Irish, you had asked me to simulate a hand which you encountered in live play, but I think your original post got deleted when the forum was taken offline for a few weeks. I had done some investigating and written a response (which I saved in another document), so I present it to you now.
___________________________
I first had a look at the original hand.
S4 has AKS + A9H + AD (QS)
100,000 tested, ~77,000 get to S4 for bidding
Call alone: (7,690 / 64,689 / 4,583) [4 pts. / 1 pt. / euchred] EV = +1.12
Bid wp: (10,877 / 60,770 / 5,315) EV = +0.94
My program currently has S4 bidding alone, which appears to be the correct call.
Note that the successful loner rate is only 10%  why? The odds of both bowers residing with either partner in S2 or in the kitty are 8/18 x 7/17 = 18% (likely where the mathematician got their number). This represents the upper limit on successful lone calls, but they should almost all be realized. That would indeed be the statistic IF S4 got to bid on ALL the hands. But as it stands, 23% of the hands are bid by seats 1, 2 and 3 – in fact, 18.6% by S2. And those are specifically hands where S2 has one or both bowers*. Which messes up the overall statistics, to the point that in the remaining 73% of hands, one or both bowers reside with S1 or S3 90% of the time.
As an aside, I experimented with forcing all other seats to pass, so that S4 bid every hand alone. The loner success rate was 16.1%, which seems reasonable (S3 will be void in D 15% or the time, will be void in H 23% of the time; S1 leading to S3’s void AND S3 having a trump 1.9% of the time isn’t so far fetched).
[* S2 will bid with R+1 trump, or with L+1 trump if a) R not turned or, b) 2 or 3suited and has no good bid R2; and also with any 3+ trump]
Not sure where you get your 26% loner success rate, Irish.
Curiously, in the above test, the euchre rate was greater when S4 took partner along! This is known to be the case in certain situations, where one player needs to control the hand throughout (not have partner trump in and then lead the wrong card), so I assume this is just one of those cases.
Other initial hands would also result in S4 having the same hand, so I also tested:
S4 has KQS + A9H + AD (AS): same result
S4 has AKS + AH + AD + 9C (QS): slightly better results (understandable) but no change in what is the better call.
I then considered the slightly different hand:
S4 has AKS + AKH + 9C (QS)
A similar (very slightly lower) percentage of hands make it to S4 for a bid.
S4 bids alone: (7,864 / 53,430 / 15,063) EV = +0.70
S4 bids wp: (10,900 / 55,516 / 9,441) EV = +0.77
Here, it’s better for S4 to bid wp. Curious.
I tried the scenario twice more with slightly different dealt hands:
S4 has AKS + AK9H (QS): lone EV = +0.47; wp EV = +0.69
S4 has AKS + AKH + 9D (QS): lone EV = +0.69; wp EV = +0.78
What appears to be the issue here is the high euchre rate when S4 calls alone. I imagine S4 needs to trump any H/D lead; leads trump and opponents win with a bower; need to trump the 3rd trick with my last trump, and it’s often curtains from there. The euchre rate is not very high, but is marginally causing problems.
I then looked at the classic case of having AK + A in green suits + 3 trump, and having to decide what to discard: the K (leaving yourself 3suited but with 2 Aces) or the lone A (leaving yourself 2suited but with top 2 in the nontrump suit.) My understanding is that the conventional move is to discard the lone A.
S4 has AKS + AKH + AD (QS); discards AD.
Bid alone: (7,445 / 53,763 / 15,692) EV = +0.68
Bid wp: (11,003 / 60,603 / 5,294) EV = +0.94
Same hand, but S4 discards KH.
Bid alone: (7,382 / 65,033 / 4,683) EV = +1.11
Bid wp: (10,787 / 60,893 / 5,418) EV = +0.93
Looks like it’s better to keep the 2 aces! Once again, it’s the euchre rate that comes into play: keep the AKH, and you’ll probably end up having to trump a D lead, using one of your precious 3 trump and tilting the advantage to taking partner along. And less chance (with this particular hand) of drawing out all the other trump to make your AKH sure winners.
What if you have a better hand?
S4 has AKS + AKH + AD (JS)
Discard AD: lone EV = +2.58; wp EV = +1.56
Discard KH: lone EV = +2.67; wp EV = +1.56
Still slightly better to discard the KH (but definitely go alone)
A really good hand:
S4 has AS + AKH + AD + JC (JS)
Discard KH, go alone: EV = +3.55
Discard AD, go alone: EV = +3.60
[exception to the rule, and not worth memorizing]
A weaker hand:
910S + AKH + AD (QS) [must bid this hand, wp]:
Discard KH: EV = +0.65
Discard AD: EV = +0.36
I tested a few more hands, and resolved to the following rule:
If you have 3 trump (including a bower) + AK green suit (as dealer, after discard*), go alone.
If you don’t have that bower, take partner along.
[*note that in the AK + A scenario, both green suits, the K should be discarded, defying conventional wisdom. In all cases, except the super power hand, where the difference is marginal].
What if the AK combo (offsuit) is in next?
S4 has AKS + AKC + 9D (QS)
Bid alone: EV = +0.56
Bid wp: EV = +0.72
S4 has AKS + AKC + AD (QS):
Discard KH: lone EV = +1.04; wp EV = +0.85
Discard AD: lone EV = +0.52; wp EV = +0.66
It still looks better to bid alone, with the caveat that you must keep the 2 Aces.
I imagine there will be some backlash to these findings. And I accept that my simulator is not perfect, and therefore its results cannot be taken as gospel truth. Additionally, it is fundamentally based on the belief that all players are playing optimally, which is obviously not the case.
Nonetheless, I think its usefulness resides in the fact that it is nearly impossible to calculate CONDITIONAL odds in one’s head (though I admit one may have an intuitive feel). In the initial example, the calculated odds gave one result, but the CONDITIONAL odds gave a very different one. I stand by those, while realizing live game play incorporates “reading” your opponents’ proclivities.
I hope the theoretical results prove useful.
Bid alone or with partner?

 Posts: 1321
 Joined: Tue Apr 24, 2018 9:33 pm
Hey Ray,
I disagree with your calculation here about bowers: Note that the successful loner rate is only 10%  why? The odds of both bowers residing with either partner in S2 or in the kitty are 8/18 x 7/17 = 18% INCORRECT, IMO
Here is how I arrived at the answer of 26% opponents do NOT have either bower:
2C0 16C5 18C5 = 4368/8568 = 50.98% NEITHER BOWER FOR ONE OPPONENT AND THIS HAS TO OCCUR WITH BOTH OPPONENTS.
50.98% X 50.98% = 25.9898%
But that is not all which I did not do initially:
I did not go on to factor in S1 leading to S3 void. Has to be to either Ace. Okay so let's do that now. Suppose either ace is a Green ace. So S1 has to have that suit to lead but he could also have one, two, three or four (or even five) of that particular suit. And simultaneously S3 must be void and have at least one trump. But also consider that about 33% of the time is likely to lead to the Dealer void as well. That number is around 12%.
So 26% minus 10%  I project the loner rate will be 12%  14$ SUCCESSFUL LONERS. Depends on lead H vs D.About a 2% difference.
Reason so variable is how many of a suit S1 but also consider that he will be saving a doubleton and probably leading a Singleton King or Queen first. And will he lead a H or D? So considering that 76% x 23.4% = 18%. But have to adjust for wrong lead 33% to S3 void. So 18% x 33% = 6% so the number should be 12%.
So 26%  12% = 14% successful loners by my projection. So we are in the range of variability, I guess.
Irish
I disagree with your calculation here about bowers: Note that the successful loner rate is only 10%  why? The odds of both bowers residing with either partner in S2 or in the kitty are 8/18 x 7/17 = 18% INCORRECT, IMO
Here is how I arrived at the answer of 26% opponents do NOT have either bower:
2C0 16C5 18C5 = 4368/8568 = 50.98% NEITHER BOWER FOR ONE OPPONENT AND THIS HAS TO OCCUR WITH BOTH OPPONENTS.
50.98% X 50.98% = 25.9898%
But that is not all which I did not do initially:
I did not go on to factor in S1 leading to S3 void. Has to be to either Ace. Okay so let's do that now. Suppose either ace is a Green ace. So S1 has to have that suit to lead but he could also have one, two, three or four (or even five) of that particular suit. And simultaneously S3 must be void and have at least one trump. But also consider that about 33% of the time is likely to lead to the Dealer void as well. That number is around 12%.
So 26% minus 10%  I project the loner rate will be 12%  14$ SUCCESSFUL LONERS. Depends on lead H vs D.About a 2% difference.
Reason so variable is how many of a suit S1 but also consider that he will be saving a doubleton and probably leading a Singleton King or Queen first. And will he lead a H or D? So considering that 76% x 23.4% = 18%. But have to adjust for wrong lead 33% to S3 void. So 18% x 33% = 6% so the number should be 12%.
So 26%  12% = 14% successful loners by my projection. So we are in the range of variability, I guess.
Irish

 Posts: 1321
 Joined: Tue Apr 24, 2018 9:33 pm
RAY,
This (see below) is what I want to comment on. I am in agreement with keeping the two Aces without having both bowers when going alone. Why, because with a weaker hand Left or Ace high, (A K Q A A vs A K Q AK)the dealer needs to keep both Aces as it 'extends' the strength of your hand by him not having to use a trump. Thus the Euchre ate will be lower, and loner rate should be a little higher too. Or with Right without the Left  Keep both Aces as you have a stronger hand not to have to use a trump. The loner reate will be higher. So I agree with you here. Shouldn't be much if any Push Back!
Once again, it’s the euchre rate that comes into play: keep the AKH, and you’ll probably end up having to trump a D lead, using one of your precious [3] trump . . . I totally agree. The only two hands I disagree with are the really two strong hands: J J A and J J K. So I tend to disagree here that it is tie! I say discard an Ace singleton only with those two powerful hands. WHY? With J J A or J J K You give S1 & S3 one shot (suit), 33% to hit your suit vs 66% when holding two Aces the chance of spoiling is slightly higher. There is a slight advantage. Funny the simulator did not find that?
But interesting! OMG, we are agreeing more and more!
Peace Bro!
IRISH

Ray said, I tested a few more hands, and resolved to the following rule:
If you have 3 trump (including a bower) + AK green suit (as dealer, after discard*), go alone.
If you don’t have that bower, take partner along.
[*note that in the AK + A scenario, both green suits, the K should be discarded, defying conventional wisdom. In all cases, except the super power hand, where the difference is marginal].
What if the AK combo (offsuit) is in next?
S4 has AKS + AKC + 9D (QS)
Bid alone: EV = +0.56
Bid wp: EV = +0.72
S4 has AKS + AKC + AD (QS):
Discard KH: lone EV = +1.04; wp EV = +0.85
Discard AD: lone EV = +0.52; wp EV = +0.66
It still looks better to bid alone, with the caveat that you must keep the 2 Aces.
I imagine there will be some backlash to these findings. And I accept that my simulator is not perfect, and therefore its results cannot be taken as gospel truth. Additionally, it is fundamentally based on the belief that all players are playing optimally, which is obviously not the case.
This (see below) is what I want to comment on. I am in agreement with keeping the two Aces without having both bowers when going alone. Why, because with a weaker hand Left or Ace high, (A K Q A A vs A K Q AK)the dealer needs to keep both Aces as it 'extends' the strength of your hand by him not having to use a trump. Thus the Euchre ate will be lower, and loner rate should be a little higher too. Or with Right without the Left  Keep both Aces as you have a stronger hand not to have to use a trump. The loner reate will be higher. So I agree with you here. Shouldn't be much if any Push Back!
Once again, it’s the euchre rate that comes into play: keep the AKH, and you’ll probably end up having to trump a D lead, using one of your precious [3] trump . . . I totally agree. The only two hands I disagree with are the really two strong hands: J J A and J J K. So I tend to disagree here that it is tie! I say discard an Ace singleton only with those two powerful hands. WHY? With J J A or J J K You give S1 & S3 one shot (suit), 33% to hit your suit vs 66% when holding two Aces the chance of spoiling is slightly higher. There is a slight advantage. Funny the simulator did not find that?
But interesting! OMG, we are agreeing more and more!
Peace Bro!
IRISH

Ray said, I tested a few more hands, and resolved to the following rule:
If you have 3 trump (including a bower) + AK green suit (as dealer, after discard*), go alone.
If you don’t have that bower, take partner along.
[*note that in the AK + A scenario, both green suits, the K should be discarded, defying conventional wisdom. In all cases, except the super power hand, where the difference is marginal].
What if the AK combo (offsuit) is in next?
S4 has AKS + AKC + 9D (QS)
Bid alone: EV = +0.56
Bid wp: EV = +0.72
S4 has AKS + AKC + AD (QS):
Discard KH: lone EV = +1.04; wp EV = +0.85
Discard AD: lone EV = +0.52; wp EV = +0.66
It still looks better to bid alone, with the caveat that you must keep the 2 Aces.
I imagine there will be some backlash to these findings. And I accept that my simulator is not perfect, and therefore its results cannot be taken as gospel truth. Additionally, it is fundamentally based on the belief that all players are playing optimally, which is obviously not the case.
Last edited by irishwolf on Fri Jul 29, 2022 10:31 am, edited 1 time in total.

 Posts: 1321
 Joined: Tue Apr 24, 2018 9:33 pm
As to going alone vs with pard, all DEPENDS on the score!
Your results and conclusion is good when score is tied or with in one point. However, say you are down 4 to 8 or similar wide spread losing. You have no choice, IMO, to go alone. You are going to lose so you have to Swing for the fence. Score dictates a lot as you are playing THIS Game not 10,000. This is the game that counts not the next 9,999. Just wanted to put that in perspective, and that applies generally speaking.
IRISH
I then considered the slightly different hand:
S4 has AKS + AKH + 9C (QS)
A similar (very slightly lower) percentage of hands make it to S4 for a bid.
S4 bids alone: (7,864 / 53,430 / 15,063) EV = +0.70
S4 bids wp: (10,900 / 55,516 / 9,441) EV = +0.77
Here, it’s better for S4 to bid wp. Curious.
Your results and conclusion is good when score is tied or with in one point. However, say you are down 4 to 8 or similar wide spread losing. You have no choice, IMO, to go alone. You are going to lose so you have to Swing for the fence. Score dictates a lot as you are playing THIS Game not 10,000. This is the game that counts not the next 9,999. Just wanted to put that in perspective, and that applies generally speaking.
IRISH
I then considered the slightly different hand:
S4 has AKS + AKH + 9C (QS)
A similar (very slightly lower) percentage of hands make it to S4 for a bid.
S4 bids alone: (7,864 / 53,430 / 15,063) EV = +0.70
S4 bids wp: (10,900 / 55,516 / 9,441) EV = +0.77
Here, it’s better for S4 to bid wp. Curious.

 Posts: 260
 Joined: Thu Sep 16, 2021 6:56 pm
I posted a reply but it didn't seem to go through. I'll try again.
Your probability calculation is faulty, Irish. With 18 unknown cards, the odds that a given player (say, S1, who has 5 cards) doesn't have either of 2 specific cards (in the case, the two bowers) is indeed approx 51%. But when we stipulate that second player (say, S3, also with 5 cards) ALSO doesn't have either bower, those conditional odds are lower. Since we already established that S1 has no bowers, there are only 13 cards left, including both bowers. The odds that S3 doesn't either in his hand is about 36% (I'll leave the exact calculation to you).
Another way to do the calculation would be to combine S1 and S3 and simply say that those 10 cards do not include either bower. A third way is what I did, which was to say that the OTHER 8 cards (in S2's hand and the kitty) contained BOTH bowers (that's the 8/18 x 7/17).
_____________________________
And a comment on a different topic...
S4 has JS + AKH + AD + JC (AS turned). Dealer picks up and goes alone  discard KH or AD?
Irish, you correctly note that keeping the two red aces gives S1/S3 approx. twice the chance of trumping one of your aces on the first trick (which argues for keeping the AK instead). But there is also some chance that one of the opponents has 3 trump, and if S4 is forced to trump in to win the first trick, the opponents will have a stopper. S4 will be approx. twice as likely to need to trump the first trick if they keep the AK than if they keep both aces (which argues for keeping both aces instead).
These are both lowlikelihood yet important scenarios; which is actually more likely? I am going to run a monte carlo simulation and simply tally the instances  it's simply too easy to mess up probability calculations (I mess them up plenty myself). I'll let you know what I find.
I'll run the other simulations next week as well.
Your probability calculation is faulty, Irish. With 18 unknown cards, the odds that a given player (say, S1, who has 5 cards) doesn't have either of 2 specific cards (in the case, the two bowers) is indeed approx 51%. But when we stipulate that second player (say, S3, also with 5 cards) ALSO doesn't have either bower, those conditional odds are lower. Since we already established that S1 has no bowers, there are only 13 cards left, including both bowers. The odds that S3 doesn't either in his hand is about 36% (I'll leave the exact calculation to you).
Another way to do the calculation would be to combine S1 and S3 and simply say that those 10 cards do not include either bower. A third way is what I did, which was to say that the OTHER 8 cards (in S2's hand and the kitty) contained BOTH bowers (that's the 8/18 x 7/17).
_____________________________
And a comment on a different topic...
S4 has JS + AKH + AD + JC (AS turned). Dealer picks up and goes alone  discard KH or AD?
Irish, you correctly note that keeping the two red aces gives S1/S3 approx. twice the chance of trumping one of your aces on the first trick (which argues for keeping the AK instead). But there is also some chance that one of the opponents has 3 trump, and if S4 is forced to trump in to win the first trick, the opponents will have a stopper. S4 will be approx. twice as likely to need to trump the first trick if they keep the AK than if they keep both aces (which argues for keeping both aces instead).
These are both lowlikelihood yet important scenarios; which is actually more likely? I am going to run a monte carlo simulation and simply tally the instances  it's simply too easy to mess up probability calculations (I mess them up plenty myself). I'll let you know what I find.
I'll run the other simulations next week as well.

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 Joined: Tue Apr 24, 2018 9:33 pm
RAY, RAY, RAY: No Sir, I am Right on this 26% over Thousands of hands dealt which is what your simulator is doing. So Read on!
Of course what you are saying is true FOR ONE HAND DEALT. But this is a statistical calculation for many (HUNDREDS 400 FOR A 95% PROBABILITY that it is indeed 26% that neither will have a bower.)
What you are saying here is incorrect Ray. Why, because you are giving an example of ONE hand sample as dealt which is true I agree. But you have to look at a minimum of 400 hands 18 cards dealt to the three players simultaneously which will prove that I am correct: "With 18 unknown cards, the odds that a given player (say, S1, who has 5 cards) doesn't have either of 2 specific cards (in the case, the two bowers) is indeed approx 51%. But when we stipulate that second player (say, S3, also with 5 cards) ALSO doesn't have either bower, those conditional odds are lower. NOT FOR MANY HANDS DEALT ONLY TRUE FOR ONE HAND SAMPLE!
"Since we already established that S1 has no bowers, ONE HAND NOT FOR HUNDREDS OF HANDS there are only 13 cards left, ONLY TRUE FOR ONE HAND OR AS YOU LOOK AT A PARTICULAR HAND. including both bowers. The odds that S3 doesn't either in his hand is about 36% YES FOR THAT ONE HAND ONLY, NOT THOUSANDS AS DEALT (I'll leave the exact calculation to you)."
WRONG  BECAUSE YOU ARE TAKING ONE HAND, ONE DEAL! I am saying true for statistical sample of thousands of hands which is what your simulator is doing.
I conclude with this: When you look at the JH 18 cards the statistics 55.6% either S1 or S3 has the JH and 44.4% to Talon/S2. Those statistics are not going to change for THE LAW OF LARGE NUMBERS. Of course if you deal 5 to just one then 5 another it matters how you are doing that.
IRISH
Of course what you are saying is true FOR ONE HAND DEALT. But this is a statistical calculation for many (HUNDREDS 400 FOR A 95% PROBABILITY that it is indeed 26% that neither will have a bower.)
What you are saying here is incorrect Ray. Why, because you are giving an example of ONE hand sample as dealt which is true I agree. But you have to look at a minimum of 400 hands 18 cards dealt to the three players simultaneously which will prove that I am correct: "With 18 unknown cards, the odds that a given player (say, S1, who has 5 cards) doesn't have either of 2 specific cards (in the case, the two bowers) is indeed approx 51%. But when we stipulate that second player (say, S3, also with 5 cards) ALSO doesn't have either bower, those conditional odds are lower. NOT FOR MANY HANDS DEALT ONLY TRUE FOR ONE HAND SAMPLE!
"Since we already established that S1 has no bowers, ONE HAND NOT FOR HUNDREDS OF HANDS there are only 13 cards left, ONLY TRUE FOR ONE HAND OR AS YOU LOOK AT A PARTICULAR HAND. including both bowers. The odds that S3 doesn't either in his hand is about 36% YES FOR THAT ONE HAND ONLY, NOT THOUSANDS AS DEALT (I'll leave the exact calculation to you)."
WRONG  BECAUSE YOU ARE TAKING ONE HAND, ONE DEAL! I am saying true for statistical sample of thousands of hands which is what your simulator is doing.
I conclude with this: When you look at the JH 18 cards the statistics 55.6% either S1 or S3 has the JH and 44.4% to Talon/S2. Those statistics are not going to change for THE LAW OF LARGE NUMBERS. Of course if you deal 5 to just one then 5 another it matters how you are doing that.
IRISH

 Posts: 1321
 Joined: Tue Apr 24, 2018 9:33 pm
RAY,
So you do not believe me. So where is what you can do to PROVE to yourself about this issue.
Give the Dealer this hand AS KS 10S 9S 9H with upcard QS. With five (5) trumps to S4 that leaves 2 bowers and accounts for all the trump. 18 unknown cards of which there is the JS & JC as random + 16 other suit cards.
Run tests of 430, 10,000 and 100,000 as three tests. The 430 will tell you if my 95% is correct to account for when S2 orders with two bowers to get approximately 400 hands for S4 going alone.
Okay  S1 & S3 should never order but that will validate your program to see if they do as they will get euchred each time.
S2 will order with JS & JC combo about 6.5% and we will know that number is part of the 26%, thus will reduce S4 loner rate. Just have to factor that in.
Now the loners for S4 will be the 26% minus what hands S2 ordered. 26% is in total of all hand samples. That will tell us if your or myself is correct. It will also tell you if your Dealer is truly randomly dealing the cards as well.
Not Rocket science but I think this is a critical test as you have called my integrity into question on the 26%.
IRISH
So you do not believe me. So where is what you can do to PROVE to yourself about this issue.
Give the Dealer this hand AS KS 10S 9S 9H with upcard QS. With five (5) trumps to S4 that leaves 2 bowers and accounts for all the trump. 18 unknown cards of which there is the JS & JC as random + 16 other suit cards.
Run tests of 430, 10,000 and 100,000 as three tests. The 430 will tell you if my 95% is correct to account for when S2 orders with two bowers to get approximately 400 hands for S4 going alone.
Okay  S1 & S3 should never order but that will validate your program to see if they do as they will get euchred each time.
S2 will order with JS & JC combo about 6.5% and we will know that number is part of the 26%, thus will reduce S4 loner rate. Just have to factor that in.
Now the loners for S4 will be the 26% minus what hands S2 ordered. 26% is in total of all hand samples. That will tell us if your or myself is correct. It will also tell you if your Dealer is truly randomly dealing the cards as well.
Not Rocket science but I think this is a critical test as you have called my integrity into question on the 26%.
IRISH

 Posts: 260
 Joined: Thu Sep 16, 2021 6:56 pm
I ran the simulation with the original hand (the exact cards that S4 holds not important, as long as both bowers are still unaccounted for). I simply checked if S1 or S3 had any bowers, and tallied the hands where one or the other (or both) had no bowers. Also irrelevant who bids  this is a simple probability distribution test.
400 hands: S1 has no bower 177 hands; S3 has no bower 218 hands; neither has a bower 69 hands
10,000 hands: S1 has no bower 5,050 hands; S3 has no bower 5,104 hands; either has a bower 1,800 hands
100,000 hands: S1 has no bower 51,005 hands; S3 has no bower 50,962 hands; neither has a bower 18,333 hands
400 hands: S1 has no bower 177 hands; S3 has no bower 218 hands; neither has a bower 69 hands
10,000 hands: S1 has no bower 5,050 hands; S3 has no bower 5,104 hands; either has a bower 1,800 hands
100,000 hands: S1 has no bower 51,005 hands; S3 has no bower 50,962 hands; neither has a bower 18,333 hands

 Posts: 1321
 Joined: Tue Apr 24, 2018 9:33 pm
You did not run the test I suggested.
10,000 hands: S1 has no bower 5,050 hands; S3 has no bower 5,104 hands = 50.77% not bower and my calculation as 51%.
100,000 hands: S1 has no bower 51,005 hands; S3 has no bower 50,962 hands vs 51% Expected.
Something is questionable ?? here with 177 / 400?
400 hands look at the difference of a random dealing 218 hands to S3 & 177 to S1. 177 / 400 = 44.25 below 95% of 51%. Sometime is wrong there. Still 54.5 + 44.25 = 49.375% is within 95% of 51%.
10,000 hands: S1 has no bower 5,050 hands; S3 has no bower 5,104 hands = 50.77% not bower and my calculation as 51%.
100,000 hands: S1 has no bower 51,005 hands; S3 has no bower 50,962 hands vs 51% Expected.
Something is questionable ?? here with 177 / 400?
400 hands look at the difference of a random dealing 218 hands to S3 & 177 to S1. 177 / 400 = 44.25 below 95% of 51%. Sometime is wrong there. Still 54.5 + 44.25 = 49.375% is within 95% of 51%.

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 Joined: Thu Sep 16, 2021 6:56 pm
Indeed, when looking at 10,000 or 100,000 sample hands, the observed % of hands where S1 or S3 has no bower aligns quite well with the calculated % of 51.0% Let me also add that the observed % of hands where neither S1 nor S3 has a bower also aligns quite well with the calculated % of 18.3%
When looking at only 400 hands, the sample size is rather small and the results will be more volatile. I ran 3 more simulations of 400 hands and found:
194 / 205 / 65 [S1 no bowers / S3 no bowers / neither has bowers]
206 / 201 / 68
198 / 206 / 68
So it appears that the first simulation I ran was a real outlier, which can happen. That's why it's best to use a larger sample size of hands. I ran the 100,000 sample hand size again and found:
51,131 / 50,792 / 18,241
Once again, consistent with expectations.
__________________________________________
I ran the simulation again with the following hand:
S4 has AK109S + 9S (QS turned)
I ran 100,000 hands and here are the stats:
S2 called 6,457 hands (indeed, every time they had both bowers, and consistent with the 6.5% of times we would expect this to occur)
S1 never called
S3 called 552 hands (with both bowers and some other good cards; I have to dig deeper to tell you exactly what. I'll also have to confirm that it is reasonable for S3 to call in these situations  they clearly won't face a S4 with 5 trump very often!)
S4 called the rest of the hands alone (92,991)
When S3 called they were euchred, obviously.
When S4 called:
S1 no bowers 43,774
S3 no bowers 44,409
S1 and S3 both no bowers 11,467
S4 loner success rate = 11,467/92,991 = 12.3% (successful loner whenever neither opponent held a bower, obviously)
When S2 called, the stats for S1 no bowers, S3 no bowers and S1+S3 no bowers were identical: 6,457 (S2 had both bowers in these situations, so this just confirms my counting was correct).
Combining these 3 cases yields:
S1 no bowers 50,783
S2 no bowers 50,866
both no bowers 17,924
Again, consistent with predictions
I ran the simulation again, forcing S2 and S3 to pass, so S4 went alone every hand:
S1 no bowers 50,957
S3 no bowers 50,981
both no bowers 18,430
loner success rate = 18,430/100,000 = 18.43%
Again, no surprises
Irish, I don't know what else I can do.
When looking at only 400 hands, the sample size is rather small and the results will be more volatile. I ran 3 more simulations of 400 hands and found:
194 / 205 / 65 [S1 no bowers / S3 no bowers / neither has bowers]
206 / 201 / 68
198 / 206 / 68
So it appears that the first simulation I ran was a real outlier, which can happen. That's why it's best to use a larger sample size of hands. I ran the 100,000 sample hand size again and found:
51,131 / 50,792 / 18,241
Once again, consistent with expectations.
__________________________________________
I ran the simulation again with the following hand:
S4 has AK109S + 9S (QS turned)
I ran 100,000 hands and here are the stats:
S2 called 6,457 hands (indeed, every time they had both bowers, and consistent with the 6.5% of times we would expect this to occur)
S1 never called
S3 called 552 hands (with both bowers and some other good cards; I have to dig deeper to tell you exactly what. I'll also have to confirm that it is reasonable for S3 to call in these situations  they clearly won't face a S4 with 5 trump very often!)
S4 called the rest of the hands alone (92,991)
When S3 called they were euchred, obviously.
When S4 called:
S1 no bowers 43,774
S3 no bowers 44,409
S1 and S3 both no bowers 11,467
S4 loner success rate = 11,467/92,991 = 12.3% (successful loner whenever neither opponent held a bower, obviously)
When S2 called, the stats for S1 no bowers, S3 no bowers and S1+S3 no bowers were identical: 6,457 (S2 had both bowers in these situations, so this just confirms my counting was correct).
Combining these 3 cases yields:
S1 no bowers 50,783
S2 no bowers 50,866
both no bowers 17,924
Again, consistent with predictions
I ran the simulation again, forcing S2 and S3 to pass, so S4 went alone every hand:
S1 no bowers 50,957
S3 no bowers 50,981
both no bowers 18,430
loner success rate = 18,430/100,000 = 18.43%
Again, no surprises
Irish, I don't know what else I can do.

 Posts: 1321
 Joined: Tue Apr 24, 2018 9:33 pm
RAY  "I don't know what else I can do."
Not a thing, very convincing.
IRISH
Not a thing, very convincing.
IRISH

 Posts: 260
 Joined: Thu Sep 16, 2021 6:56 pm
Circling back to a couple of other comments I made:
1) when S3 has both bowers and no other trump, they need at least 2 offsuit aces to call trump [how my simulator is coded]. That did indeed occur a few times in the simulation I did for Irish, with S4 having 5 trump, resulting in S3 getting euchred. But I think that is clearly an outlier case, and it is otherwise a good call. Very good chance of making a point (or even 2), and a significant chance that S4 passes, with S2 getting a chance to bid a green suit, R2.
2) S4 is dealt AS + AKH + AD + JC (JS turned). S4 calls alone with the 3 highest trump, and the question is whether to discard the AD or the KH. Irish pointed out that the threat to having the sweep spoiled lies with S1 leading a suit that S4 has but S3 is void in (assuming S3 also has trump), and that this is about twice as likely if S4 hangs on to both aces. This would make keeping the AKH the better play. I pointed out that if either opponent has 3 trump, forcing S4 to trump the first trick will give the opponent and extra trump and thus the stopper. This would make keeping the two aces the better play. So which scenario is more likely?
I ran 100,000 hands, and found the following:
S1 has 3 trump: 4,277
S1 has 4 trump: 152 [just curious; loner is stopped no matter what is discarded]
S2 has 3 or 4 trump: 4,397
S3 has 3 trump: 4,194
S3 has 4 trump: 170 [ditto]
S2 will always bid with 3 or 4 trump.
S1 and S3 will always bid with 4 trump, will never bid with 3 trump (since the R is turned) [again, how my simulator is coded]
So in the 8,471 hands where one of the opponents holds 3 trump, S4 will always trump a C lead (probably thwarting their loner), they will never trump a H lead (fate is the same no matter which card had been discarded), but if a D is lead the outcome will depend on whether the AD or KH was discarded. For simplicity, we can say that about 2,500 successful loners will be sacrificed if S4 had discarded the AD (they trump this trick, are left with just 2 trump vs. the 3 that an opponent holds). There could of course be overlap with the other scenario, as S3 may have 3 trump and be void in D; they trump, S4 overtrumps; loner still intact. So this is just a rough estimate.
I also tallied how often S1 led a D, with S3 being void in D AND having at least 1 trump: 4,676 [of these, S3 had 3 trump 675 times  the overlap I mentioned earlier]. In these cases, when S3 has 1 or 2 trump (4,001 times), keeping the AKH will allow S4 to trump the D lead and keep their successful loner intact.
So, in the cases that matter, it's seems better to keep the AKH in about 4% of cases and better to keep both aces in about 2.5% of cases. When I looked at the actual difference in successful loners when one or the card were discarded, I found 1,884 additional successful loners when the AD was discarded [the above results would have predicted 1,500, so not bad].
Irish, with this very strong hand it is indeed better to keep the AKH and get 2suited, but that is the result I found as well (see my original post). I was simply saying that the difference was so small that it was perhaps worth not memorizing an extra exception to the general rule of "keep the 2 aces". You provided a concrete example of why it was better to do so, I presented an opposite scenario, and then tried to find the relative frequency of these scenarios occurring.
So no disagreement here, I just wanted to have a little fun exploring this particular hand.
1) when S3 has both bowers and no other trump, they need at least 2 offsuit aces to call trump [how my simulator is coded]. That did indeed occur a few times in the simulation I did for Irish, with S4 having 5 trump, resulting in S3 getting euchred. But I think that is clearly an outlier case, and it is otherwise a good call. Very good chance of making a point (or even 2), and a significant chance that S4 passes, with S2 getting a chance to bid a green suit, R2.
2) S4 is dealt AS + AKH + AD + JC (JS turned). S4 calls alone with the 3 highest trump, and the question is whether to discard the AD or the KH. Irish pointed out that the threat to having the sweep spoiled lies with S1 leading a suit that S4 has but S3 is void in (assuming S3 also has trump), and that this is about twice as likely if S4 hangs on to both aces. This would make keeping the AKH the better play. I pointed out that if either opponent has 3 trump, forcing S4 to trump the first trick will give the opponent and extra trump and thus the stopper. This would make keeping the two aces the better play. So which scenario is more likely?
I ran 100,000 hands, and found the following:
S1 has 3 trump: 4,277
S1 has 4 trump: 152 [just curious; loner is stopped no matter what is discarded]
S2 has 3 or 4 trump: 4,397
S3 has 3 trump: 4,194
S3 has 4 trump: 170 [ditto]
S2 will always bid with 3 or 4 trump.
S1 and S3 will always bid with 4 trump, will never bid with 3 trump (since the R is turned) [again, how my simulator is coded]
So in the 8,471 hands where one of the opponents holds 3 trump, S4 will always trump a C lead (probably thwarting their loner), they will never trump a H lead (fate is the same no matter which card had been discarded), but if a D is lead the outcome will depend on whether the AD or KH was discarded. For simplicity, we can say that about 2,500 successful loners will be sacrificed if S4 had discarded the AD (they trump this trick, are left with just 2 trump vs. the 3 that an opponent holds). There could of course be overlap with the other scenario, as S3 may have 3 trump and be void in D; they trump, S4 overtrumps; loner still intact. So this is just a rough estimate.
I also tallied how often S1 led a D, with S3 being void in D AND having at least 1 trump: 4,676 [of these, S3 had 3 trump 675 times  the overlap I mentioned earlier]. In these cases, when S3 has 1 or 2 trump (4,001 times), keeping the AKH will allow S4 to trump the D lead and keep their successful loner intact.
So, in the cases that matter, it's seems better to keep the AKH in about 4% of cases and better to keep both aces in about 2.5% of cases. When I looked at the actual difference in successful loners when one or the card were discarded, I found 1,884 additional successful loners when the AD was discarded [the above results would have predicted 1,500, so not bad].
Irish, with this very strong hand it is indeed better to keep the AKH and get 2suited, but that is the result I found as well (see my original post). I was simply saying that the difference was so small that it was perhaps worth not memorizing an extra exception to the general rule of "keep the 2 aces". You provided a concrete example of why it was better to do so, I presented an opposite scenario, and then tried to find the relative frequency of these scenarios occurring.
So no disagreement here, I just wanted to have a little fun exploring this particular hand.

 Posts: 1321
 Joined: Tue Apr 24, 2018 9:33 pm
COOL  a lot of work for a small advantage.
IRISH
IRISH