JB
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JB
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June 23rd, 2014 at 5:46:25 PM permalink
Quote: BobbyMac

I noticed the strategy has "joker+Ace" ranked above "4 to a Flush having one joker and one Ace". If this is accurate then one would never hold "4 to a Flush having one joker and one Ace" for the first draw. Im also confused as to why TJQA is better than 7Ts but 36s is better than A235. I realize many of these hands run very close and change with straight/flush penalties. So given 36s,A25 what is the best hold? How about 47s,68Q or KJs,AT3? The expected value of first draw hands would be a great addition as well if you have them.

ps. I just found this forum and am very thankful i did. Great work JB


The strategy is somewhat inaccurate because it is based solely on the average return of each play in hands where that play was the best play; it doesn't take into account hands where it exists but was not the best play, because that would have been an enormous undertaking.

So for example, in all of the hands where Joker+Ace was the best play, the EV is 2.01265; in all of the hands where 4 to a Flush including a Joker+Ace was the best play, the EV is 1.98682 or lower. Since I sorted the strategy list by EV, this puts Joker+Ace higher than 4 to a Flush w/Joker+Ace, but in reality, Joker+Ace is the better play only if you don't also have 4 to a Flush.

With 36 suited vs. unsuited A235 in the same hand, such as 3♣ 6♣ 2♦ 5♥ A♠, holding A235 is the better play.
beachbumbabs
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beachbumbabs
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June 23rd, 2014 at 5:50:39 PM permalink
Quote: BobbyMac

I noticed the strategy has "joker+Ace" ranked above "4 to a Flush having one joker and one Ace". If this is accurate then one would never hold "4 to a Flush having one joker and one Ace" for the first draw. Im also confused as to why TJQA is better than 7Ts but 36s is better than A235. I realize many of these hands run very close and change with straight/flush penalties. So given 36s,A25 what is the best hold? How about 47s,68Q or KJs,AT3? The expected value of first draw hands would be a great addition as well if you have them.

ps. I just found this forum and am very thankful i did. Great work JB



Welcome to the forum, BobbyMac! My understanding (having played this game, not as a mathematician) of the joker+A hold is (and the rest, though I think this is the best example), you have to take into account the paytables/return on the holds, not just the likelihood of filling them. There are 4 aces +2 jokers, and 5 aces pays extremely well compared to a flush, even though it's harder to fill. So the strategy is, in the long run, you will make more money holding for the 5 aces (and lesser hands that can result from holding that way, because you're drawing 3, then 1 if you need it), than holding for a flush, (assuming you don't have a straight flush draw, where you're drawing 1, then 1 if you need it), because even if you make the flush more often, it's the only hand you're drawing to, and it's one of the lower-paying ones.

EDIT: answered the same time as JB, but letting it stand anyway. Thanks, JB!
If the House lost every hand, they wouldn't deal the game.
BobbyMac
BobbyMac
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June 24th, 2014 at 1:34:32 AM permalink
I see, thankyou very much for explaining. I was simply organizing a workable strategy based on your findings. I can now rank a W+A56s type hand over W+A as it will always be of greater value. Just as an inside straight will always be better than a 3 straight although a 3 straight usually has a higher value if the inside straight were`nt an option. Most everything else is fairly straight forward with a few exceptions. Most notably the 25s,47s,KJs type hands if they were typically better than inside straight draws given a choice. Although i can only make educated guesses without EV`s for each hand.

Anyways, here are some hand ranks for everyone 2ND DRAW...
Values are based on a 4 unit bet
8.1) WW+A2s
7.1) WW+8Ts
6.0) FLUSH
5.6) W+678s
5.2) WW+KTs
5.0) STRAIGHT
3.7) WW+59s
3.6) WW+55
3.3) W+AKQs
-.32) W+A56s
-.88) W+567
-1.6) W+679
-1.7) W+479s
-2.8) PAIR
-2.9) STR4i
beachbumbabs
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June 24th, 2014 at 4:29:24 AM permalink
Quote: JB

The strategy is somewhat inaccurate because it is based solely on the average return of each play in hands where that play was the best play; it doesn't take into account hands where it exists but was not the best play, because that would have been an enormous undertaking.

So for example, in all of the hands where Joker+Ace was the best play, the EV is 2.01265; in all of the hands where 4 to a Flush including a Joker+Ace was the best play, the EV is 1.98682 or lower. Since I sorted the strategy list by EV, this puts Joker+Ace higher than 4 to a Flush w/Joker+Ace, but in reality, Joker+Ace is the better play only if you don't also have 4 to a Flush.



Darn it, JB, now you've got me confused. If Joker+Ace has an EV of 2.01265, and the best 4 to a flush including those cards is 1.98682, what is the element that counteracts this and makes 4TAF the better play?
If the House lost every hand, they wouldn't deal the game.
JB
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JB
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June 24th, 2014 at 1:47:39 PM permalink
Quote: beachbumbabs

Darn it, JB, now you've got me confused. If Joker+Ace has an EV of 2.01265, and the best 4 to a flush including those cards is 1.98682, what is the element that counteracts this and makes 4TAF the better play?


I will try to explain it better. When my analyzer outputted the strategy results, it listed the best play and its EV for each possible starting hand. From that list, I averaged the EV's of each type of "best play" made.

The end result is that the average EV of Joker+Ace is 2.01265, because it only considers hands where Joker+Ace was the best play. In other words, the output did not list Joker+Ace as the best play in hands where 4 to a Flush was the best play, therefore the value of Joker+Ace in 4-to-a-Flush hands are excluded from the Joker+Ace average.

In generic terms, the EV for "Play A" does not take into consideration the value of "Play A" in hands where "Play B" is the better play. Ideally it should, in terms of developing the strategy, but for Double Draw Poker this would have been a huge undertaking. So the strategy just shows the plays made sorted by their EV. The result is that there are some anomalies such as the Joker+Ace vs. 4-to-a-Flush scenario.

The same situation occurs when developing video poker strategies, but I spent a great deal of time making the VP strategy generator on WizardOfOdds.com take this into consideration, so that the resulting strategies are as accurate as possible.
beachbumbabs
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June 24th, 2014 at 3:33:10 PM permalink
I get it now, thanks so much!
If the House lost every hand, they wouldn't deal the game.
JamesGarnerOK
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July 22nd, 2014 at 7:43:22 AM permalink
Quote: tringlomane

...

And the 2nd draw strategy is pretty much...do you have a chance to win? Call.



It seems to me that this isn't quite right. For example, one can show that you should fold an inside straight draw if you've already discarded one of your "outs" from the deck. Here's a way this could happen in the game: You are dealt 4c, 5h, 6h, 7h, 9d. On the first draw, optimal play is to keep the 5,6,7 suited and toss the 4 and 9. Then you draw 2d, 3d. For the second draw, the only possible play is to hold either 2,3, 5, 6 or 3, 5, 6, 7 and try for the inside straight. But since you've tossed one of your fours, then you have 5 ways of completing the straight (three fours + two joker), with 47 cards remaining in the deck. The EV of the second draw is thus
5 * (5/47) + (-4) * (42/47) = -3.04. But the EV of folding is -3, and so this hand should be folded.

The situation above is probably rare, but there might be others where folding is right. Does anyone know the percentage of 2nd-draw situations in which there's a positive probability of winning that instead should be folded, for a single player at the table playing all decisions optimally?
beachbumbabs
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July 22nd, 2014 at 7:54:19 AM permalink
That would be a really hard calculation, IMO, since you have to factor back in what you've discarded.

If it helps, the Bally rep at the trade show I first played this said a little more than 97% of hands should stay in for the first draw, and a little more than 92% should stay in for the 2nd.
If the House lost every hand, they wouldn't deal the game.
JamesGarnerOK
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July 22nd, 2014 at 8:23:58 PM permalink
I ran a 1-player simulation of the game to determine how frequently you would fold on the 2nd draw even though you have a chance to win. Out of 10,000 games played, there were 58 hands that were folded on the 2nd draw with a positive probability of winning something. 57 of the 58 hands were exactly the situation described in my previous post, in which the player had an inside-straight draw (or A234 or JQKA) with no jokers in the hand but in which one of the cards needed to fill the straight had been discarded. That is, there were 5 outs instead of the full 6. One hand of the 58 was a situation in which you held a pair and two other ranks to try for a two-pair or three-of-a-kind, but in which each of the two offcards had a card of that rank discarded. That is, there were only 4 ways to fill out a 2-pair instead of the usual 6.

For each of the 58 hands, the error if you didn't fold was less than 0.04 units and was typically around 0.02 units. Since these are so rare, a strategy of always playing a hand in which you have a chance of winning will be very nearly optimal for the 2nd draw.
Nickmush1
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August 26th, 2014 at 8:24:37 AM permalink
No double draw games in Detriot

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