012422 OE Monday  got euchred #1
 Dlan
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S3 should play the AC on first street. If S2 is guarded this ploy will allow S3 to strip him everytime on 2nd street for the easy point. If S4 is the one who has the Left, S3 playing the AC can't really hurt him as that means S4 is guarded so he was gonna get a trick in this hand somewhere anyways no matter what S3 did. If the Left is in the kitty then this strategy also nets an easy point everytime after S3 leads the Right on 2nd street.
After S3 incorrectly plays the Right on first street, leading trump again is suicidal without knowing where the Left is. Gotta go fishing with offsuit at that point. And we also do know one thing about the where the Left could be. It can't be in S1's hand cuz he would've led it on 1st street. It's either in enemy hands or the kitty.
Incidentally no matter what S3 does in this hand he should be euchred if S2S4 play really well.
After S3 incorrectly plays the Right on first street, leading trump again is suicidal without knowing where the Left is. Gotta go fishing with offsuit at that point. And we also do know one thing about the where the Left could be. It can't be in S1's hand cuz he would've led it on 1st street. It's either in enemy hands or the kitty.
Incidentally no matter what S3 does in this hand he should be euchred if S2S4 play really well.

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Big mistake not playing the AC to force the Left on trick 1. But if the Dealer leads the KD euchred anyway.
This might be better to pass on this hand!
IRISH
This might be better to pass on this hand!
IRISH

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With an amateur partner I know we're both autocalling. So the question is, is this a +EV call with a good P we can trust. With a P I can trust this is a default call for me but I wouldn't be surprised one bit if Ray's simulator showed that passing is better. 3SR1 is such a mindf**k!!
 LeftyK
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 Location: North Carolina
Wes: I'm pretty sure I called it and yeah with the ammy partners I had (I only got to p was DLan last hand which we won handily) (again I went 3/10 with two of those games costly from my P play/pass bid). I understand now why to only play the Ace trumps there. I was thinking I would have the 7th trump no matter what and my P would pick up a trick but it didn't happen. No one's posted the me 3rd chair and we are winning 96 and my P 1st round passes with no cover and s2r1 calls loner and steals the win. that's the type of basic play that was going on all night. Plus, numerous chancy loners that only made 1 pt each time. (if I recall)

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I ran this scenario for 100,000 hands. The raw results showed an EV for S1/S3 of +0.681 for passing and +0.683 for bidding  so virtually identical.
Looking more closely at how the hands were played when S3 bid, I saw many plays which were debatable and perhaps not optimal. But impossible to say without doing extensive testing of various scenarios for a plethora of hands. Just too time consuming for a relatively rare scenario. While I do want to optimize my program, I am looking more for the obvious errors  the "lowhanging fruit", as it were.
So at this point I would just conclude that it's a tossup whether it's better to bid or pass this hand from S3, R1. What's much more important is the quality of your  and partner's  play.
All that said, I then ran a simulation for 100,000 hands to test what S3 should play after a trump lead by S1 on the first trick. 1st scenario is that S3 always plays the R. 2nd scenario is that S3 will play the 9C if S1 led the KC, otherwise will play the AC. [S3 will always play the 9C if S1 leads L]
1) [play R]: (9,919 / 49,482 / 4,191) [2 pts / 1 pt / euchred] EV = +0.96
2) [play AC]: (10,471 / 47,213 / 5,908) EV = +0.89
So I find that playing the R in this case is better. It leads to fewer 2 pt wins, but also far fewer euchres. That said, although my results are statistically significant, they do still rely on correct play of the hand by my program. I find this to be a complicated hand to play out, so I can't guarantee my program plays everything correctly. But the mistakes are also equally likely to be on offense as on defense. I would opt to play the R after seeing these results. Why? I find it important to be able to explain the results I arrive at, and not simply blindly follow them.
So what is going on with this hand? S3 has 3 trump, S1 has at least 1 trump [we are looking at the scenario where they lead trump, 1st trick], and S4 has at least 1 trump [picked up by dealer]. So 2 more trump out there, including L [in this particular hand, S2 followed suit, and so accounted for the 6th trump, but I am assuming a more general case]. If S2 has the guarded L, it would be best to play the AC from S3  but this is a very rare case [one particular seat has both the unaccounted for trump]. If S4 has 3 trump [can't be true in this particular hand, as S2 followed suit, but can be true in the more general scenario I tested], things look dire for friends in S1/S3. But this, again, is a very rare case.
[I will also assume that S1 doesn't have a 2nd trump, as this would be an equally rare case.]
So most likely, L is either buried or sits with S4. If it's buried, S3 has 2 boss trump and is almost assured of making a point; the uncertainty of where it lies (and subsequent decisions on how to play) may explain the difference in 2pt wins. What appears critical here is the case where S4 has 2 trump and guarded L, which happens to also be this particular case.
To euchre S1/S3, the opponents need 3 tricks. Say S3 wins the 1st trick with the R, S4 plays their low trump (QC), and S3 leads a low D. S2/S4 can euchre if they have 2 boss Diamonds (and coordinate their play appropriately), otherwise it's a bit up in the air. If S3 plays the AC on the 1st trick and lets S4 win with the L, S2/S4 can win a 2nd trick with a D lead (if they have the boss card  and even possibly a 3rd), but also set up S4 to have the choice of overtrumping S3 on a later trick, or playing off, saving their trump for later. I thinks it's this flexibility  having a trump and playing after the strong opponent  that gives S4 the edge.
My reasoning may be flawed  I'm sure Irish will let me know if it is ! And by the way, regarding the "rare" cases, one favors S1/S2, one favors S2/S4, and the 3rd is just dire for S1/S3 either way  so a wash.
Looking more closely at how the hands were played when S3 bid, I saw many plays which were debatable and perhaps not optimal. But impossible to say without doing extensive testing of various scenarios for a plethora of hands. Just too time consuming for a relatively rare scenario. While I do want to optimize my program, I am looking more for the obvious errors  the "lowhanging fruit", as it were.
So at this point I would just conclude that it's a tossup whether it's better to bid or pass this hand from S3, R1. What's much more important is the quality of your  and partner's  play.
All that said, I then ran a simulation for 100,000 hands to test what S3 should play after a trump lead by S1 on the first trick. 1st scenario is that S3 always plays the R. 2nd scenario is that S3 will play the 9C if S1 led the KC, otherwise will play the AC. [S3 will always play the 9C if S1 leads L]
1) [play R]: (9,919 / 49,482 / 4,191) [2 pts / 1 pt / euchred] EV = +0.96
2) [play AC]: (10,471 / 47,213 / 5,908) EV = +0.89
So I find that playing the R in this case is better. It leads to fewer 2 pt wins, but also far fewer euchres. That said, although my results are statistically significant, they do still rely on correct play of the hand by my program. I find this to be a complicated hand to play out, so I can't guarantee my program plays everything correctly. But the mistakes are also equally likely to be on offense as on defense. I would opt to play the R after seeing these results. Why? I find it important to be able to explain the results I arrive at, and not simply blindly follow them.
So what is going on with this hand? S3 has 3 trump, S1 has at least 1 trump [we are looking at the scenario where they lead trump, 1st trick], and S4 has at least 1 trump [picked up by dealer]. So 2 more trump out there, including L [in this particular hand, S2 followed suit, and so accounted for the 6th trump, but I am assuming a more general case]. If S2 has the guarded L, it would be best to play the AC from S3  but this is a very rare case [one particular seat has both the unaccounted for trump]. If S4 has 3 trump [can't be true in this particular hand, as S2 followed suit, but can be true in the more general scenario I tested], things look dire for friends in S1/S3. But this, again, is a very rare case.
[I will also assume that S1 doesn't have a 2nd trump, as this would be an equally rare case.]
So most likely, L is either buried or sits with S4. If it's buried, S3 has 2 boss trump and is almost assured of making a point; the uncertainty of where it lies (and subsequent decisions on how to play) may explain the difference in 2pt wins. What appears critical here is the case where S4 has 2 trump and guarded L, which happens to also be this particular case.
To euchre S1/S3, the opponents need 3 tricks. Say S3 wins the 1st trick with the R, S4 plays their low trump (QC), and S3 leads a low D. S2/S4 can euchre if they have 2 boss Diamonds (and coordinate their play appropriately), otherwise it's a bit up in the air. If S3 plays the AC on the 1st trick and lets S4 win with the L, S2/S4 can win a 2nd trick with a D lead (if they have the boss card  and even possibly a 3rd), but also set up S4 to have the choice of overtrumping S3 on a later trick, or playing off, saving their trump for later. I thinks it's this flexibility  having a trump and playing after the strong opponent  that gives S4 the edge.
My reasoning may be flawed  I'm sure Irish will let me know if it is ! And by the way, regarding the "rare" cases, one favors S1/S2, one favors S2/S4, and the 3rd is just dire for S1/S3 either way  so a wash.

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Just to add a bit to an already long post, it seems the rational behind playing the AC and potentially being overtrumped by the L by S4 is to "draw out" that L. But being left with the boss trump is only useful if we have a natural entry into our hand to then lead that R and eliminate the remaining trump. But we don't. S3 has 2 low Diamonds, so the only entry is via trump. And entry via the 9C is fraught because S4 could always potentially overtrump (and using the R as the entry defeats the whole purpose).

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YEP, clearly wrong to play the Right to that first trick to force out the Left if the Dealer has it, or if S2 has it! But I should qualify that statement: Unless S3 has an off suit ace or K/Q doubleton.
"My reasoning may be flawed  I'm sure Irish will let me know if it is !
IRISH
"My reasoning may be flawed  I'm sure Irish will let me know if it is !
IRISH

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Now I'm confused!
Playing the R at S3 DOESN'T force out the L if dealer has it, nor does it force out the L from the player who already played  S2. And S3 doesn't have an offsuit A or K/Q doubleton either. That isn't the point of playing it. The point is to win the hand. So are you saying it is simply wrong to play the R from S3?
What is your rationale? I just ran a simulation which showed it is BETTER to play the R in this situation, and even went to great lengths to explain why.
Your use of the word "clearly" is like mathematicians sarcastically saying QED !
Playing the R at S3 DOESN'T force out the L if dealer has it, nor does it force out the L from the player who already played  S2. And S3 doesn't have an offsuit A or K/Q doubleton either. That isn't the point of playing it. The point is to win the hand. So are you saying it is simply wrong to play the R from S3?
What is your rationale? I just ran a simulation which showed it is BETTER to play the R in this situation, and even went to great lengths to explain why.
Your use of the word "clearly" is like mathematicians sarcastically saying QED !

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If S3 has Right, Ace 10 + Off suit Ace (King up card)  Then there is no need to Play the A of trump because, S3 plays the Right, then the Off suit Ace, and this will force the Dealer's other trump. In addition if he has K/Q off suit, still better to play the Right and get your King off suit played.
If he has poor off suit, better to force that Left if the Dealer has it by playing the Ace of trump as he needs to get that Left out of play.
It all about forcing the Left and looking ahead to the next tricks, depending on what has been played. If S2 has no trump, you want to keep the Right to protect a small trump as the Dealer will be afraid to lead trump. If you win the trick and only have Ace + small trump and lead an off suit the dealer wins the trick or sloughs to get behind S3.
There is also the issue that S3 does not know where that Left is. It could be buried, with S2 or with S4 (Dealer). So the advantage of playing the AC flushes it from the dealer and if it is with S2, you are now behind him. Additionally, if it is with S2 or buried, and Eldest win the trick, he can lead an ace if he has it.
Its not sarcasm. Proof is in the pudding!
If you think you are correct  give me some hands that prove hour point where it is better to play the Right other than the two example I gave.
IRISH
If he has poor off suit, better to force that Left if the Dealer has it by playing the Ace of trump as he needs to get that Left out of play.
It all about forcing the Left and looking ahead to the next tricks, depending on what has been played. If S2 has no trump, you want to keep the Right to protect a small trump as the Dealer will be afraid to lead trump. If you win the trick and only have Ace + small trump and lead an off suit the dealer wins the trick or sloughs to get behind S3.
There is also the issue that S3 does not know where that Left is. It could be buried, with S2 or with S4 (Dealer). So the advantage of playing the AC flushes it from the dealer and if it is with S2, you are now behind him. Additionally, if it is with S2 or buried, and Eldest win the trick, he can lead an ace if he has it.
Its not sarcasm. Proof is in the pudding!
If you think you are correct  give me some hands that prove hour point where it is better to play the Right other than the two example I gave.
IRISH

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 Joined: Tue Apr 24, 2018 9:33 pm
Back to your comment and simulator: (the Hand S3 had above)
Ray said, "All that said, I then ran a simulation for 100,000 hands to test what S3 should play after a trump lead by S1 on the first trick. 1st scenario is that S3 always plays the R. 2nd scenario is that S3 will play the 9C if S1 led the KC, otherwise will play the AC. [S3 will always play the 9C if S1 leads L]
1) [play R]: (9,919 / 49,482 / 4,191) [2 pts / 1 pt / euchred] EV = +0.96
2) [play AC]: (10,471 / 47,213 / 5,908) EV = +0.89 "
S3 has JC AC 9C JD 10D, QC UPCARD S3 orders and S1 ALWAYS LEADS THE 10C and only has ONE Trump. However there are only Seven Known cards and 17 are randomly dealt.
If S1 has the KC or the Left (JS) he must always play it and S3 plays the AC (unless S2 played the Left).
I doubt that your Simulator can even simulate this type of combination. That said, it has to be played correctly. If S2 plays the Left and S3 must over trump and he now has the boss trump (AC).
Secondly, if S2 plays the KC, or is void in trump, S3 always plays the AC. This forces Left (JS). If S4 has the Left, S3 still has JC/9C and can get back in to lead the Right as he has end position. If S4 leads something S1 wins the trick he can slough if he chooses so.
Is that the way your simulator played to get the results where you said it is BEST to play the Right.
This is critical to the results. I am seriously doubting your conclusion(s).
IRISH
P.S. Consider that if S3 always plays the JC (Right) under this hand, S4 or S2 will have the Left 72% of the time (28% buried in Stock). But if S3 plays AC 36% S4 has the Left and 36% S2 with S3 behind S2.
Ray said, "All that said, I then ran a simulation for 100,000 hands to test what S3 should play after a trump lead by S1 on the first trick. 1st scenario is that S3 always plays the R. 2nd scenario is that S3 will play the 9C if S1 led the KC, otherwise will play the AC. [S3 will always play the 9C if S1 leads L]
1) [play R]: (9,919 / 49,482 / 4,191) [2 pts / 1 pt / euchred] EV = +0.96
2) [play AC]: (10,471 / 47,213 / 5,908) EV = +0.89 "
S3 has JC AC 9C JD 10D, QC UPCARD S3 orders and S1 ALWAYS LEADS THE 10C and only has ONE Trump. However there are only Seven Known cards and 17 are randomly dealt.
If S1 has the KC or the Left (JS) he must always play it and S3 plays the AC (unless S2 played the Left).
I doubt that your Simulator can even simulate this type of combination. That said, it has to be played correctly. If S2 plays the Left and S3 must over trump and he now has the boss trump (AC).
Secondly, if S2 plays the KC, or is void in trump, S3 always plays the AC. This forces Left (JS). If S4 has the Left, S3 still has JC/9C and can get back in to lead the Right as he has end position. If S4 leads something S1 wins the trick he can slough if he chooses so.
Is that the way your simulator played to get the results where you said it is BEST to play the Right.
This is critical to the results. I am seriously doubting your conclusion(s).
IRISH
P.S. Consider that if S3 always plays the JC (Right) under this hand, S4 or S2 will have the Left 72% of the time (28% buried in Stock). But if S3 plays AC 36% S4 has the Left and 36% S2 with S3 behind S2.

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RAY  did you revisit this important issue?
Please address the the question to the Simulator results: (SEE BELOW) Is that the way your simulator played to get the results where you said it is BEST to play the Right?
i would hate to leaves this hanging in the air.
IRISH
DISCUSSION:
1) [play R]: (9,919 / 49,482 / 4,191) [2 pts / 1 pt / euchred] EV = +0.96
2) [play AC]: (10,471 / 47,213 / 5,908) EV = +0.89 "
S3 has JC AC 9C JD 10D, QC UPCARD S3 orders and S1 ALWAYS LEADS THE 10C and only has ONE Trump. However there are only Seven Known cards and 17 are randomly dealt.
If S1 has the KC or the Left (JS) he must always play it and S3 plays the AC (unless S2 played the Left).
I doubt that your Simulator can even simulate this type of combination. That said, it has to be played correctly. If S2 plays the Left and S3 must over trump and he now has the boss trump (AC).
Secondly, if S2 plays the KC, or is void in trump, S3 always plays the AC. This forces Left (JS). If S4 has the Left, S3 still has JC/9C and can get back in to lead the Right as he has end position. If S4 leads something S1 wins the trick he can slough if he chooses so.
Is that the way your simulator played to get the results where you said it is BEST to play the Right.
This is critical to the results. I am seriously doubting your conclusion(s).
IRISH
Please address the the question to the Simulator results: (SEE BELOW) Is that the way your simulator played to get the results where you said it is BEST to play the Right?
i would hate to leaves this hanging in the air.
IRISH
DISCUSSION:
1) [play R]: (9,919 / 49,482 / 4,191) [2 pts / 1 pt / euchred] EV = +0.96
2) [play AC]: (10,471 / 47,213 / 5,908) EV = +0.89 "
S3 has JC AC 9C JD 10D, QC UPCARD S3 orders and S1 ALWAYS LEADS THE 10C and only has ONE Trump. However there are only Seven Known cards and 17 are randomly dealt.
If S1 has the KC or the Left (JS) he must always play it and S3 plays the AC (unless S2 played the Left).
I doubt that your Simulator can even simulate this type of combination. That said, it has to be played correctly. If S2 plays the Left and S3 must over trump and he now has the boss trump (AC).
Secondly, if S2 plays the KC, or is void in trump, S3 always plays the AC. This forces Left (JS). If S4 has the Left, S3 still has JC/9C and can get back in to lead the Right as he has end position. If S4 leads something S1 wins the trick he can slough if he chooses so.
Is that the way your simulator played to get the results where you said it is BEST to play the Right.
This is critical to the results. I am seriously doubting your conclusion(s).
IRISH

 Posts: 251
 Joined: Thu Sep 16, 2021 6:56 pm
Irish, sorry for the delay. I very much appreciate your insights, and have actually looked into this further, but was waylaid when I discovered some issue with my program.
First of all, the scenario I set up was the following:
1) S3 gets AJ9C + J10D;
2) QC turned;
3) all other cards randomly distributed.
I then looked at ONLY those hands where S1 led the K or 10 C and S2 did NOT play the L. This turned out to be about 23% of the hands. I did it this way to make the scenario more general, and perhaps better reflect the actual odds of various specific cases arising. This is how S3 sees things.
So in those 23% of cases, S3 was left with a decision:
Case A: S3 always plays the R, hoping the L is buried or at least getting the lead, trick 2;
Case B: S3 plays the AC if S1 plays the 10C (and S2 plays the KC or is void in trump), or plays the 9C if S1 plays the KC (and S2 plays the 10C or is void in trump).
I looked specific for cases where Case A gives a better outcome than Case B, and that's where I hit a problem. The first few cases I found where playing the R yielded a point and playing the AC yielded a euchre were not played out properly by my program. I then had to generalize the case, test it, tweak it, retest it, and finally update my program. I did this for a couple of situations, and that took me several days. In the end, the changes I made arose in less than 10% of hands, and affected the outcome in less than 1% (net aggregate difference) of those 10% of hands. But those situations arise much more frequently for the hand being simulated here.
Anyway, I made those changes and reran my simulation on 100,000 hands (as described above). The new results are:
Case A (play R): EV = +0.37
Case B (play 9C or AC): EV = +0.32
The absolute value of these results changed, but not the fact that playing the R still appears more favorable.
At this point I plan of continuing to investigate if there are other particular situations where my program is playing incorrectly, but it is very time consuming to uncover those instances.
In the meantime, I did find a couple of hands where I believe S1/S3 get a point when S3 plays the R but get euchred if S3 plays the AC or 9C. I'd like you to check them out and see if I've analyzed them correctly (as if I needed to ask!)
Hand 1:
S1: QS + A10H + 9D + KC
S2: KS + KQH + AQD
S3: [see above]
S4: AJ109S + JH
(QC turned)
Hand 2:
S1: QJ9S + K10H
S2: QJH + KQD + 10C
S3: [see above]
S4: AKS + AH + AKD
(QC turned)
I also found a hand where S3 playing the R yields 1 pt but playing the 9C yields 2 pts:
S1: KC + AS + AH + 2 other cards
S2: 10C + AD + 3 other cards
S3: [see above]
S4: any 5 cards (no trump)
(QC turned; JS buried)
First of all, the scenario I set up was the following:
1) S3 gets AJ9C + J10D;
2) QC turned;
3) all other cards randomly distributed.
I then looked at ONLY those hands where S1 led the K or 10 C and S2 did NOT play the L. This turned out to be about 23% of the hands. I did it this way to make the scenario more general, and perhaps better reflect the actual odds of various specific cases arising. This is how S3 sees things.
So in those 23% of cases, S3 was left with a decision:
Case A: S3 always plays the R, hoping the L is buried or at least getting the lead, trick 2;
Case B: S3 plays the AC if S1 plays the 10C (and S2 plays the KC or is void in trump), or plays the 9C if S1 plays the KC (and S2 plays the 10C or is void in trump).
I looked specific for cases where Case A gives a better outcome than Case B, and that's where I hit a problem. The first few cases I found where playing the R yielded a point and playing the AC yielded a euchre were not played out properly by my program. I then had to generalize the case, test it, tweak it, retest it, and finally update my program. I did this for a couple of situations, and that took me several days. In the end, the changes I made arose in less than 10% of hands, and affected the outcome in less than 1% (net aggregate difference) of those 10% of hands. But those situations arise much more frequently for the hand being simulated here.
Anyway, I made those changes and reran my simulation on 100,000 hands (as described above). The new results are:
Case A (play R): EV = +0.37
Case B (play 9C or AC): EV = +0.32
The absolute value of these results changed, but not the fact that playing the R still appears more favorable.
At this point I plan of continuing to investigate if there are other particular situations where my program is playing incorrectly, but it is very time consuming to uncover those instances.
In the meantime, I did find a couple of hands where I believe S1/S3 get a point when S3 plays the R but get euchred if S3 plays the AC or 9C. I'd like you to check them out and see if I've analyzed them correctly (as if I needed to ask!)
Hand 1:
S1: QS + A10H + 9D + KC
S2: KS + KQH + AQD
S3: [see above]
S4: AJ109S + JH
(QC turned)
Hand 2:
S1: QJ9S + K10H
S2: QJH + KQD + 10C
S3: [see above]
S4: AKS + AH + AKD
(QC turned)
I also found a hand where S3 playing the R yields 1 pt but playing the 9C yields 2 pts:
S1: KC + AS + AH + 2 other cards
S2: 10C + AD + 3 other cards
S3: [see above]
S4: any 5 cards (no trump)
(QC turned; JS buried)

 Posts: 1291
 Joined: Tue Apr 24, 2018 9:33 pm
RAY thanks for responding and for your efforts.
But I respectfully disagree.
First is that you moved the goal Posts when you took the 10C as a random card and randomized 18 as this results in S1 having no trump (statistically) 35% of the time (3 unknown). When that occurs, there are 3 trumps out against S3 (10C JS KC) instead of 2. This Scenario is that S1 has the 10C and always leads the 10C. Thus there are 17 unknown card with S1 having 10C. And that I suspect is difficult for you to program has S1 has to have 4 random nontrump cards (only 15 to S1 with 17 to S2&S4) with the 2 randomly split between Stock, S2 & S4, 17 cards. Now that is a tough task to Program. But be as it may, let's move on to what I have to say next:
Here is a quick review, I might not have all the combinations that S3 gets euchred when he has JC AC 9C QD 10D, QC UP & S1 has 10C. But most anyway. Here they are:
Hands that S3 gets euchred – five major ones without looking deeper. And look at when get euchred JC vs AC. No contest.
1) S4 HAS JS QC AD + KD OR QD ore both. (PLAYING AC OR JC = EUCHRE)
2) S4 HAS JS QC – S2 HAS AD + KD or QD (PLAYING AC OR JC = EUCHRE) BUT IF PLAYS AC, DEALER HAS TO LEAD 9D – MORE DIFFICULT COMBINATION.
3)S4 HAS QC – BUT S2 HAS JS & KC AND HAS AD + KD or QD or BOTH (PLAYING JS = EUCHRE BUT NOT AC)
4) S4 HAS QD, AD KD or QD or BOTH AND S2 HAS JS & KC. (PLAYING JS = EUCHRE BUT NOT THE AC)
5) S4 HAS QC – S2 HAS JS & KC & LEADS AD THEN 9D AND S4 HAS TO HAVE KD & QD. (PLAYING JC FIRST, S3 IS IN BIG TROUBLE NOW = EUCHRED PLAYING THE JC BUT NOT THE AC.)
So look at those card holdings, and I don't need 100,000 to see the advantage/disadvantage. When S3 always plays the JC if S2 does not play it, he can never lead trump to trick 2 regardless if it is buried or with S4. Very few sweeps will occur. It not good euchre to blindly leading when you have the boss trump out and unknown even if buried (17%).
Now look to the lethal off suit combinations. You are MORE vulnerable playing the JC than the AC. S4 will have the JS 28% of the time. Even if he has it, he now leads ans S1 can potentially win a trick for you but seldom if you play the JC. But look at the probability of being euchred playing the AC vs JC. That Euchre rate will be around 14%.
THE LETHAL COMBINATION(S): There are 4 Diamonds out and it requires a Combination of AD/KD, AD/QD, AD/KD/QD, KD/QD (if AD is buried). That combination is about 14% chance. And that is even when S3 plays the JC. It (euchre) is less than that when the AC is played as S4 (in particular) has to have it or lead a Diamond and S2 has to have the lethal combination.
The above is my position  better to play the AC when the JS has not been played.
IRISH
You gave me three hands and I will answer those in another post. Shortly!
But I respectfully disagree.
First is that you moved the goal Posts when you took the 10C as a random card and randomized 18 as this results in S1 having no trump (statistically) 35% of the time (3 unknown). When that occurs, there are 3 trumps out against S3 (10C JS KC) instead of 2. This Scenario is that S1 has the 10C and always leads the 10C. Thus there are 17 unknown card with S1 having 10C. And that I suspect is difficult for you to program has S1 has to have 4 random nontrump cards (only 15 to S1 with 17 to S2&S4) with the 2 randomly split between Stock, S2 & S4, 17 cards. Now that is a tough task to Program. But be as it may, let's move on to what I have to say next:
Here is a quick review, I might not have all the combinations that S3 gets euchred when he has JC AC 9C QD 10D, QC UP & S1 has 10C. But most anyway. Here they are:
Hands that S3 gets euchred – five major ones without looking deeper. And look at when get euchred JC vs AC. No contest.
1) S4 HAS JS QC AD + KD OR QD ore both. (PLAYING AC OR JC = EUCHRE)
2) S4 HAS JS QC – S2 HAS AD + KD or QD (PLAYING AC OR JC = EUCHRE) BUT IF PLAYS AC, DEALER HAS TO LEAD 9D – MORE DIFFICULT COMBINATION.
3)S4 HAS QC – BUT S2 HAS JS & KC AND HAS AD + KD or QD or BOTH (PLAYING JS = EUCHRE BUT NOT AC)
4) S4 HAS QD, AD KD or QD or BOTH AND S2 HAS JS & KC. (PLAYING JS = EUCHRE BUT NOT THE AC)
5) S4 HAS QC – S2 HAS JS & KC & LEADS AD THEN 9D AND S4 HAS TO HAVE KD & QD. (PLAYING JC FIRST, S3 IS IN BIG TROUBLE NOW = EUCHRED PLAYING THE JC BUT NOT THE AC.)
So look at those card holdings, and I don't need 100,000 to see the advantage/disadvantage. When S3 always plays the JC if S2 does not play it, he can never lead trump to trick 2 regardless if it is buried or with S4. Very few sweeps will occur. It not good euchre to blindly leading when you have the boss trump out and unknown even if buried (17%).
Now look to the lethal off suit combinations. You are MORE vulnerable playing the JC than the AC. S4 will have the JS 28% of the time. Even if he has it, he now leads ans S1 can potentially win a trick for you but seldom if you play the JC. But look at the probability of being euchred playing the AC vs JC. That Euchre rate will be around 14%.
THE LETHAL COMBINATION(S): There are 4 Diamonds out and it requires a Combination of AD/KD, AD/QD, AD/KD/QD, KD/QD (if AD is buried). That combination is about 14% chance. And that is even when S3 plays the JC. It (euchre) is less than that when the AC is played as S4 (in particular) has to have it or lead a Diamond and S2 has to have the lethal combination.
The above is my position  better to play the AC when the JS has not been played.
IRISH
You gave me three hands and I will answer those in another post. Shortly!

 Posts: 1291
 Joined: Tue Apr 24, 2018 9:33 pm
Now to those three hands you gave as example:
In the meantime, I did find a couple of hands where I believe S1/S3 get a point when S3 plays the R but get euchred if S3 plays the AC or 9C. I'd like you to check them out and see if I've analyzed them correctly (as if I needed to ask!)
Hand 1:
S1: QS + A10H + 9D + KC
S2: KS + KQH + AQD
S3: [see above]
S4: AJ109S + JH
(QC turned)
A EUCHRE PLAYING JC OR AC. WHY, S3 PLAYS THE JC HE HAS TO NOW PLAY A DIAMOND AND IF S4 SLOUGHS RESULTS IS THE SAME. MATTERS LITTLE WHAT S3 PLAYS.
Hand 2:
S1: QJ9S + K10H
S2: QJH + KQD + 10C
S3: [see above]
S4: AKS + AH + AKD
(QC turned)
NOT A EUCHRE HERE AS S1 LEADS THE LEFT  S3 ONLY PLAY IS THE 9C AND THEN HAS TWO BOSS TRUMPS.
I also found a hand where S3 playing the R yields 1 pt but playing the 9C yields 2 pts:
S1: KC + AS + AH + 2 other cards
S2: 10C + AD + 3 other cards
S3: [see above]
S4: any 5 cards (no trump)
(QC turned; JS buried)
SAME RESULT AS S3 MUST PLAY THE 9C HERE AS HE HAS THE JC/AC TO FORCE THE LEFT WITH THE KC LEAD. SO SAME RESULT.
Like I said in an earlier post, 36% of the time S4 will have that left (when S1 does not have it), and 64% of the time it's buried or with S2. In that case when it is, S3 now leads the JC and has his point. S1 then has to have the AD for a sweep.
My conclusion on your examples, NO ADVANTAGE playing the JC. But you moved the goals posts from the original hand.
As to S1 playing KC or JS, S3 ALWAYS plays according to S1 lead.
S3 PLAYS JC: With your simulator running 100,000 hands S1 plays the the 10C, you certainly should be able to test ALL HANDS with S3 playing the JC. So you should have 100,000.
S3 PLAYING AC: You won't net 100,000 for AC, initially. Why, because testing the AC you have pull out those ~36% S2 has JS (statistically where S2 played the JS, unguarded) of those hands where S2 played the JS. Except those where he has it guarded with KC he does not and S3 plays AC. Guarded will be somewhere about 7  8%. All those hands S3 of course would play the JC. It would make no sense to play the AC if S2 played the JS. If S3 wins the trick with the AC, he then leads the JC to get S2's JS if he has it. Then he leads 10D and he still has 10C for a point.
NOTE: Those hands pulled out but those get added back to S3 strategy (EV) of playing the AC as it is all part of how this hand is played. And should be noted. Now you have 100,000 one way vs another.
S3 can be Euchred playing when Dealer has JS KC QC (7%)  possible but has to have help from S2 and S3 can also win a trick. Just as bad playing the JC as the KC. S3 could he forego some sweeps playing this way? Yes, that is when statistically S1 does not have the AD or KD when AD is buried but S1 has to also have something else (boss card) to go with it. In euchre, always play to get your point first, and a secondly go for a sweep. Playing the JC, always you open yourself up for more Euchres. And most importantly, S1 still has to have the AD.
To test S1 having KC, that test, S3 would always play the 9C. No need to play the AC or JC to force S4 to play it if he has it. But you could just like above, always play the JC (not advised by me).
For a fair test, your simulator would have to play like this and account for the above issues.
IRISH
In the meantime, I did find a couple of hands where I believe S1/S3 get a point when S3 plays the R but get euchred if S3 plays the AC or 9C. I'd like you to check them out and see if I've analyzed them correctly (as if I needed to ask!)
Hand 1:
S1: QS + A10H + 9D + KC
S2: KS + KQH + AQD
S3: [see above]
S4: AJ109S + JH
(QC turned)
A EUCHRE PLAYING JC OR AC. WHY, S3 PLAYS THE JC HE HAS TO NOW PLAY A DIAMOND AND IF S4 SLOUGHS RESULTS IS THE SAME. MATTERS LITTLE WHAT S3 PLAYS.
Hand 2:
S1: QJ9S + K10H
S2: QJH + KQD + 10C
S3: [see above]
S4: AKS + AH + AKD
(QC turned)
NOT A EUCHRE HERE AS S1 LEADS THE LEFT  S3 ONLY PLAY IS THE 9C AND THEN HAS TWO BOSS TRUMPS.
I also found a hand where S3 playing the R yields 1 pt but playing the 9C yields 2 pts:
S1: KC + AS + AH + 2 other cards
S2: 10C + AD + 3 other cards
S3: [see above]
S4: any 5 cards (no trump)
(QC turned; JS buried)
SAME RESULT AS S3 MUST PLAY THE 9C HERE AS HE HAS THE JC/AC TO FORCE THE LEFT WITH THE KC LEAD. SO SAME RESULT.
Like I said in an earlier post, 36% of the time S4 will have that left (when S1 does not have it), and 64% of the time it's buried or with S2. In that case when it is, S3 now leads the JC and has his point. S1 then has to have the AD for a sweep.
My conclusion on your examples, NO ADVANTAGE playing the JC. But you moved the goals posts from the original hand.
As to S1 playing KC or JS, S3 ALWAYS plays according to S1 lead.
S3 PLAYS JC: With your simulator running 100,000 hands S1 plays the the 10C, you certainly should be able to test ALL HANDS with S3 playing the JC. So you should have 100,000.
S3 PLAYING AC: You won't net 100,000 for AC, initially. Why, because testing the AC you have pull out those ~36% S2 has JS (statistically where S2 played the JS, unguarded) of those hands where S2 played the JS. Except those where he has it guarded with KC he does not and S3 plays AC. Guarded will be somewhere about 7  8%. All those hands S3 of course would play the JC. It would make no sense to play the AC if S2 played the JS. If S3 wins the trick with the AC, he then leads the JC to get S2's JS if he has it. Then he leads 10D and he still has 10C for a point.
NOTE: Those hands pulled out but those get added back to S3 strategy (EV) of playing the AC as it is all part of how this hand is played. And should be noted. Now you have 100,000 one way vs another.
S3 can be Euchred playing when Dealer has JS KC QC (7%)  possible but has to have help from S2 and S3 can also win a trick. Just as bad playing the JC as the KC. S3 could he forego some sweeps playing this way? Yes, that is when statistically S1 does not have the AD or KD when AD is buried but S1 has to also have something else (boss card) to go with it. In euchre, always play to get your point first, and a secondly go for a sweep. Playing the JC, always you open yourself up for more Euchres. And most importantly, S1 still has to have the AD.
To test S1 having KC, that test, S3 would always play the 9C. No need to play the AC or JC to force S4 to play it if he has it. But you could just like above, always play the JC (not advised by me).
For a fair test, your simulator would have to play like this and account for the above issues.
IRISH
Last edited by irishwolf on Thu Feb 10, 2022 2:43 pm, edited 1 time in total.

 Posts: 1291
 Joined: Tue Apr 24, 2018 9:33 pm
Ray, I went back to your post on Feb 04  Did you pull out those hands where S2 played JS and S3 over trumped with JC? Then add those back to how the hand should be played AC strategy?
IRISH
"Fri Feb 04, 2022 8:49 am
Back to your comment and simulator: (the Hand S3 had above)
Ray said, "All that said, I then ran a simulation for 100,000 hands to test what S3 should play after a trump lead by S1 on the first trick. 1st scenario is that S3 always plays the R. 2nd scenario is that S3 will play the 9C if S1 led the KC, otherwise will play the AC.
[S3 will always play the 9C if S1 leads L]
1) [play R]: (9,919 / 49,482 / 4,191) [2 pts / 1 pt / euchred] EV = +0.96
2) [play AC]: (10,471 / 47,213 / 5,908) EV = +0.89 "
IRISH
"Fri Feb 04, 2022 8:49 am
Back to your comment and simulator: (the Hand S3 had above)
Ray said, "All that said, I then ran a simulation for 100,000 hands to test what S3 should play after a trump lead by S1 on the first trick. 1st scenario is that S3 always plays the R. 2nd scenario is that S3 will play the 9C if S1 led the KC, otherwise will play the AC.
[S3 will always play the 9C if S1 leads L]
1) [play R]: (9,919 / 49,482 / 4,191) [2 pts / 1 pt / euchred] EV = +0.96
2) [play AC]: (10,471 / 47,213 / 5,908) EV = +0.89 "

 Posts: 1291
 Joined: Tue Apr 24, 2018 9:33 pm
RAY,
I have another question. In 1) & 2), 100,000 hands but only 63,592 hands (63.6%) make it to S3 who orders. So S1 can't order, S2 then must be ordering 35.4% (35,000 or so). I estimate he will only order with JS/KC but that is only 7  9% so it seems like ~93,000 hands should make it to S3. Can you explain what missing here, 63,592 vs 93,000?
IRISH
"All that said, I then ran a simulation for 100,000 hands to test what S3 should play after a trump lead by S1 on the first trick. 1st scenario is that S3 always plays the R. 2nd scenario is that S3 will play the 9C if S1 led the KC, otherwise will play the AC. [S3 will always play the 9C if S1 leads L]
1) [play R]: (9,919 / 49,482 / 4,191) [2 pts / 1 pt / euchred] EV = +0.96
2) [play AC]: (10,471 / 47,213 / 5,908) EV = +0.89"
I have another question. In 1) & 2), 100,000 hands but only 63,592 hands (63.6%) make it to S3 who orders. So S1 can't order, S2 then must be ordering 35.4% (35,000 or so). I estimate he will only order with JS/KC but that is only 7  9% so it seems like ~93,000 hands should make it to S3. Can you explain what missing here, 63,592 vs 93,000?
IRISH
"All that said, I then ran a simulation for 100,000 hands to test what S3 should play after a trump lead by S1 on the first trick. 1st scenario is that S3 always plays the R. 2nd scenario is that S3 will play the 9C if S1 led the KC, otherwise will play the AC. [S3 will always play the 9C if S1 leads L]
1) [play R]: (9,919 / 49,482 / 4,191) [2 pts / 1 pt / euchred] EV = +0.96
2) [play AC]: (10,471 / 47,213 / 5,908) EV = +0.89"

 Posts: 251
 Joined: Thu Sep 16, 2021 6:56 pm
FYI, I'm looking into all your questions, just keep getting sidetracked with other issues (this is proving very fruitful at optimizing my program!) and have finite time to spend on this. I WILL post a reply eventually.

 Posts: 1291
 Joined: Tue Apr 24, 2018 9:33 pm
Ray,
Take your time. But I am confident in my statement(s) that playing the AC to S1 leads of 10C (his only trump) will yield more sweeps, fewer euchres with a better +EV. Has to because if Dealer has JS, S3 cannot get it. And if S2 has it, S3 is behind S2 and now leads the JC to pick up the Left. This is a big advantage of playing the AC.
Furthermore, when the JC is played, S3 now has to commit to his weak off suit and there can only be sweeps if JS is buried because the JC was led. When AC is played, dealer has three choices and S3 has 2 diamonds so this favors a heart or spade lead and S1 can (sometimes) help but seldom if S3 played the Right.
When S3 wins the trick with the AC he must lead the JC then he plays his JD or 10D. He saves his 9C for later. S3 is vulnerable when Dealer has JS + KC + QC (upcard). He still has a chance as S1 might still win a trick. That is about 7/8%. S3 can still get euchred if S4 has the Left win the 1st trick but that boils down to who can win S3 JD/10D lead. For S3 to get euchred, it requires the diamond combinations I indicated earlier.
But difficult task you have is the hand has to be set up that S1 has 10C and only the 10C in trump with 4 other random cards while the KC/JS + all the other cards are random between S2, S4 and Stock. S3 has the hand as discussed. I think that is challenging and I don't know how difficult that is for you.
IRISH
Take your time. But I am confident in my statement(s) that playing the AC to S1 leads of 10C (his only trump) will yield more sweeps, fewer euchres with a better +EV. Has to because if Dealer has JS, S3 cannot get it. And if S2 has it, S3 is behind S2 and now leads the JC to pick up the Left. This is a big advantage of playing the AC.
Furthermore, when the JC is played, S3 now has to commit to his weak off suit and there can only be sweeps if JS is buried because the JC was led. When AC is played, dealer has three choices and S3 has 2 diamonds so this favors a heart or spade lead and S1 can (sometimes) help but seldom if S3 played the Right.
When S3 wins the trick with the AC he must lead the JC then he plays his JD or 10D. He saves his 9C for later. S3 is vulnerable when Dealer has JS + KC + QC (upcard). He still has a chance as S1 might still win a trick. That is about 7/8%. S3 can still get euchred if S4 has the Left win the 1st trick but that boils down to who can win S3 JD/10D lead. For S3 to get euchred, it requires the diamond combinations I indicated earlier.
But difficult task you have is the hand has to be set up that S1 has 10C and only the 10C in trump with 4 other random cards while the KC/JS + all the other cards are random between S2, S4 and Stock. S3 has the hand as discussed. I think that is challenging and I don't know how difficult that is for you.
IRISH

 Posts: 251
 Joined: Thu Sep 16, 2021 6:56 pm
OK, I finally have some results to share.
I "moved back the goal posts" to look ONLY at hands where S3 calls up the QC, R1, holding the AJ9C + 10JD, and the first 2 cards played are the 10C by S1 and the KC by S2. To do this, I forced the 5 stated cards to be in S3's hand, the 10C to be in S1's hand, the KC to be in S2's hand and the QC to be turned. The distribution of all other cards was random.
First, I played 100,000 hands to ensure that the JS was indeed randomly distributed among the 3 other players and the kitty. This proved to be the case. Here are the frequencies:
S1: 24,838 (expect 4/16)
S2: 24,908 (expect 4/16)
S4: 31,509 (expect 5/16)
buried: 18,745 (expect 3/16)
I also found that S2 actually bid 4,620 hands (always when they had the JS + KC), because just seeing their hand and the turn, it made sense to do so (though they got crushed with this particular hand). I in fact spent quite a bit of time exploring when S2 should bid holding L+1 trump  that will be the subject of a separate post.
I then broke the current hand down into 3 distinct scenarios:
A) S2 holds the JS
B) S4 holds the JS
C) JS is buried
Each time I ran a batch of 100,000 hands, I tested if the JS was where I wanted it to be, and only used the results from hands where this was the case. This was easier than forcing the JS to be in a certain place, and ensured the relative frequencies worked out correctly.
Remember, in all these hands the first 2 cards played are the 10C by S1 and the KC by S2, so I will just show what S3 plays. I also assumed that S2 would play the KC on the first trick when holding both the KC and the JS.
Scenario A: (100,000 hands; 20,319 counted  I didn't count the hands where S2 was declarer, so less than 25%)
play JC, then 10D: EV = +0.35
play AC, then JC, then 9C: EV = +1.29
play AC, then 10D: EV = +1.12
play AC, then JC, then 10D: EV = +1.37
So if S2 holds the JS it is best to play the AC, then lead the JC on trick 2, then lead the 10D on trick 3
Scenario B: (100,000 hands; 31,348 counted)
play JC, then 10D: EV = +0.24
play JC, then AC: EV = 0.01
play AC: EV = 0.36
Here I found that it is best to play the JC from S3 when S4 holds the L (and the QC). I know Irish will not be comfortable with this result; here's an example hand where S3 makes a point playing the JC and gets euchred playing the AC.
S1:Q109S + QD + 10C
S2: A9H + K9D + KC
S3: AJ9C + J10D
S4: AKJS + QJH [discards JH when picks up QC]
And by the way, regarding the examples I gave in an earlier post, I see now that the 1st example is no good, and in the 2nd example I accidentally shifted the hands by 1 position so it doesn't make sense. I hope this example here is good!
Scenario C: (100,000 hands; 18,808 counted)
play JC, then 10D: EV = +1.28
play AC, then JC, then 9C: EV = +1.24
play AC, then 10D: EV = +1.30
play AC, then JC, then 10D: EV = +1.30
Very little to distinguish these hands, which is perhaps not surprising given that the JC and AC are identical since the JS is buried.
At this point I took a weighted average of the results, using the fact that the JS is more likely to sit with S4 than S2, and there is still some probability of it being buried. That yielded the following:
play JC, then 10D: EV = +0.55
play AC, then JC (if win), then 10D: EV = +0.58
So a slight advantage to playing the AC [but statistical significance?]
I then ran the whole shebang with 100,000 hands, not counting the hands where S1 had the JS or where S2 bid. The results matched up well with the piecedtogether results above:
play JC, then 10D: EV = +0.56
play AC, then JC (if win), then 10D: EV = +0.58
For me, there is not a clearcut winner in this case, but there is still the very real possibility that my program is not playing the hands optimally and that is biasing the results.
I "moved back the goal posts" to look ONLY at hands where S3 calls up the QC, R1, holding the AJ9C + 10JD, and the first 2 cards played are the 10C by S1 and the KC by S2. To do this, I forced the 5 stated cards to be in S3's hand, the 10C to be in S1's hand, the KC to be in S2's hand and the QC to be turned. The distribution of all other cards was random.
First, I played 100,000 hands to ensure that the JS was indeed randomly distributed among the 3 other players and the kitty. This proved to be the case. Here are the frequencies:
S1: 24,838 (expect 4/16)
S2: 24,908 (expect 4/16)
S4: 31,509 (expect 5/16)
buried: 18,745 (expect 3/16)
I also found that S2 actually bid 4,620 hands (always when they had the JS + KC), because just seeing their hand and the turn, it made sense to do so (though they got crushed with this particular hand). I in fact spent quite a bit of time exploring when S2 should bid holding L+1 trump  that will be the subject of a separate post.
I then broke the current hand down into 3 distinct scenarios:
A) S2 holds the JS
B) S4 holds the JS
C) JS is buried
Each time I ran a batch of 100,000 hands, I tested if the JS was where I wanted it to be, and only used the results from hands where this was the case. This was easier than forcing the JS to be in a certain place, and ensured the relative frequencies worked out correctly.
Remember, in all these hands the first 2 cards played are the 10C by S1 and the KC by S2, so I will just show what S3 plays. I also assumed that S2 would play the KC on the first trick when holding both the KC and the JS.
Scenario A: (100,000 hands; 20,319 counted  I didn't count the hands where S2 was declarer, so less than 25%)
play JC, then 10D: EV = +0.35
play AC, then JC, then 9C: EV = +1.29
play AC, then 10D: EV = +1.12
play AC, then JC, then 10D: EV = +1.37
So if S2 holds the JS it is best to play the AC, then lead the JC on trick 2, then lead the 10D on trick 3
Scenario B: (100,000 hands; 31,348 counted)
play JC, then 10D: EV = +0.24
play JC, then AC: EV = 0.01
play AC: EV = 0.36
Here I found that it is best to play the JC from S3 when S4 holds the L (and the QC). I know Irish will not be comfortable with this result; here's an example hand where S3 makes a point playing the JC and gets euchred playing the AC.
S1:Q109S + QD + 10C
S2: A9H + K9D + KC
S3: AJ9C + J10D
S4: AKJS + QJH [discards JH when picks up QC]
And by the way, regarding the examples I gave in an earlier post, I see now that the 1st example is no good, and in the 2nd example I accidentally shifted the hands by 1 position so it doesn't make sense. I hope this example here is good!
Scenario C: (100,000 hands; 18,808 counted)
play JC, then 10D: EV = +1.28
play AC, then JC, then 9C: EV = +1.24
play AC, then 10D: EV = +1.30
play AC, then JC, then 10D: EV = +1.30
Very little to distinguish these hands, which is perhaps not surprising given that the JC and AC are identical since the JS is buried.
At this point I took a weighted average of the results, using the fact that the JS is more likely to sit with S4 than S2, and there is still some probability of it being buried. That yielded the following:
play JC, then 10D: EV = +0.55
play AC, then JC (if win), then 10D: EV = +0.58
So a slight advantage to playing the AC [but statistical significance?]
I then ran the whole shebang with 100,000 hands, not counting the hands where S1 had the JS or where S2 bid. The results matched up well with the piecedtogether results above:
play JC, then 10D: EV = +0.56
play AC, then JC (if win), then 10D: EV = +0.58
For me, there is not a clearcut winner in this case, but there is still the very real possibility that my program is not playing the hands optimally and that is biasing the results.

 Posts: 1291
 Joined: Tue Apr 24, 2018 9:33 pm
Thanks Ray for your work.
Okay you put the KC at S2 (JS at S2 33.3%, and 41.7% at S4 & buried 25%). My question is what did your simulator have as frequency for each for the JS?
This means JS void at S4 58.3%. I see you did it a little differently, pulling out the S1  JS. I will have to think about that!
I need time to digest the rest of your post.
IRISH
Okay you put the KC at S2 (JS at S2 33.3%, and 41.7% at S4 & buried 25%). My question is what did your simulator have as frequency for each for the JS?
This means JS void at S4 58.3%. I see you did it a little differently, pulling out the S1  JS. I will have to think about that!
I need time to digest the rest of your post.
IRISH

 Posts: 1291
 Joined: Tue Apr 24, 2018 9:33 pm
Ray,
Okay is my beef, and it with Scenario B  the magnitude of your number EV = 0.36 and EV = +0.24 of JC (I think those are closer as S4 with JS behind S3 is a big issue for him:
Scenario B JS at S4: (100,000 hands; 31,348 counted)
play JC, then 10D: EV = +0.24
play JC, then AC: EV = 0.01
play AC: EV = 0.36

S3 makes a point playing the JC and gets euchred playing the AC.
So follow my logic, I say not Probable:
E = 40% =  80 + 60 = 0.20 = EV Euchre rate has to be 45%! Questionable!
E = 45% = 90 + 55 = 0.35 = EV
Dealer wins AC with JS – now must lead off suit. 2 of 3 leads will be trumped by S3 (Spades or Hearts). That is when his QC will be depleted.
33% or 1/3 of the time will lead a Diamond (but there are 5 Spades & 6 Hearts & 4 Diamonds) and this is when S3 is vulnerable. S1 will have the AD equal to S2 or S4, 33%. Each at this point has 4 cards and S4 has QC so S1  4, S2  4 & S4  3 for the AD. (or KD if AD is buried. If S2 has AD his might lead a Heart or Spade and S3 ducks unless it's an AS or AH. It is when S2 has the AD dblt is when S3 is vulnerable. Just so tough to get that euchre. Is your program ducking when S2 has the AD and lead a non Ace Spade or Heart as that is when S4 must have a void. But also S3 can play the JC then lead his other Diamond and not trump with the 9C because he knows S4 has the QC.
Thus I serious doubt the euchred will exceed 40% when S4 has the JS. And that will be no different than when S3 plays the JC because S3 is still vulnerable to the Diamonds dblt. Is your Simulator playing optimally? I have played with type of hand testing and come up with Better to play the AC.
I think Scenario B Euchre rate is between 33 & 40%. Hard for me accept higher but I need to do some hands myself to look harder at Scenario B.
So I agree, discount Scenario C  a tie on EV. It's B that I have issues with. And I have at JS should be 41.7 at S4 and 33.3 at S2. Your number both their favor the JC play. Surprised, 12 cards: 4 to S2, 5 to S4 & 3 to Stock = 12. The discard cannot be the JS. QC (up) + 10C + KC + 5 (S3) = 8 16 Cards, 4 to S1.
The issue is bigger than this hand tho. As I recommend for all hand combinations if S3 ordered 9c to KC and he had JC AC + xC. Also consider the Universe of holdings at S3  Play the AC when he has dblt of K/9, K/10, K/Q, K/J etc. etc. Some weaker some stronger. My point is a whole array of combinations, that includes two singletons. S3 gets two bites of the apple playing the AC, in leading those two suits when he plays ACE most of the time. Playing the JC eliminates his options that favor the opponents to save a doubleton.
I appreciate your efforts. Feel free to shoot this down! I am only looking for the truth.
IRISH
Okay is my beef, and it with Scenario B  the magnitude of your number EV = 0.36 and EV = +0.24 of JC (I think those are closer as S4 with JS behind S3 is a big issue for him:
Scenario B JS at S4: (100,000 hands; 31,348 counted)
play JC, then 10D: EV = +0.24
play JC, then AC: EV = 0.01
play AC: EV = 0.36

S3 makes a point playing the JC and gets euchred playing the AC.
So follow my logic, I say not Probable:
E = 40% =  80 + 60 = 0.20 = EV Euchre rate has to be 45%! Questionable!
E = 45% = 90 + 55 = 0.35 = EV
Dealer wins AC with JS – now must lead off suit. 2 of 3 leads will be trumped by S3 (Spades or Hearts). That is when his QC will be depleted.
33% or 1/3 of the time will lead a Diamond (but there are 5 Spades & 6 Hearts & 4 Diamonds) and this is when S3 is vulnerable. S1 will have the AD equal to S2 or S4, 33%. Each at this point has 4 cards and S4 has QC so S1  4, S2  4 & S4  3 for the AD. (or KD if AD is buried. If S2 has AD his might lead a Heart or Spade and S3 ducks unless it's an AS or AH. It is when S2 has the AD dblt is when S3 is vulnerable. Just so tough to get that euchre. Is your program ducking when S2 has the AD and lead a non Ace Spade or Heart as that is when S4 must have a void. But also S3 can play the JC then lead his other Diamond and not trump with the 9C because he knows S4 has the QC.
Thus I serious doubt the euchred will exceed 40% when S4 has the JS. And that will be no different than when S3 plays the JC because S3 is still vulnerable to the Diamonds dblt. Is your Simulator playing optimally? I have played with type of hand testing and come up with Better to play the AC.
I think Scenario B Euchre rate is between 33 & 40%. Hard for me accept higher but I need to do some hands myself to look harder at Scenario B.
So I agree, discount Scenario C  a tie on EV. It's B that I have issues with. And I have at JS should be 41.7 at S4 and 33.3 at S2. Your number both their favor the JC play. Surprised, 12 cards: 4 to S2, 5 to S4 & 3 to Stock = 12. The discard cannot be the JS. QC (up) + 10C + KC + 5 (S3) = 8 16 Cards, 4 to S1.
The issue is bigger than this hand tho. As I recommend for all hand combinations if S3 ordered 9c to KC and he had JC AC + xC. Also consider the Universe of holdings at S3  Play the AC when he has dblt of K/9, K/10, K/Q, K/J etc. etc. Some weaker some stronger. My point is a whole array of combinations, that includes two singletons. S3 gets two bites of the apple playing the AC, in leading those two suits when he plays ACE most of the time. Playing the JC eliminates his options that favor the opponents to save a doubleton.
I appreciate your efforts. Feel free to shoot this down! I am only looking for the truth.
IRISH

 Posts: 1291
 Joined: Tue Apr 24, 2018 9:33 pm
RAY,
Check this out, using your Data as follows:
Comparing PLAY JC PLAY AC
SCENARIO A: +0.35 +1.37
SCENARIO B: +0.24 0.36
SCENARIO C: +1.28 +1.30
ADDING EVs: +1.87 + 2.31
NET DIFFERENCE AC 2.31 minus JC 1.87 = + 0.44 IN FAVOR OF PLAYING THE AC.
A EV of +0.58 vs 0.55 is Statistically Significant. Your own work proves my point.
WEIGHTED AVERAGE JC (EV = +0.55) vs AC (EV = +0.58) 70,745 hands that is still Statistically Significant. Of course if S1 leads KC, it becomes even better or if S3 has KD/9D or QD/9D.
However, I tested Scenario B, 100 hands played both ways and got Euchre rate of 28% playing JC (EV = +0.16 vs Euchre rate of 36% playing AC for a EV = 0.08. Of course negative when JS is at Dealer spot but playing like this with Scenario A & C more than off sets this. So we differ there on the Euchre rates. I suspect more on how I would play the Hands vs your Simulator.
Thanks for working on this, I hope not too painful!
irish
Check this out, using your Data as follows:
Comparing PLAY JC PLAY AC
SCENARIO A: +0.35 +1.37
SCENARIO B: +0.24 0.36
SCENARIO C: +1.28 +1.30
ADDING EVs: +1.87 + 2.31
NET DIFFERENCE AC 2.31 minus JC 1.87 = + 0.44 IN FAVOR OF PLAYING THE AC.
A EV of +0.58 vs 0.55 is Statistically Significant. Your own work proves my point.
WEIGHTED AVERAGE JC (EV = +0.55) vs AC (EV = +0.58) 70,745 hands that is still Statistically Significant. Of course if S1 leads KC, it becomes even better or if S3 has KD/9D or QD/9D.
However, I tested Scenario B, 100 hands played both ways and got Euchre rate of 28% playing JC (EV = +0.16 vs Euchre rate of 36% playing AC for a EV = 0.08. Of course negative when JS is at Dealer spot but playing like this with Scenario A & C more than off sets this. So we differ there on the Euchre rates. I suspect more on how I would play the Hands vs your Simulator.
Thanks for working on this, I hope not too painful!
irish
Last edited by irishwolf on Wed Feb 16, 2022 7:56 pm, edited 2 times in total.

 Posts: 251
 Joined: Thu Sep 16, 2021 6:56 pm
Irish, I will look at the play of the hands again to see if S3 (or other players) are making mistakes. My understanding is that you do not disagree that, if S4 holds the JS, it is better for the S3 to play the JC on the first trick. BUT, you think that playing the AC is not as unfavorable as my program makes it out to be.
That is not something I have thought too much about. My reasoning (to explain my findings) is simply that, if S3 wins the first trick (with the JC), they need to win just 2 more of the 4 tricks which remain, and one of those is guaranteed since they are the only player with two trump. Leading the 10D, trick 2, will sometimes win (if partner has the boss diamond), and will often set up their JD for a trick later. Partner in S2 will also have the AS and/or AH a decent amount of the time  just so many ways of making that necessary 3rd trick.
If S3 plays the AC, however, and S4 wins with the JS, S3 must now win 3 of the remaining 4 tricks, which will be tough given they have 2 losing diamonds AND DON'T HAVE THE LEAD. If S4 leads an A on the 2nd trick (or partner in S2 has the A or boss card in that suit), S3 will need to trump in, leaving them with just 1 trump, albeit the boss. S3 will need to lead then trump (2nd trick won) but then a crap diamond card. Many ways to lose the next 2 tricks  I would say even probable they lose them both: euchred!
As far as the likelihood of the JS lying with S2, S4 or buried, I agree that the raw probabilities are 4/12, 5/12 and 3/12, respectively. [let's convert that to 33, 42 and 25 times out of 100]. BUT, S2 will (or should) call trump almost 20% of the time, in which case the scenario we are studying here doesn't occur. So reduce the S2 occurrences of holding the JS by 18%, or from 33 to 27. The relative frequencies are now 27/94, 42/94 and 25/94: 28% held by S2, 45% held by S4 and 27% buried [rounded figures] This makes the scenario we would hope for (S2 holding the JS, because we can finesse it) a bit less likely. That explains why scenario A gets a bit less importance, and why my probabilities don't jive with what you were expecting.
_______________________
Am just seeing your most recent post. Please correct Scenario C, Play JC to read 1.28 rather than 1.23. Also, weighted average is key (using weights as explained in my previous paragraph, no the raw weights of where the JS is likely to be dealt  perhaps you already did this).
I already did this calculation, and, like you, found it slightly more favorable to play the AC. But really a sucky "barbell" situation: one play is clearly more favorable in one scenario, the other play clearly more favorable in the other scenario. So it comes down to accessing the relative advantage of playing one card versus playing the other card in a given scenario [which is where you currently disagree with my program] and the relative probability of either scenario occurring. [I'm considering "JS buried" as a virtual wash here]
Like I said, I'll look again at how my program plays the cards. I'm always interested in getting my algorithm to play better!
That is not something I have thought too much about. My reasoning (to explain my findings) is simply that, if S3 wins the first trick (with the JC), they need to win just 2 more of the 4 tricks which remain, and one of those is guaranteed since they are the only player with two trump. Leading the 10D, trick 2, will sometimes win (if partner has the boss diamond), and will often set up their JD for a trick later. Partner in S2 will also have the AS and/or AH a decent amount of the time  just so many ways of making that necessary 3rd trick.
If S3 plays the AC, however, and S4 wins with the JS, S3 must now win 3 of the remaining 4 tricks, which will be tough given they have 2 losing diamonds AND DON'T HAVE THE LEAD. If S4 leads an A on the 2nd trick (or partner in S2 has the A or boss card in that suit), S3 will need to trump in, leaving them with just 1 trump, albeit the boss. S3 will need to lead then trump (2nd trick won) but then a crap diamond card. Many ways to lose the next 2 tricks  I would say even probable they lose them both: euchred!
As far as the likelihood of the JS lying with S2, S4 or buried, I agree that the raw probabilities are 4/12, 5/12 and 3/12, respectively. [let's convert that to 33, 42 and 25 times out of 100]. BUT, S2 will (or should) call trump almost 20% of the time, in which case the scenario we are studying here doesn't occur. So reduce the S2 occurrences of holding the JS by 18%, or from 33 to 27. The relative frequencies are now 27/94, 42/94 and 25/94: 28% held by S2, 45% held by S4 and 27% buried [rounded figures] This makes the scenario we would hope for (S2 holding the JS, because we can finesse it) a bit less likely. That explains why scenario A gets a bit less importance, and why my probabilities don't jive with what you were expecting.
_______________________
Am just seeing your most recent post. Please correct Scenario C, Play JC to read 1.28 rather than 1.23. Also, weighted average is key (using weights as explained in my previous paragraph, no the raw weights of where the JS is likely to be dealt  perhaps you already did this).
I already did this calculation, and, like you, found it slightly more favorable to play the AC. But really a sucky "barbell" situation: one play is clearly more favorable in one scenario, the other play clearly more favorable in the other scenario. So it comes down to accessing the relative advantage of playing one card versus playing the other card in a given scenario [which is where you currently disagree with my program] and the relative probability of either scenario occurring. [I'm considering "JS buried" as a virtual wash here]
Like I said, I'll look again at how my program plays the cards. I'm always interested in getting my algorithm to play better!

 Posts: 1291
 Joined: Tue Apr 24, 2018 9:33 pm
Ray, Here are my comments on your last post. I will put my comments in Blue and before I do that, let's cut to the chase. I think we can agree the overall EV values of playing the JC vs AC (of course if S2 played the JS if he had it S3 would use the AC The values being EV = +0.55 (JC) and EV = + 0.58 (AC). We might disagree if that 0.03 is just a little bit It's not 3/55 = 5.45%. I did a tstatistic on that with 70,745 hands. The tstatistic = 3.97 P= 0.0001, Statistically Significant at the 99% level that the AC is a better play. I think it speaks for itself.
Where are we in agreement? When JS is at the Dealer seat EV will be better playing the JC. Where we differ is what that frequency is at S4 or at S2. So here are my comments below commenting on your Post below.
Ray said:
Irish, I will look at the play of the hands again to see if S3 (or other players) are making mistakes. My understanding is that you do not disagree that, if S4 holds the JS, it is better for the S3 to play the JC on the first trick. . No I disagree, I want to force Dealer to play the Left. Even though I have to still win 3 tricks, I am handcuffed with the Left behind me. Since the Left will win a trick anyway, I am better off to force the Left as his lead most likely will be a Spade or Heart and I can pick up his QC. Yes, I have to get a trick from my partner or get lucky that neither opponent has a Diamond dblt.
BUT, you think that playing the AC is not as unfavorable as my program makes it out to be. Yes, true.
That is not something I have thought too much about. My reasoning (to explain my findings) is simply that, if S3 wins the first trick (with the JC), they need to win just 2 more of the 4 tricks which remain, and one of those is guaranteed since they are the only player with two trump. Leading the 10D, trick 2, will sometimes win (if partner has the boss diamond),True, and this is the same Playing the JC or AC. and will often set up their JD for a trick later. Partner in S2 will also have the AS and/or AH a decent amount of the time  just so many ways of making that necessary 3rd trick. True, playing either JC or AC.
If S3 plays the AC, however, and S4 wins with the JS, S3 must now win 3 of the remaining 4 tricks, which will be tough given they have 2 losing diamonds AND DON'T HAVE THE LEAD. If S4 leads an A on the 2nd trick (or partner in S2 has the A or boss card in that suit), S3 will need to trump in, leaving them with just 1 trump, albeit the boss. S3 will need to lead then trump (2nd trick won) but then a crap diamond card. Many ways to lose the next 2 tricks  I would say even probable they lose them both: euchred! Consider when playing the JC, the Dealer now has the JS and is be hind S3 AC/9C. Dealer will most likely have a Void and can over trump and lead his best. I see this as being more of a disadvantage than playing the AC. When the Dealer has the Left, it will always boil down to S1 winning a trick or if S3 gets lucky with 4 unknown diamonds, AD or KD or QD could be buried. Sometimes hjis JD/10D will win a trick. More likely than playing the JC, because one of the opponents if get forced with Ace/Diamond, can better play S3 for his doubleton diamond, IMO.
As far as the likelihood of the JS lying with S2, S4 or buried, I agree that the raw probabilities are 4/12, 5/12 and 3/12, respectively. [let's convert that to 33, 42 and 25 times out of 100]. BUT, S2 will (or should) call trump almost 20% of the time, in which case the scenario we are studying here doesn't occur. So reduce the S2 occurrences of holding the JS by 18%, or from 33 to 27. The relative frequencies are now 27/94, 42/94 and 25/94: 28% held by S2, 45% held by S4 and 27% buried [rounded figures] This makes the scenario we would hope for (S2 holding the JS, because we can finesse it) a bit less likely. That explains why scenario A gets a bit less importance, and why my probabilities don't jive with what you were expecting. . I think here I differ with your thinking. When will S2 order up the QC? If he holds KC/JS + AH or AD no spade. He should not not when he holds KC/JS + AS, or without an ace, or when he holds KC/JS + 9S, 10S. Why, to lay on Next. So I have major objection to your frequency of S2 having JS. Best I can determine JS at S2 36%, S4 36% and rest buried. Big issue is that at S3 I cannot trust when that occurs. Thus, he needs to play the AC, my reasoning. So we differ on the Importance of Scenaroio A.
_______________________
Am just seeing your most recent post. Please correct Scenario C, Play JC to read 1.28 rather than 1.23. Also, weighted average is key (using weights as explained in my previous paragraph, no the raw weights of where the JS is likely to be dealt  perhaps you already did this).
I already did this calculation, and, like you, found it slightly more favorable to play the AC. But really a sucky "barbell" situation: one play is clearly more favorable in one scenario, the other play clearly more favorable in the other scenario. So it comes down to accessing the relative advantage of playing one card versus playing the other card in a given scenario [which is where you currently disagree with my program] and the relative probability of either scenario occurring. [I'm considering "JS buried" as a virtual wash here] I agree![/color]
Like I said, I'll look again at how my program plays the cards. I'm always interested in getting my algorithm to play better!
Thanks again Ray. I am going look more at Scenaro B, both ways.
IRISH
Where are we in agreement? When JS is at the Dealer seat EV will be better playing the JC. Where we differ is what that frequency is at S4 or at S2. So here are my comments below commenting on your Post below.
Ray said:
Irish, I will look at the play of the hands again to see if S3 (or other players) are making mistakes. My understanding is that you do not disagree that, if S4 holds the JS, it is better for the S3 to play the JC on the first trick. . No I disagree, I want to force Dealer to play the Left. Even though I have to still win 3 tricks, I am handcuffed with the Left behind me. Since the Left will win a trick anyway, I am better off to force the Left as his lead most likely will be a Spade or Heart and I can pick up his QC. Yes, I have to get a trick from my partner or get lucky that neither opponent has a Diamond dblt.
BUT, you think that playing the AC is not as unfavorable as my program makes it out to be. Yes, true.
That is not something I have thought too much about. My reasoning (to explain my findings) is simply that, if S3 wins the first trick (with the JC), they need to win just 2 more of the 4 tricks which remain, and one of those is guaranteed since they are the only player with two trump. Leading the 10D, trick 2, will sometimes win (if partner has the boss diamond),True, and this is the same Playing the JC or AC. and will often set up their JD for a trick later. Partner in S2 will also have the AS and/or AH a decent amount of the time  just so many ways of making that necessary 3rd trick. True, playing either JC or AC.
If S3 plays the AC, however, and S4 wins with the JS, S3 must now win 3 of the remaining 4 tricks, which will be tough given they have 2 losing diamonds AND DON'T HAVE THE LEAD. If S4 leads an A on the 2nd trick (or partner in S2 has the A or boss card in that suit), S3 will need to trump in, leaving them with just 1 trump, albeit the boss. S3 will need to lead then trump (2nd trick won) but then a crap diamond card. Many ways to lose the next 2 tricks  I would say even probable they lose them both: euchred! Consider when playing the JC, the Dealer now has the JS and is be hind S3 AC/9C. Dealer will most likely have a Void and can over trump and lead his best. I see this as being more of a disadvantage than playing the AC. When the Dealer has the Left, it will always boil down to S1 winning a trick or if S3 gets lucky with 4 unknown diamonds, AD or KD or QD could be buried. Sometimes hjis JD/10D will win a trick. More likely than playing the JC, because one of the opponents if get forced with Ace/Diamond, can better play S3 for his doubleton diamond, IMO.
As far as the likelihood of the JS lying with S2, S4 or buried, I agree that the raw probabilities are 4/12, 5/12 and 3/12, respectively. [let's convert that to 33, 42 and 25 times out of 100]. BUT, S2 will (or should) call trump almost 20% of the time, in which case the scenario we are studying here doesn't occur. So reduce the S2 occurrences of holding the JS by 18%, or from 33 to 27. The relative frequencies are now 27/94, 42/94 and 25/94: 28% held by S2, 45% held by S4 and 27% buried [rounded figures] This makes the scenario we would hope for (S2 holding the JS, because we can finesse it) a bit less likely. That explains why scenario A gets a bit less importance, and why my probabilities don't jive with what you were expecting. . I think here I differ with your thinking. When will S2 order up the QC? If he holds KC/JS + AH or AD no spade. He should not not when he holds KC/JS + AS, or without an ace, or when he holds KC/JS + 9S, 10S. Why, to lay on Next. So I have major objection to your frequency of S2 having JS. Best I can determine JS at S2 36%, S4 36% and rest buried. Big issue is that at S3 I cannot trust when that occurs. Thus, he needs to play the AC, my reasoning. So we differ on the Importance of Scenaroio A.
_______________________
Am just seeing your most recent post. Please correct Scenario C, Play JC to read 1.28 rather than 1.23. Also, weighted average is key (using weights as explained in my previous paragraph, no the raw weights of where the JS is likely to be dealt  perhaps you already did this).
I already did this calculation, and, like you, found it slightly more favorable to play the AC. But really a sucky "barbell" situation: one play is clearly more favorable in one scenario, the other play clearly more favorable in the other scenario. So it comes down to accessing the relative advantage of playing one card versus playing the other card in a given scenario [which is where you currently disagree with my program] and the relative probability of either scenario occurring. [I'm considering "JS buried" as a virtual wash here] I agree![/color]
Like I said, I'll look again at how my program plays the cards. I'm always interested in getting my algorithm to play better!
Thanks again Ray. I am going look more at Scenaro B, both ways.
IRISH

 Posts: 251
 Joined: Thu Sep 16, 2021 6:56 pm
Okay, I found my error:
1)When S3 plays the AC ...
2) and S4 overtrumps with the L ...
3) and S3 wins the 2nd trick with the 9C ...
S3 should then play the R to draw out S4's remaining trump, then lead a low diamond. I had it stuck in my head that this was usually going to lead to losing the last 2 tricks, so not the way to go  I was leading a low D on that 3rd trick. But when I had S3 lead the R on trick 3, I found that the EV rose from 0.36 (45% euchres) to 0.18 (39% euchres). So in fact, many ways to eke out another trick on the last two, and thus a point for the hand.
I also take your point about S2 perhaps not calling as often as I have simulated, when holding L+1 trump (it will happen with some frequency, but hard to predict, and likely less often than 18%).
So in the end, I agree that it is better to play the AC in this exact scenario (or even if we take the 9, 10, Q and KC to be fungible, and allocate them any which way). Understanding that if S1 leads the KC, S3 can play a low trump (if S2 hasn't proffered the L), and that S2 does indeed play trump.
I will adapt my program to take into account that small range of scenarios [S3 calls trump, R1, with R + A + x of trump; S1 and S2 play trump, 1st trick; S3 to play A of trump rather than R if sufficient to take lead  and of course play low trump if S1 is winning with the K]. But what if S2 shows themselves to be void in trump? Still best to play A trump, 1st trick; but if S3 loses the 1st trick and then trumps in to win 2nd trick, lead R from S3?? S4 may have 3 trump. Or S1 may have the extra trump card. Don't want to give the advantage to S4, or have my partner in S1 waste their trump. From the scenario investigated initially in this thread, IF S4 has L + 1 trump, it's better if S3 plays the R on the first trick and leads back a low D. The difference here is that S4 may have L + 2 trump.
I'd be interested in hearing how you'd play, Irish, to narrow down the scenarios I test.
1)When S3 plays the AC ...
2) and S4 overtrumps with the L ...
3) and S3 wins the 2nd trick with the 9C ...
S3 should then play the R to draw out S4's remaining trump, then lead a low diamond. I had it stuck in my head that this was usually going to lead to losing the last 2 tricks, so not the way to go  I was leading a low D on that 3rd trick. But when I had S3 lead the R on trick 3, I found that the EV rose from 0.36 (45% euchres) to 0.18 (39% euchres). So in fact, many ways to eke out another trick on the last two, and thus a point for the hand.
I also take your point about S2 perhaps not calling as often as I have simulated, when holding L+1 trump (it will happen with some frequency, but hard to predict, and likely less often than 18%).
So in the end, I agree that it is better to play the AC in this exact scenario (or even if we take the 9, 10, Q and KC to be fungible, and allocate them any which way). Understanding that if S1 leads the KC, S3 can play a low trump (if S2 hasn't proffered the L), and that S2 does indeed play trump.
I will adapt my program to take into account that small range of scenarios [S3 calls trump, R1, with R + A + x of trump; S1 and S2 play trump, 1st trick; S3 to play A of trump rather than R if sufficient to take lead  and of course play low trump if S1 is winning with the K]. But what if S2 shows themselves to be void in trump? Still best to play A trump, 1st trick; but if S3 loses the 1st trick and then trumps in to win 2nd trick, lead R from S3?? S4 may have 3 trump. Or S1 may have the extra trump card. Don't want to give the advantage to S4, or have my partner in S1 waste their trump. From the scenario investigated initially in this thread, IF S4 has L + 1 trump, it's better if S3 plays the R on the first trick and leads back a low D. The difference here is that S4 may have L + 2 trump.
I'd be interested in hearing how you'd play, Irish, to narrow down the scenarios I test.

 Posts: 1291
 Joined: Tue Apr 24, 2018 9:33 pm
RAY,
We probably have bored all the followers on this hand  ANALYSIS TO PARALYSIS! LOL
COOL, glad you found an issue and now we are in agreement on Scenario B Playing the AC. Yes S3 must lead the JC on trick 3 if he wins the trick with 9C. Then pray for Help.
I was still stuck on why Scenario B was so negative for Playing AC and the magnitude of the difference between the two. As far as Scenario B Playing the JC. I think that EV is too high. To get your EV for JC that means the Euchre rate is 25% and too low So I think your Simulator not playing as I would play. Any time an opponent at S2 has AD dblt should result in a euchre. Any time AD buried, and either opponent holds KD/QD results in a euchre. Or when one has AD/KD, AD/QD, those are also all euchres. I think leading the JC euchre rate for S3 is about 6 to 8% lower than playing the AC when S4 has Left guarded. Still when S4 has JS/QC the EV will be positive and greater than playing the AC. I did another 50 hands adding to the 100 I did. When I do this I get to see many things and patterns as well as how I should be playing. With Combinatorics, I can project the correct frequencies to zero in on the correct combinations and approximate EVs. It's a hobby and as much a Rubik Cube. Nothing is perfect. Did you know that Euchre has its probabilities, and combinations but what Players do or don't do in playing their hand FAR exceeds what the cards display. That is a whole added dimension. It's fluid and dynamic.
As far as your simulator, I have no problem helping in looking best Practices.
Peace bro!
IRISH
====================
Okay, I found my error:
1)When S3 plays the AC ...
2) and S4 overtrumps with the L ...
3) and S3 wins the 2nd trick with the 9C ...
S3 should then play the R to draw out S4's remaining trump, then lead a low diamond. I had it stuck in my head that this was usually going to lead to losing the last 2 tricks, so not the way to go  I was leading a low D on that 3rd trick. But when I had S3 lead the R on trick 3, I found that the EV rose from 0.36 (45% euchres) to 0.18 (39% euchres). So in fact, many ways to eke out another trick on the last two, and thus a point for the hand.
We probably have bored all the followers on this hand  ANALYSIS TO PARALYSIS! LOL
COOL, glad you found an issue and now we are in agreement on Scenario B Playing the AC. Yes S3 must lead the JC on trick 3 if he wins the trick with 9C. Then pray for Help.
I was still stuck on why Scenario B was so negative for Playing AC and the magnitude of the difference between the two. As far as Scenario B Playing the JC. I think that EV is too high. To get your EV for JC that means the Euchre rate is 25% and too low So I think your Simulator not playing as I would play. Any time an opponent at S2 has AD dblt should result in a euchre. Any time AD buried, and either opponent holds KD/QD results in a euchre. Or when one has AD/KD, AD/QD, those are also all euchres. I think leading the JC euchre rate for S3 is about 6 to 8% lower than playing the AC when S4 has Left guarded. Still when S4 has JS/QC the EV will be positive and greater than playing the AC. I did another 50 hands adding to the 100 I did. When I do this I get to see many things and patterns as well as how I should be playing. With Combinatorics, I can project the correct frequencies to zero in on the correct combinations and approximate EVs. It's a hobby and as much a Rubik Cube. Nothing is perfect. Did you know that Euchre has its probabilities, and combinations but what Players do or don't do in playing their hand FAR exceeds what the cards display. That is a whole added dimension. It's fluid and dynamic.
As far as your simulator, I have no problem helping in looking best Practices.
Peace bro!
IRISH
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Okay, I found my error:
1)When S3 plays the AC ...
2) and S4 overtrumps with the L ...
3) and S3 wins the 2nd trick with the 9C ...
S3 should then play the R to draw out S4's remaining trump, then lead a low diamond. I had it stuck in my head that this was usually going to lead to losing the last 2 tricks, so not the way to go  I was leading a low D on that 3rd trick. But when I had S3 lead the R on trick 3, I found that the EV rose from 0.36 (45% euchres) to 0.18 (39% euchres). So in fact, many ways to eke out another trick on the last two, and thus a point for the hand.

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 Joined: Tue Apr 24, 2018 9:33 pm
Ray, I final comment on Scenario B. I indicated playing JC to Eldest's lead, you got EV = +0.24, is too high. I will toss out this idea why I think that number should be lower as played by your Simulator. When Dealer (S4) holds this type of hand JS QC QH 9D QD, or JS QC 9D KD (any singleton off suit less than a King) it should never lead the QD or KD. Always the singleton or the 9D. And the 9D is always preferred. Why, you are trying to catch either a doubleton Diamond and it is the same for a Singleton Diamond. The low Diamond is a better lead in that if S2 or S1 has the AD it gives you a preference to Slough the Singleton, trump or over play S3.
I am not sure what your Simulator does but you might want to verify. The combination of this 9D/QD OR 9D/KD can possibly account for the differences between what you and I are getting. Approximately on the magnitude of 7 to 14% more euchres playing the JC and leading to trick 2. I think the Euchre rate should be higher than your _26% to get EV = +0.24.
But the play at S1 is different, IMO, always cover the dealer's lead with Queen but not with the King. Why, because S2 & S3 still have to play and the Ace most likely will be played. And the Dealer is most likely trying to set up his higher card but not true if he had the Ace. This only applies on R1. Especially when the Dealer is the maker, not ordered.
You might want to check it out as it could have implications on past and future EVs, simulatons.
IRISH
RAY said, " I found that the EV rose from 0.36 (45% euchres) to 0.18 (39% euchres). So in fact, many ways to eke out another trick on the last two, and thus a point for the hand." I think we are in close agreement, my data indicates 36 to 39%. Close enough!
SCENARIO B: +0.24 0.36
I am not sure what your Simulator does but you might want to verify. The combination of this 9D/QD OR 9D/KD can possibly account for the differences between what you and I are getting. Approximately on the magnitude of 7 to 14% more euchres playing the JC and leading to trick 2. I think the Euchre rate should be higher than your _26% to get EV = +0.24.
But the play at S1 is different, IMO, always cover the dealer's lead with Queen but not with the King. Why, because S2 & S3 still have to play and the Ace most likely will be played. And the Dealer is most likely trying to set up his higher card but not true if he had the Ace. This only applies on R1. Especially when the Dealer is the maker, not ordered.
You might want to check it out as it could have implications on past and future EVs, simulatons.
IRISH
RAY said, " I found that the EV rose from 0.36 (45% euchres) to 0.18 (39% euchres). So in fact, many ways to eke out another trick on the last two, and thus a point for the hand." I think we are in close agreement, my data indicates 36 to 39%. Close enough!
SCENARIO B: +0.24 0.36