NEWBE,
Below is what you asked (see below).
First you look what are the trump cards unknown. There are three JD AH 9H. What are the probabilities, with 18 unknown cards in total, for opponents?
1 trump card  47%
2 trump cards  16%
3 trump cards  ~1.5%
0 trump cards  35% (I rounded but needs to add to 100%)
So let's assume you always play the hand correctly by leading JH first followed by the KH to the 2nd trick.
To stop your loner one opponent has to have JD/9D, or JD/AH or AH/9H or all three. Any 2 of the 3 unknown will stop your loner  that 16%. But you have two opponents and that number is doubled 32% probability that your loner will be stopped just by those unknown trumps.
What about that QS lead to the last trump  that is the turned down suit and quite difficult to calculate an EXACT chance it goes through. You partner could have AS or KS or they in combination could be with S3 or buried. And sometimes the opponents get forced with AS/AC, AS/AD, or KS/AC, ect etc. But chance of that going through with 3 off suits is quite high 50  60% (estimate, guessing)
So you are looking at being euchred  3%. I can't believe your statement about getting euchred FREQUENTLY? An opponent has to have 3 trumps?
For a successful loner  32% get stopped by trump, and 40  50% with AS/KS. Those are not additive as could hold at same time but I estimate a loner is successful 30  35? Definitely go alone as you will score more points than taking your partner along. All that depends on score as well  how bad do you need a loner? Spades is the turned down suit and favors those critical cards being buried and/or with S3. And it is always more difficult for opponents to stop a loner when they have to decide with three (3) off suits. SWAG on making a loner 30  36%. That score way more points than taking your partner along.
Can my partner help me? Yes if he has the JD guarded or AH guarded, with an off suit ace. Or has AS or KS if AS is buried if an opponent only has one trump. But that is less less than is around 40% (estimate, guessing). You can't give exact probabilities on that a march.
I hope that helps. And there might be other opinions but I think I am pretty close and what I would do. Others can jump in and comment.
I know of know chart on probabilities. You would have to learn how to calculate probabilities if interested.
P.S. (I should have also said  that 36% is about stopping you.)
For example consider that each opponent may have just one trump each  47% x 47% = about 24% of the time; and some times one has 1 and 0  47% x 35% = about 12% x 2 = 24% (doubled as 0 & 1 and 1 & 0); and sometimes 0 & 0  35% x 35% = ~11% and those add to 24 + 24 + 11 = 64% (without rounding add to about 64%.) Has to add to (64 + 36 =) 100% all probabilities. When that is true then only the AS/KS remains to stop you. Gets complex!
IRISH
Newbe said, "Second of all, I am stuck thinking about a specific hand and the actual probabilities of each outcome. I'm pretty sure I've been euchred going alone on this exact hand in the past, and just want to see the numbers laid out to help me wrap my head around why this is not a good choice.
My Hand:
I am pretty sure a was passed on by all and flipped, prior to me being "stuck" and calling Hearts.
Is there an easy way to calculate my odds of taking all if I go alone?
Odds of getting at least 3?
Odds of getting Euchred?
Odds of getting all if I bring my partner?
Odds of getting Euchred if I bring my partner?
How much does taking my partner with me actually benefit me?"
NEWBE's POSSIBLER LONE HAND

 Posts: 3
 Joined: Wed Nov 24, 2021 12:11 am
Irish,
I greatly appreciate you taking the time to craft such a response.
Thank you for your help.
I greatly appreciate you taking the time to craft such a response.
Thank you for your help.

 Posts: 1273
 Joined: Tue Apr 24, 2018 9:33 pm
Newbie, I like math, statistics and how it applies to Euchre. Sometimes I go gut intuition and forget the stats! lol
And you got the full Eddie!
You're welcome!
IRISH
And you got the full Eddie!
You're welcome!
IRISH

 Posts: 238
 Joined: Thu Sep 16, 2021 6:56 pm
I ran several simulations of this hand and have lots of data to share.
The first thing I noticed is that, at a table of solid euchre players, this chances of this situation cropping up are exceedingly rare. I simulated 1,000,000 hands, and it was passed around to the dealer, R2, only 1400 times, or 0.14%. Multiply that by the rarity of dealer being dealt a hand similar to the one in question (4 of a suit that is not the turn suit) and it is something you may encounter once in maybe 50,000 hands.
I ultimately found that it was marginally better to take your partner along on this hand, but in reality it could go either way. The subset of possible hands held by the other 3 players is so whittled down by eliminating the hands on which they would have bid (before it ever got around to dealer again) that it is not really possible to calculate odds of someone having the left bower, or AS. Irish's logic, and statistics, are impeccable, but they assume we are looking at ALL 100% of possible card distributions, which is not even close to being true. My program may not always bid exactly as an experienced player would bid, so the 0.14% of hands I ended up with may be wrong subset. I would intuitively say it's correct to bid this hand alone, S4, R2, but it is not possible (in my opinion) to actually mathematically justify that choice.
The statistics:
1,000,000 hands tested
dealer bids alone: 82 / 1285 / 30 (4 pts / 1 pt / 2 pts) EV = 1.11
dealer bid w/p: 261 / 1100 / 36 EV = 1.11
_____________________________
I also decided to change the scenario slightly and analyze how it should be played if it arose in the first round of bidding. To do that I simply switched the 9S and the 10H. So dealer is dealt , with being the turn card. Should dealer call alone or with partner?
I simulated 100,000 hands, 95.6% of which made it around to dealer to bid (3.1% were bid by S2, 1.3% by S1 or S3). When dealer bids alone, they are successful 13.7% of the time, get 1 pt. 85.4% of the time, and get euchred 0.9% of the time. [for the sake of completeness, this led to an EV of 1.39 bidding alone and 1.30 bidding with partner; bidding w/p led to about 30% fewer euchres and 120% more 5card wins, but this couldn't overcome the advantage of 4pt successful loners].
I was very surprised that the odds of a successful loner, at 13.6%, were far lower than the 3036% predicted by Irish, and strove to discover why. Initially, four possible contributing factors came to mind:
1) When S2 bids, this takes away hands that would likely have been successful loners for dealer [I tested this and found that if S2 passed these hands, dealer would bid successful loners on 25% of them];
2) If S1 or S3 has Lx or Ax of trump, that reduces the chances of them having the AS or KS stopper [only 8 possible cards left in their hands], so those two methods of stopping have some correlation;
3) If S1 leads a Spade, S3 could trump it;
4) Maybe the KS is a stopper more often than we think (as Irish mentioned, a difficult thing to accurately predict).
In terms of numbers:
loner successful (4 pts): 13,010 hands
3 or 4 tricks (1 pt): 81,642 hands
2 tricks (euchred): 860 hands
Next, I looked at all the tricks won by S1 or S3 in these 100,000 hands (there were 111,510 of them), and recorded what card they won with.
AS: 44%
JD: 22%
KS: 19%
AH: 13%
9H: 1.5%
some other card: 0.5% (likely that third trick won when euchring)
Since the loner bid is thwarted 86.2% of the time, I can multiply the above frequencies by .862 to approximate how often each of those cards is the critical stopper.
This data seems to suggest that the opponents thwart the loner with Lx of trump 19% of the time, with Ax of trump 11% of the time, with the AS 38% of the time, with the KS 16% of the time and with the 9H 2% of the time [the scenario of S1 leading Spades and S3 trumping]. How do these compare with Irish's predictions? Lx + Ax = 30%, very close. AS + KS = 54%, a bit higher than predicted. 9S = 2% (not considered). Critically, this assumes perfect additivity of the "win with trump" v. "win with high Spade" factors, which can't be true [sometimes the opponents win 2 tricks, one each way]. So maybe the odds of winning with each card is a shade higher. And maybe they are much more additive than Irish suggested.
Of course, the final possibility is that my program is not playing the hands correctly. It is not possible for me to manually check all 100,000 hands (I have looked at a bunch of them), but honestly, these hands are not so hard to play. I could admit that the success rate could be a percent or two higher, but not 150% higher. Maybe this is just a case where detailed simulations can inform statistical estimates.
Most importantly, we both come to the same conclusion: dealer should bid alone!
___________________________
Finally, I also tried changing the offsuit Q to a J or K (in the second scenario  dealer bidding R1), and the results were predictable:
100,000 hands played
with KS as offsuit:
alone: 24,959 / 69,861 / 825 (4pts / 1pt / 2pts) EV = 1.76 *
w/p: 32,770 / 62,275 / 600 EV = 1.32
with JS as offsuit:
alone: 8,099 / 86,435 / 1027 (4pts / 1pt / 2pts) EV = 1.22
w/p: 29,840 / 65,149 / 572 EV = 1.29 *
The first thing I noticed is that, at a table of solid euchre players, this chances of this situation cropping up are exceedingly rare. I simulated 1,000,000 hands, and it was passed around to the dealer, R2, only 1400 times, or 0.14%. Multiply that by the rarity of dealer being dealt a hand similar to the one in question (4 of a suit that is not the turn suit) and it is something you may encounter once in maybe 50,000 hands.
I ultimately found that it was marginally better to take your partner along on this hand, but in reality it could go either way. The subset of possible hands held by the other 3 players is so whittled down by eliminating the hands on which they would have bid (before it ever got around to dealer again) that it is not really possible to calculate odds of someone having the left bower, or AS. Irish's logic, and statistics, are impeccable, but they assume we are looking at ALL 100% of possible card distributions, which is not even close to being true. My program may not always bid exactly as an experienced player would bid, so the 0.14% of hands I ended up with may be wrong subset. I would intuitively say it's correct to bid this hand alone, S4, R2, but it is not possible (in my opinion) to actually mathematically justify that choice.
The statistics:
1,000,000 hands tested
dealer bids alone: 82 / 1285 / 30 (4 pts / 1 pt / 2 pts) EV = 1.11
dealer bid w/p: 261 / 1100 / 36 EV = 1.11
_____________________________
I also decided to change the scenario slightly and analyze how it should be played if it arose in the first round of bidding. To do that I simply switched the 9S and the 10H. So dealer is dealt , with being the turn card. Should dealer call alone or with partner?
I simulated 100,000 hands, 95.6% of which made it around to dealer to bid (3.1% were bid by S2, 1.3% by S1 or S3). When dealer bids alone, they are successful 13.7% of the time, get 1 pt. 85.4% of the time, and get euchred 0.9% of the time. [for the sake of completeness, this led to an EV of 1.39 bidding alone and 1.30 bidding with partner; bidding w/p led to about 30% fewer euchres and 120% more 5card wins, but this couldn't overcome the advantage of 4pt successful loners].
I was very surprised that the odds of a successful loner, at 13.6%, were far lower than the 3036% predicted by Irish, and strove to discover why. Initially, four possible contributing factors came to mind:
1) When S2 bids, this takes away hands that would likely have been successful loners for dealer [I tested this and found that if S2 passed these hands, dealer would bid successful loners on 25% of them];
2) If S1 or S3 has Lx or Ax of trump, that reduces the chances of them having the AS or KS stopper [only 8 possible cards left in their hands], so those two methods of stopping have some correlation;
3) If S1 leads a Spade, S3 could trump it;
4) Maybe the KS is a stopper more often than we think (as Irish mentioned, a difficult thing to accurately predict).
In terms of numbers:
loner successful (4 pts): 13,010 hands
3 or 4 tricks (1 pt): 81,642 hands
2 tricks (euchred): 860 hands
Next, I looked at all the tricks won by S1 or S3 in these 100,000 hands (there were 111,510 of them), and recorded what card they won with.
AS: 44%
JD: 22%
KS: 19%
AH: 13%
9H: 1.5%
some other card: 0.5% (likely that third trick won when euchring)
Since the loner bid is thwarted 86.2% of the time, I can multiply the above frequencies by .862 to approximate how often each of those cards is the critical stopper.
This data seems to suggest that the opponents thwart the loner with Lx of trump 19% of the time, with Ax of trump 11% of the time, with the AS 38% of the time, with the KS 16% of the time and with the 9H 2% of the time [the scenario of S1 leading Spades and S3 trumping]. How do these compare with Irish's predictions? Lx + Ax = 30%, very close. AS + KS = 54%, a bit higher than predicted. 9S = 2% (not considered). Critically, this assumes perfect additivity of the "win with trump" v. "win with high Spade" factors, which can't be true [sometimes the opponents win 2 tricks, one each way]. So maybe the odds of winning with each card is a shade higher. And maybe they are much more additive than Irish suggested.
Of course, the final possibility is that my program is not playing the hands correctly. It is not possible for me to manually check all 100,000 hands (I have looked at a bunch of them), but honestly, these hands are not so hard to play. I could admit that the success rate could be a percent or two higher, but not 150% higher. Maybe this is just a case where detailed simulations can inform statistical estimates.
Most importantly, we both come to the same conclusion: dealer should bid alone!
___________________________
Finally, I also tried changing the offsuit Q to a J or K (in the second scenario  dealer bidding R1), and the results were predictable:
100,000 hands played
with KS as offsuit:
alone: 24,959 / 69,861 / 825 (4pts / 1pt / 2pts) EV = 1.76 *
w/p: 32,770 / 62,275 / 600 EV = 1.32
with JS as offsuit:
alone: 8,099 / 86,435 / 1027 (4pts / 1pt / 2pts) EV = 1.22
w/p: 29,840 / 65,149 / 572 EV = 1.29 *

 Posts: 1273
 Joined: Tue Apr 24, 2018 9:33 pm
RAY, RAY, RAY: You said:
My comments in RED! (WOLF)
raydog » Sun Nov 28, 2021 10:29 am
I ran several simulations of this hand and have lots of data to share.
The first thing I noticed is that, at a table of solid euchre players, this chances of this situation cropping up are exceedingly rare. I simulated 1,000,000 hands, and it was passed around to the dealer, R2, only 1400 times, or 0.14%. Multiply that by the rarity of dealer being dealt a hand similar to the one in question (4 of a suit that is not the turn suit) and it is something you may encounter once in maybe 50,000 hands.
S1 has JH KH 10H 9H QS  is is now the chance only knowing that the 9S was the upcard. Now you have 23 unknown cards and 7 are trumps in the suit of Hearts. Thus 7c4 x 16c1 // 23c5 = 2.37% assuming 9s turned down. It is true I am not allowing for those hands that any one had strong enough hands to order up the 9S. Even if factoring out 50% that is Way different .14% vs 2.37% = (6%) or 17 times. Can't buy that. Thus my number is NO any way close to your 1,000,000 hands of 0.14%. No clue what your program is doing.
I think YOUR PROGRAM IS FLAWED about chance of having the hand in question. THE LAW OF LARGE NUMBERS, you should be very very close to the theoretical calculated number! However we do know that the dealer will have fewer Spades and more likely to have the JD than S3. And S3 will have more spades but those are UNKNOWN. Still that does not give S1/S3 a better chance for a sweep, IMO. I say no contest on that!
The statistics:
1,000,000 hands tested
dealer bids alone: 82 / 1285 / 30 (4 pts / 1 pt / 2 pts) EV = 1.11
dealer bid w/p: 261 / 1100 / 36 EV = 1.11 This is wrong, because you having the DEALER BIDDING ALONE? I don't follow as S1 is the one going alone and he has an an advantage. I estimate a 50 point difference going alone vs taking your partner alone.
.
But to be fair to you and me, I am going to test this giving S1 the hand, and and all PASSING ON 9S because I need to know! How many hands I need to test to confirm or refute EV = 1.11 from S1 going alone vs taking his partner alone. I will only test if you give numbers for S1, Eldest, going alone vs taking his partner along? Waste of my time for the Dealer going alone. And EV needs to be based per 100 hands. And give the Euchre rate, Sweep rate, 1 point rate. If you can do that, I am done!
I stand by my original post and this one as well and will stop there.
IRSHWOLF
My comments in RED! (WOLF)
raydog » Sun Nov 28, 2021 10:29 am
I ran several simulations of this hand and have lots of data to share.
The first thing I noticed is that, at a table of solid euchre players, this chances of this situation cropping up are exceedingly rare. I simulated 1,000,000 hands, and it was passed around to the dealer, R2, only 1400 times, or 0.14%. Multiply that by the rarity of dealer being dealt a hand similar to the one in question (4 of a suit that is not the turn suit) and it is something you may encounter once in maybe 50,000 hands.
S1 has JH KH 10H 9H QS  is is now the chance only knowing that the 9S was the upcard. Now you have 23 unknown cards and 7 are trumps in the suit of Hearts. Thus 7c4 x 16c1 // 23c5 = 2.37% assuming 9s turned down. It is true I am not allowing for those hands that any one had strong enough hands to order up the 9S. Even if factoring out 50% that is Way different .14% vs 2.37% = (6%) or 17 times. Can't buy that. Thus my number is NO any way close to your 1,000,000 hands of 0.14%. No clue what your program is doing.
I think YOUR PROGRAM IS FLAWED about chance of having the hand in question. THE LAW OF LARGE NUMBERS, you should be very very close to the theoretical calculated number! However we do know that the dealer will have fewer Spades and more likely to have the JD than S3. And S3 will have more spades but those are UNKNOWN. Still that does not give S1/S3 a better chance for a sweep, IMO. I say no contest on that!
The statistics:
1,000,000 hands tested
dealer bids alone: 82 / 1285 / 30 (4 pts / 1 pt / 2 pts) EV = 1.11
dealer bid w/p: 261 / 1100 / 36 EV = 1.11 This is wrong, because you having the DEALER BIDDING ALONE? I don't follow as S1 is the one going alone and he has an an advantage. I estimate a 50 point difference going alone vs taking your partner alone.
.
But to be fair to you and me, I am going to test this giving S1 the hand, and and all PASSING ON 9S because I need to know! How many hands I need to test to confirm or refute EV = 1.11 from S1 going alone vs taking his partner alone. I will only test if you give numbers for S1, Eldest, going alone vs taking his partner along? Waste of my time for the Dealer going alone. And EV needs to be based per 100 hands. And give the Euchre rate, Sweep rate, 1 point rate. If you can do that, I am done!
I stand by my original post and this one as well and will stop there.
IRSHWOLF

 Posts: 1273
 Joined: Tue Apr 24, 2018 9:33 pm
RAY,
Do you not realize the Dealer is not the lone RUNNER? 2R S1 is is going alone. So of course your results will be different?? Enough to give me heartburn.
I was very surprised that the odds of a successful loner, at 13.6%, were far lower than the 3036% predicted by Irish, and strove to discover why. Initially, four possible contributing factors came to mind:
1) When S2 bids, this takes away hands that would likely have been successful loners for dealer [no no no] [I tested this and found that if S2 passed these hands, dealer would bid successful loners on 25% of them];
2) If S1 or S3 has Lx or Ax of trump, that reduces the chances of them having the AS or KS stopper [only 8 possible cards left in their hands], so those two methods of stopping have some correlation;
3) If S1 leads a Spade, S3 could trump it;
4) Maybe the KS is a stopper more often than we think (as Irish mentioned, a difficult thing to accurately predict).
Do you not realize the Dealer is not the lone RUNNER? 2R S1 is is going alone. So of course your results will be different?? Enough to give me heartburn.
I was very surprised that the odds of a successful loner, at 13.6%, were far lower than the 3036% predicted by Irish, and strove to discover why. Initially, four possible contributing factors came to mind:
1) When S2 bids, this takes away hands that would likely have been successful loners for dealer [no no no] [I tested this and found that if S2 passed these hands, dealer would bid successful loners on 25% of them];
2) If S1 or S3 has Lx or Ax of trump, that reduces the chances of them having the AS or KS stopper [only 8 possible cards left in their hands], so those two methods of stopping have some correlation;
3) If S1 leads a Spade, S3 could trump it;
4) Maybe the KS is a stopper more often than we think (as Irish mentioned, a difficult thing to accurately predict).

 Posts: 238
 Joined: Thu Sep 16, 2021 6:56 pm
Irish, you and I have interpreted the initial post differently. Newbie said that the 9S was turned and he/she was "stuck" calling hearts. To me that means they are the dealer, and playing STD  that's the only place you can get stuck. You interpreted the "stuck" as being sarcastic, and that they are playing from S1. Obviously a HUGE difference.
We are comparing apples and oranges....
Newbie, which is it? I fear I did all that work for nought.
We are comparing apples and oranges....
Newbie, which is it? I fear I did all that work for nought.

 Posts: 795
 Joined: Sun Jun 16, 2019 9:14 pm
 Location: Las Vegas
I went back to look. It's definitely a s4r2 hand after 9s was turned down and passed by all.
Tbolt65
Edward
Tbolt65
Edward

 Posts: 1273
 Joined: Tue Apr 24, 2018 9:33 pm
You are CORRECT RAY it was the Dealer. I stand corrected. What I posted was only for Eldest going alone after all passed on R1. Sorry! My bad!
So having the stopper now increases and being squeezed is much reduced (that's about 7%. Yes fewer loners, more euchres because if a spade is led and S3 wins it the dealer can now be over trumped and can be euchred by S3 having 2 trumps. I can see that. How much, not going to guess.
But I do strongly suspect that going alone will have a higher EV than taking partner along. But what makes that very difficult conclude is 7 PASSES. In fact I cannot see how your Program can do that either? Or any programmer. Too many assumptions and variables to contend with. You have to pull out all those biddable hands for all suits for S1, S2 & S3 and S4 R1. Not to mention how and what S1 leads in given situations.
But I can conclude that if I were the dealer, I am going alone.
Irish
So having the stopper now increases and being squeezed is much reduced (that's about 7%. Yes fewer loners, more euchres because if a spade is led and S3 wins it the dealer can now be over trumped and can be euchred by S3 having 2 trumps. I can see that. How much, not going to guess.
But I do strongly suspect that going alone will have a higher EV than taking partner along. But what makes that very difficult conclude is 7 PASSES. In fact I cannot see how your Program can do that either? Or any programmer. Too many assumptions and variables to contend with. You have to pull out all those biddable hands for all suits for S1, S2 & S3 and S4 R1. Not to mention how and what S1 leads in given situations.
But I can conclude that if I were the dealer, I am going alone.
Irish

 Posts: 238
 Joined: Thu Sep 16, 2021 6:56 pm
I freely admit that isolating the 0.14% of hands that survive 7 passes is entirely dependent on assumptions and can be given little credence. Which is why I said that I didn't think that it could ever be definitively demonstrated, mathematically, that going alone is the better option. And like you, I simply believe from experience that going alone with those cards is the better move  it's what I would do, too.
_______________________________________
I also looked at the case where S1 bids, R2, with those same cards, and discovered something interesting, though not surprising.
I simulated 1,000,000 hands, with S1 holding , and turned. Here are the raw results:
bid alone bid with partner
4pts. 1pt. 2pts.  2pts. 1pt. 2pts.
S1,R1: (passes)
S2,R1: 25,088 28,373 3,935  59,133 121,292 28,052
S3,R1: 5,073 20,987 862  5,630 37,732 7,686
S4,R1: 90,687 88,341 3,381  40,301 169,579 38,810

S1,R2: 59,261 161,551 4,246  83,949 138,244 2,865
NOTE: for S1,R2, I tried the same 1,000,000 hands with S1 bidding H alone and with partner, and put the results on the same line above  it's one or the other, not both! The EV is higher when S1 bids alone (1.73) than w/p (1.33).
Also, in the case where S1 bid H alone, R2, I tallied which cards won each trick, the results of which are:
10H: 223,718 KS: 39,054
QS: 90,262 AS: 93,546
QH: 221,450 AH: 35,848
KH: 131,592 JD: 64,762
JH: 225,058
A total of 225,058 hands were passed by everyone R1 and allowed S1 to bid R2, which amounts to less than a quarter of all hands. The choices made by my program of what hands are bid by S2, S3 and S4 are somewhat subjective, but I think that they are largely correct, so that the 225,058 I actually played are very similar in characteristic to the hands that would actually be played if this situation were to arise around a table of very good players.
Also, this hand is very easy to play, especially since S1 gets first lead. S1 leads the JH first (and wins), then the KH, and subsequently always leads their highest trump, playing the QS as late as possible in the hand.
Some statistics:
 with 6 cards fixed, there are 18 cards to distribute among the other 3 players + the kitty. There are 18!/(5!*5!*5!*3!) nonordered ways of doing this, or
617,512,896. My program randomly picked 1 million of these;
 looking at those million hands, we can calculate how often we expect an opponent (S2 or S4) to have the JD + 1 or 2 other trumps. For the JD it's 10/18 =
.556. For the extra trump we calculate the odds of that player having NO other trump, and subtract from 1. We get 1  (15/17)*(14/16)*(13/15)*(12/14) = .426;
So the odds of having the JD AND at least one extra trump is (.556) *(.426) =
0.237, or about 24%;
 if an opponent has the JD + another trump (guarded left), they will a win a trick with it. We can see from the data above that this happens 64,762 times out of
225,058 hands, which amounts to about 29% of the time;
Why the discrepancy? The secret lies in the fact that we have taken away almost 78% of the hands, and it is precisely THOSE hands which will tend to have marginally more Spades (since they bid with a S turned), and thus marginally less other cards  like Hearts or Diamonds. The hands remaining  the ones passed, on the other hand, will tend to contain Hearts and the JD marginally more often.
For that reason, the loner is less successful than it would be if S1 were able to call the exact same hand R1. This explains why the loner success rate, at slightly over 26%, is lower than what Irish predicted. And it's the same reasoning employed when we say that S1, R2 should call next rather than a green suit: if S2 and S4 passed R1, they are marginally less likely to hold the bowers in the turned suit, which means my partner in S3 is marginally MORE likely to hold them, making next our more likely strong suit.
_______________________________________
I also looked at the case where S1 bids, R2, with those same cards, and discovered something interesting, though not surprising.
I simulated 1,000,000 hands, with S1 holding , and turned. Here are the raw results:
bid alone bid with partner
4pts. 1pt. 2pts.  2pts. 1pt. 2pts.
S1,R1: (passes)
S2,R1: 25,088 28,373 3,935  59,133 121,292 28,052
S3,R1: 5,073 20,987 862  5,630 37,732 7,686
S4,R1: 90,687 88,341 3,381  40,301 169,579 38,810

S1,R2: 59,261 161,551 4,246  83,949 138,244 2,865
NOTE: for S1,R2, I tried the same 1,000,000 hands with S1 bidding H alone and with partner, and put the results on the same line above  it's one or the other, not both! The EV is higher when S1 bids alone (1.73) than w/p (1.33).
Also, in the case where S1 bid H alone, R2, I tallied which cards won each trick, the results of which are:
10H: 223,718 KS: 39,054
QS: 90,262 AS: 93,546
QH: 221,450 AH: 35,848
KH: 131,592 JD: 64,762
JH: 225,058
A total of 225,058 hands were passed by everyone R1 and allowed S1 to bid R2, which amounts to less than a quarter of all hands. The choices made by my program of what hands are bid by S2, S3 and S4 are somewhat subjective, but I think that they are largely correct, so that the 225,058 I actually played are very similar in characteristic to the hands that would actually be played if this situation were to arise around a table of very good players.
Also, this hand is very easy to play, especially since S1 gets first lead. S1 leads the JH first (and wins), then the KH, and subsequently always leads their highest trump, playing the QS as late as possible in the hand.
Some statistics:
 with 6 cards fixed, there are 18 cards to distribute among the other 3 players + the kitty. There are 18!/(5!*5!*5!*3!) nonordered ways of doing this, or
617,512,896. My program randomly picked 1 million of these;
 looking at those million hands, we can calculate how often we expect an opponent (S2 or S4) to have the JD + 1 or 2 other trumps. For the JD it's 10/18 =
.556. For the extra trump we calculate the odds of that player having NO other trump, and subtract from 1. We get 1  (15/17)*(14/16)*(13/15)*(12/14) = .426;
So the odds of having the JD AND at least one extra trump is (.556) *(.426) =
0.237, or about 24%;
 if an opponent has the JD + another trump (guarded left), they will a win a trick with it. We can see from the data above that this happens 64,762 times out of
225,058 hands, which amounts to about 29% of the time;
Why the discrepancy? The secret lies in the fact that we have taken away almost 78% of the hands, and it is precisely THOSE hands which will tend to have marginally more Spades (since they bid with a S turned), and thus marginally less other cards  like Hearts or Diamonds. The hands remaining  the ones passed, on the other hand, will tend to contain Hearts and the JD marginally more often.
For that reason, the loner is less successful than it would be if S1 were able to call the exact same hand R1. This explains why the loner success rate, at slightly over 26%, is lower than what Irish predicted. And it's the same reasoning employed when we say that S1, R2 should call next rather than a green suit: if S2 and S4 passed R1, they are marginally less likely to hold the bowers in the turned suit, which means my partner in S3 is marginally MORE likely to hold them, making next our more likely strong suit.