buzzpaff
buzzpaff
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March 29th, 2011 at 11:01:25 AM permalink
I am both mathematically and computer challenged. Where would I start ? say a SD multiplay table. 6 players, 2 spots
each. Each player has to bet minimum amount or more but equal bets on each spot. feel free to use any Vegas rules.

If after hands are dealt, dealer top card exposed, assuming perfect basic strategy, how can I possibly figure the player
advantage if I let him move his bet from one hand to the other, effectively ignoring one hand ? A complicating factor
would be giving a player the option of staying as is rather than doubling on one hand, but I am getting a headache
already
buzzpaff
buzzpaff
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March 29th, 2011 at 12:05:55 PM permalink
Not looking for free analysis, just how would I start ?????
PapaChubby
PapaChubby
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March 29th, 2011 at 12:06:23 PM permalink
Well, you'd have to start with something like the Wizard's BJ appendix 9, which shows the expected return for every combination of player's hand + dealer's up card. Then you'd have to create a 3-dimensional matrix with dealer's up card on one axis, player's hand A on a second axis, and player's hand B on a third axis. At the intersection, you'd need to enter the expected return based on the best way to play the best of the two player's hands, along with the probability that that combination would occur. (All this info can be extracted from the Wizard's table; the probability will be tricky, since there is some dependence between the hands; perhaps you ignore this for a first cut.) Then you integrate across the entire matrix to get the total expected return.

As an example, start with the first two entries in the 1D,S17 table. If the dealer is showing A and the player has 2,A and 3,A, the best play is hitting the 2,A, with an expected return of -0.0678.

OK, working thru this one example, its clear to me that you'll have to do something other than just multiplying the probabilities from the wizard's table to determine the likelihood of this event. I guess you'll have to go back to basic combinations to determine the probability that these 5 cards will be arranged in this fashion.
buzzpaff
buzzpaff
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March 29th, 2011 at 12:15:39 PM permalink
Somewhere around here I have Revere's Playing Blackjack a s a business and I seem to remember a chart showing waht hands you could see against a dealer's upcard in 100,000 hands. Should be able to factor in probability of the second hand ?
PapaChubby
PapaChubby
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March 29th, 2011 at 12:27:31 PM permalink
Quote: buzzpaff

Somewhere around here I have Revere's Playing Blackjack a s a business and I seem to remember a chart showing waht hands you could see against a dealer's upcard in 100,000 hands. Should be able to factor in probability of the second hand ?



Yeah, but you're going to have to do it manually. Consider dealer showing an A and player has A,A. This happens rarely, and will be listed in your table or the Wizard's. For two hands, the probability of A+AA+AA is zero, because there aren't that many aces in the deck. Given that the first three cards are A+AA, the probability that the first card of player's hand B is an ace is 1/49. Given all this, the probability that the final card is an ace is zero. You'll need to do this for every combination, as the composition of the first 3 cards will affect the probabilities of the final two.
PapaChubby
PapaChubby
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March 29th, 2011 at 12:45:24 PM permalink
Hey, wait, I don't get it. Considering this case of A+AA. I'd calculate that the probability the dealer will be showing an ace is 4/52. Given that the dealer is showing an ace, the probability that the player's first card is an ace is 3/51, and then the probability that the second card is an ace is 2/50. So the probability of getting this combination is 4/52 x 3/51 x 2/50 = 0.000181. The Wizard's table says the probability is 0.000122. What am I doing wrong?

Edit: I think I'm going to answer my own question. I see the 0.000181 number shows up in the table for 2+22, 3+33, etc. So I'm guessing the reduced number is based on the dealer checking for blackjack and he doesn't have one. Because if the dealer has blackjack in this situation, you don't get to make a decision.
buzzpaff
buzzpaff
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March 29th, 2011 at 12:48:42 PM permalink
Once I do this for 169 possibilities ? for each first hand versus 13 upcards, then using your example I have
reduce the second hand posibilities only is I have a pair? then move my bet to the hand that has a higher EV
and go from there !
PapaChubby
PapaChubby
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March 29th, 2011 at 2:30:58 PM permalink
I was looking at it strictly from a standpoint of one dealer hand vs. one player with two hands. In this case, the expected values are consistent with the values shown in the tables. If you want to consider all the other cards that are showing on the table (i.e. card counting) then the expected value computation and determination of the best possible play probably becomes quite complicated.

But I think the answer is "yes" to the second part of your post. Between the two hands, find the hand and play option which provides the highest EV and move your bet to that hand.
buzzpaff
buzzpaff
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March 29th, 2011 at 5:00:09 PM permalink
I was just talking about one player's two hands , that's all. Wonder with a 7 up for bealer and player having a hard
11 and 20 or 19 or 18, which hand should he play.?
Anybody want to guess player advantage overall in such a game and put me out of my misery !!
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