I am now ready to tell you the cost of playing Queens or better pays 2 instead of Jacks or better pays 2 DDS. For this discussion order counts because even though the final hand may be the same without order counting in one case you may have doubled and the other case you may not have. For example JJ234 returns 2 bets and J234J returns only 1 so your loss in playing Queens instead of Jacks DDS are 2 bets and 1 bet respectavely. Now instead of 2598960 final hands there are 2598960 times 5! = 311875200 final hands with order counting. Instead of 84480 hands of exactly 1 pair of Jacks there are 84480 times 5! = 10137600 final hands of exactly 1 pair of Jacks. There are (4C2 times 12C2 times 16 times 4!) = 152064 ways that your first 4 cards are exactly 1 pair of Jacks, (you are now doubling your bet) and 40 cards you can catch (any card of the other 10 ranks not in your 1st 4 cards) = 6082560 final hands of 1 pair of jacks where you have a pair of jacks in your first 4 cards. There are (12C4 times 4 times 4!) = 47520 ways that your first 4 cards are a 4 flush that contains a Jack. There are ((4 to the power of 4) times 4! times 3) = 18432 ways that your first 4 cards are KQJ10 or QJ(10)9 or J(10)98. If you subtract out the 288 ways that your first 4 cards are KQJ10 suited or QJ(10)9 suited or J(10)98 suited that leaves (47520+18432-288) = 65664 ways that your first 4 cards are an open ended straight or a 4 flush that contains a Jack, (you are now doubling your bet) and 3 other Jacks you can catch = 196992 final hands of 1 pair of jacks starting with a 4 card opened straight or 4 flush that contains a jack. Combining the 2 cases there are (6082560+196992) = 6279552 ways of ending up with a pair of jacks where you have doubled your original bet for a loss of (6279552 times 2) = 12559104 original bets over "Jacks" DDS. That leaves (10137600-6279552) = 3858048 final hands of exactly 1 pair of jacks that were not doubled. You lose an original bet over "Jacks" DDS in those cases. Your total loss over "Jacks" DDS is therefore (12559104+3858048) = 16417152 original bets. The loss by playing "Queens" DDS over "Jacks" DDS is therefore 16417152/311875200 = 0.05264 or 5.264%.
Interesting analysis. One error I think I found was when you started with a pair of jacks, say JJ234, you said there were 40 cards you could draw that wouldn't give you two pair. I think there are only 36 that wouldn't give you two pair or trips.
FWIW, I passed the problem by J.B., the programmer behind much of the Wizard of Odds website. His models say the game is worth 100.19% if it's Jacks or Better and 95.99% if it's Queens or Better. That comes out to a 4.20% difference. I do trust J.B.'s numbers. Nordo is clearly more competent with these calculations than I am --- but if I were forced to choose between trusting one or the other, I'm siding with J.B. with absolutely no offense intended.
I'm not going to try to reconcile the two different numbers. For my purposes, whether the game is 95% or 96% is 100% irrelevant. I'm not going to play it.
Bob
Interesting analysis. One error I think I found was when you started with a pair of jacks, say JJ234, you said there were 40 cards you could draw that wouldn't give you two pair. I think there are only 36 that wouldn't give you two pair or trips.
FWIW, I passed the problem by J.B., the programmer behind much of the Wizard of Odds website. His models say the game is worth 100.19% if it's Jacks or Better and 95.99% if it's Queens or Better. That comes out to a 4.20% difference. I do trust J.B.'s numbers. Nordo is clearly more competent with these calculations than I am --- but if I were forced to choose between trusting one or the other, I'm siding with J.B. with absolutely no offense intended.
I'm not going to try to reconcile the two different numbers. For my purposes, whether the game is 95% or 96% is 100% irrelevant. I'm not going to play it.
Bob
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Posted by: Bob Dancer <bobdancervp@hotmail.com>
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