


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ok, I've thought about it some more. Suppose that playing a hand of blackjack off the top of the deck produces a remaining deck with an unfavorable bias. I think this is very plausable. I think it's a kind of ratcheting phenomenon. Now suppose you are strictly playing basic strategy and rules give you an exactly even game off the top of the deck. If this "Hand Played Ratcheting Deck Bias Phenomenon" is true then a computer simulation should show the following: Hands played off the top of the Deck will have 0 EV. If two hands are played the overall EV will be slightly negative. And the EV will move slightly more negative as more hands are played. I'll call this the PairTheBoard Conjecture, or the Sklansky-PairTheBoard Conjecture if he likes. I don't think I've ever seen this in the Blackjack literature so it might be a small advance in the theory. It can be proved or disproved by computer simulation.

Consider the following simplified game to illustrate the point. You are dealt 1 card. If it's a Red Card you win 12/51 of your bet. If it's a Black Card you get another chance and a second card is dealt to you. If it's a Red Card you win 50 cents for every dollar you bet. Otherwise you lose your bet.

This game is 0 EV off the top of the deck.

(.5)(12/51) + (.5)(26/51)(.5) - (.5)(25/51)(1) = 0

So do you think because the game is 0 EV off the top of the deck that it's 0 EV if you play more hands into the deck? Let's see.

Just to simplify further and exagerate the principle, suppose this game is being played with a 4 Card Deck, 2 Red and 2 Black Cards. And you always play 2 hands into the deck, flat betting. To make this game 0 EV off the top we need the payoffs, 1/6 if the first card is Red and 1/4 if the first card is black but the second card is red. Then,

(.5)(1/6) + (.5)(2/3)(1/4) - (.5)(1/3)(1) = 0

and the game is 0 EV off the top of the 4 card Deck. Now do you believe that just because the game is 0 EV off the top of the deck that it's 0 EV for the player who plays 2 hands into the deck? Let's see what happens.

(.5)(1/3) of the time you will lose the first bet and automatically win (1/6) on the First Card of the Second Bet.

(.5)(2/3) of the time you will win (1/4) on the Second Card of the First Bet, then half the time win (1/6) on the First Card of the Second Bet and the rest of the time win (1/4) on the Second Card of the Second Bet.

Half the time you will win (1/6) on the First Card of the First Bet, then (1/3) of the time win (1/6) on the First Card of your Second Bet, (2/3)(.5) of the time win (1/4) on the Second Card of your Second Bet, and (2/3)(.5) lose your Second Bet.

Adding all these up we get,

(.5)(1/3)(-1) + (.5)(1/3)(1/6) +

(.5)(2/3)(1/4) + (.5)(2/3)(.5)(1/6) + (.5)(2/3)(.5)(1/4) +

(.5)(1/6) + (.5)(1/3)(1/6) + (.5)(2/3)(.5)(1/4)

- (.5)(2/3)(.5)(1)

= 0

Oh well. So much for the Sklansky-PairTheBoard conjecture.


What about for the Player who steps in and takes the Second Hand after the First has been played?

Half the time he steps into an unfavorable deck with 2 Black and 1 Red Card remaining. On average he loses by the amount,

(.5)(1/3)(1/6) + (.5)(2/3)(.5)(1/4) - (.5)(2/3)(.5)(1) =

= -7/72

Now you would expect the "Effect" to come in the other half of the time when a Black Card came first and the First Player takes another Card to "dilute" the favorable 2-Red 1-Black remaining deck. 2/3 of the time the First Player does Dilute the Favorablity of the Deck by catching a Red Card on his second try, leaving what looks like a Neutral Deck of 1-Red 1-Black for the Second Player to play. But this apparent Neutral Deck is actually highly favorable to the Second Player giving him a guaranteed win,

(.5)(2/3)(.5)(1/6) + (.5)(2/3)(.5)(1/4) =

= 5/72

Is this just a quirk of the example? Or is there something inherent about a neutral deck with fewer cards actually being favorable to the player? In Blackjack this phenomenon definitely exists. Basic Strategy off the top of a Six Deck game is about -.5% unfavorable to the player while off the top of a Single Deck it's about 0 EV.

To finish off this example, (.5)(1/3) of the time the Second Player steps into a super rich Red Card Deck and automatically wins

(.5)(1/3)(1/6) = 2/72

and thus an overall 0 EV for the Second Player taking over the Deck left by the First Player.

So it looks like my theory may be full of hooie. I'm still not convinced though. Maybe my example just needs to be tweeked.

PairTheBoard