https://worldofcardgames.com/#!replayer ... %3A1%7D%5D
Ordered 9d and got euchred
-
Dlan
- Site Admin
- Posts: 672
- Joined: Thu Apr 19, 2018 10:08 pm
- Location: Ohio
-
irishwolf
- Posts: 1307
- Joined: Tue Apr 24, 2018 9:33 pm
1st seat deserved to be euchred on this hand. Why in world with the Right and two off suit aces, you lead an Ace first? The correct play is lead the Right, then your aces. Actually, I would go alone on this hand and make it 7 out of 10 times.
IRISH
IRISH
Last edited by irishwolf on Mon Aug 15, 2022 4:09 pm, edited 1 time in total.
-
raydog
- Posts: 260
- Joined: Thu Sep 16, 2021 6:56 pm
Two mistakes on this hand:
1) not going alone;
2) leading an off-suit ace
I simulated 100,00 hands, and found that the going alone was successful 38.1% of the time, and the euchre rate was 4.2%. I haven't explored the euchre scenarios, but am confident there are some low-probability distributions where you could get euchred on this hand (going alone) But clearly rare enough that you shouldn't worry about it - it happens, and your opponents may rib you, but that doesn't mean the decision to call alone was wrong.
As far as euchre success rate, if S4 is naturally dealt the AD or JH in their 5-card hand (and pick up the 9D), they can stop the loner. How often does this happen? Well, NOT getting dealt either of those cards happens 16/18 X 15/17 X 14/16 X 13/15 X 12/14 = 51 % of the time, so 49% of the time they will get one of those cards and stop your loner. Also, if S2 is dealt 2 trump (which will necessarily include one of those stoppers), the loner is also stopped. I haven't calculated those odds [which are dependent on what S4 is dealt, so it gets more complicated], but it makes sense to see a loner success rate of 38%.
I compared the case where S1 calls with partner, and the sweep rate was lower AND the euchre rate was higher! This is the sort of hand where your partner not only can't help you, but can actually hurt you! You need to keep the lead and play the cards/suits in the proper order - don't take your partner along!
As for the off-suit A lead, this is an error I have seen so often that it actually is one of the plays that prompted me to try and develop my own euchre algorithm. Here's what S1 is thinking: I'll save my trump, and play an A, hoping it wins. If it does I'll play the R, or perhaps the other A. If both aces win, I'll play the R (only 3 trump left in hand, and that is the highest). But if you stop and compare the two scenarios (1: play and Ace, then the highest trump, vs. 2) playing the highest trump, then the Ace), the only "advantage" of the first scenario is inviting the opponents to trump in with a lone trump because they are void in your led suit. Why not collect that trump FIRST with the lead of the R??! There is NO* scenario where the first play (leading an off-Ace first) yields better results [remember, you are playing alone; but even playing w/ partner, the cases where this is better are quite rare, and far eclipsed by leading the R first].
I've tried explaining this to people on many occasions, and only rarely does the person see the logic. But this situation arises often enough that it is worth thinking through and convincing yourself that the lead of the R is the better play.
I agree with Irish: S1 deserves to be euchred!
* this may be an exaggeration, but I haven't found such a scenario. Do let me know if there is one - I'm sure it is extremely rare.
1) not going alone;
2) leading an off-suit ace
I simulated 100,00 hands, and found that the going alone was successful 38.1% of the time, and the euchre rate was 4.2%. I haven't explored the euchre scenarios, but am confident there are some low-probability distributions where you could get euchred on this hand (going alone) But clearly rare enough that you shouldn't worry about it - it happens, and your opponents may rib you, but that doesn't mean the decision to call alone was wrong.
As far as euchre success rate, if S4 is naturally dealt the AD or JH in their 5-card hand (and pick up the 9D), they can stop the loner. How often does this happen? Well, NOT getting dealt either of those cards happens 16/18 X 15/17 X 14/16 X 13/15 X 12/14 = 51 % of the time, so 49% of the time they will get one of those cards and stop your loner. Also, if S2 is dealt 2 trump (which will necessarily include one of those stoppers), the loner is also stopped. I haven't calculated those odds [which are dependent on what S4 is dealt, so it gets more complicated], but it makes sense to see a loner success rate of 38%.
I compared the case where S1 calls with partner, and the sweep rate was lower AND the euchre rate was higher! This is the sort of hand where your partner not only can't help you, but can actually hurt you! You need to keep the lead and play the cards/suits in the proper order - don't take your partner along!
As for the off-suit A lead, this is an error I have seen so often that it actually is one of the plays that prompted me to try and develop my own euchre algorithm. Here's what S1 is thinking: I'll save my trump, and play an A, hoping it wins. If it does I'll play the R, or perhaps the other A. If both aces win, I'll play the R (only 3 trump left in hand, and that is the highest). But if you stop and compare the two scenarios (1: play and Ace, then the highest trump, vs. 2) playing the highest trump, then the Ace), the only "advantage" of the first scenario is inviting the opponents to trump in with a lone trump because they are void in your led suit. Why not collect that trump FIRST with the lead of the R??! There is NO* scenario where the first play (leading an off-Ace first) yields better results [remember, you are playing alone; but even playing w/ partner, the cases where this is better are quite rare, and far eclipsed by leading the R first].
I've tried explaining this to people on many occasions, and only rarely does the person see the logic. But this situation arises often enough that it is worth thinking through and convincing yourself that the lead of the R is the better play.
I agree with Irish: S1 deserves to be euchred!
* this may be an exaggeration, but I haven't found such a scenario. Do let me know if there is one - I'm sure it is extremely rare.
-
irishwolf
- Posts: 1307
- Joined: Tue Apr 24, 2018 9:33 pm
Only if Dealer has 4 trumps and void in Hs & Ss will the dealer be euchred by leading the JD first and here it makes no difference what is led by S1 if the Dealer has JH AD QD 9D AC. Less than 1% this would ever occur. So why worry about it.
Give the Dealer this hand 9D up and now has JH AD 9D AC KC - LEADING THE JD HERE RESULTS IN A POINT IF S1 LEAD JD THEN ONE OF THE OFF SUIT ACES. CAN'T BE EUCHRED. Or if S2 has two trumps of JH AD QD unknown. Can't beat the points going alone - AGREED!
IRISH
Give the Dealer this hand 9D up and now has JH AD 9D AC KC - LEADING THE JD HERE RESULTS IN A POINT IF S1 LEAD JD THEN ONE OF THE OFF SUIT ACES. CAN'T BE EUCHRED. Or if S2 has two trumps of JH AD QD unknown. Can't beat the points going alone - AGREED!
IRISH
-
Richardb02
- Posts: 746
- Joined: Mon Dec 10, 2018 8:57 pm
- Location: Florida
Good Post!
Clear information: Maker made 2 mistakes:
Not going Alone
Not leading the Right
I am closer to Wolf's 80% than Ray's 38%.
Even at Ray's 38%:
EV=4x.38+1x.58-2x.04=2.02.
Who wants to pass up EV=+2.02?
As for the Lead, if you want 4, or even 2, you must lead the Right.
With the hand presented, obviously follow with the King.
For now, I'll just volunteer a 67% 4 point analysis (swag, scientific wild ass guess)
If interested, ask for details.
Clear information: Maker made 2 mistakes:
Not going Alone
Not leading the Right
I am closer to Wolf's 80% than Ray's 38%.
Even at Ray's 38%:
EV=4x.38+1x.58-2x.04=2.02.
Who wants to pass up EV=+2.02?
As for the Lead, if you want 4, or even 2, you must lead the Right.
With the hand presented, obviously follow with the King.
For now, I'll just volunteer a 67% 4 point analysis (swag, scientific wild ass guess)
If interested, ask for details.
-
irishwolf
- Posts: 1307
- Joined: Tue Apr 24, 2018 9:33 pm
Actually, when I said this on loner success rate (SWAG) I did not really give it a hard look saying, " Actually, I would go alone on this hand and make it 7 out of 10 times. " It is NOT 70%. Dealer just has to have Left or Ace guarded to stop the loner - that alone is 49%. You have to also consider that S2 could also have AD or JH guarded (16% S2 has 2 trumps). So I do think 38% is correct.
To get euchred, the Dealer has to have 2 of the 3 unknown trumps and 2 Clubs. Less that 16% chance but could happen somewhere around 7 to 10% (if KD is led to 2nd trick). If I were Eldest (S1) I would lead trump the 2nd time. Yes euchre rate increases but so does the successful loner rate. That increase in points FAR exceeds the rate of being euchred. It all depends on how badly you need 4 points. And it still stands, Eldest MUST lead the Right.
Here the score 8 to 9 and the maker (S1) only needs 1 point. So going alone is moot point!
IRISH
To get euchred, the Dealer has to have 2 of the 3 unknown trumps and 2 Clubs. Less that 16% chance but could happen somewhere around 7 to 10% (if KD is led to 2nd trick). If I were Eldest (S1) I would lead trump the 2nd time. Yes euchre rate increases but so does the successful loner rate. That increase in points FAR exceeds the rate of being euchred. It all depends on how badly you need 4 points. And it still stands, Eldest MUST lead the Right.
Here the score 8 to 9 and the maker (S1) only needs 1 point. So going alone is moot point!
IRISH
-
Tbolt65
- Posts: 803
- Joined: Sun Jun 16, 2019 9:14 pm
- Location: Las Vegas
Everyone is in agreement here. Leading an off Ace here is the worst lead. Any other lead, Jack/king/10 of trump is much better than leading one of the two off aces.
However, with this hand situation/scenario with it being 9-8 my team. I only need one point. I am leading the 10 of diamonds to ensure my team makes 1 point. I will set up an endplay with my right -king. I have two off aces to take tricks with or force trump out. Once I get the lead I will suck the rest of the trump out and play my remaining Ace and if some how that gets trumped I will be holding the King of diamonds for boss trump, and its good game my team. Mind you at a different score, the leads and play will vary. People may say, hey Edward you are getting way to tricky here but I disagree. I am ensuring I get my one point and that's all you need here. Nothing more.
Tbolt65
Edward
However, with this hand situation/scenario with it being 9-8 my team. I only need one point. I am leading the 10 of diamonds to ensure my team makes 1 point. I will set up an endplay with my right -king. I have two off aces to take tricks with or force trump out. Once I get the lead I will suck the rest of the trump out and play my remaining Ace and if some how that gets trumped I will be holding the King of diamonds for boss trump, and its good game my team. Mind you at a different score, the leads and play will vary. People may say, hey Edward you are getting way to tricky here but I disagree. I am ensuring I get my one point and that's all you need here. Nothing more.
Tbolt65
Edward
-
irishwolf
- Posts: 1307
- Joined: Tue Apr 24, 2018 9:33 pm
This post is not to be contradictory. But to lay out the OPTIMUM lead to minimize getting Euchred. It's all about winning the Game here. However, I disagree that leading low is better than leading the JD to trick 1.
I have already stated that S1 leading the JD to trick 1 followed by AS or AH only results in a euchre if the Dealer has four trumps: 9D JH AD QD + XC. Have all 3 of the unknown trumps is 1.2% but then must have a Club lead that wins the 5th trick. S2 could also win it. Thus, the euchre rate is at or below 1.2% but there are some blizzard hands where S2/S4 having two trumps with the proper voids and suits that could increase the euchre rate slightly but not to the rate when leading the JD. No need to spell those out.
So why get cute and concede a GIVING away a small trump? Look at all the hands that now exposes S1/S3 to getting euchred? You now increase your vulnerability by leading the 9D or any other card than the JD. If S2 has QD with AC and S4 has 9D JH AH DOUBLETON to S1 ace lead you are euchred. Or if S4 has two clubs. So what is risk that S4 has 2 of JH AD QD - 16%. The euchre will be much much higher than 1.2%.
What if S2 & S4 have two trumps each, ~8%. ? Basically, you increase your chance here too of being euchred to a greater extent than leading the JD. I say even the worse comginations you do not increase your chance of being euchred as leading the JD to trick 1. If one disagrees, lay out those hands where this is so?
Yes, leading low you will win your point the majority of the time but that is not optimizing your chances of winning the game.
Compare the two scenarios. I conclude, lead the JD then either ace, I like the AS. This forces the opponent with two trumps.
IRISH
P.S. (after giving some more thought to this)
Ray indicated a euchre rate of ..." euchre rate was 4.2%. I haven't explored the euchre scenarios, but am confident..." with many hands randomly dealt & played. That's a good number.
I calculated 4 trumps at S4 as 1.2%. x 2 = 2.4% as S2 could also have the 3 unknown trumps at the same rate. So where are the rest of those 'euchres' coming from. Well, of course when S2/S4 each have two trumps (about 8% - but also the voids and leads have to be perfect and reduce the euchres down from 8% to less than 2%) to complement their play.
I think that adds some validity to the simulator.
I have already stated that S1 leading the JD to trick 1 followed by AS or AH only results in a euchre if the Dealer has four trumps: 9D JH AD QD + XC. Have all 3 of the unknown trumps is 1.2% but then must have a Club lead that wins the 5th trick. S2 could also win it. Thus, the euchre rate is at or below 1.2% but there are some blizzard hands where S2/S4 having two trumps with the proper voids and suits that could increase the euchre rate slightly but not to the rate when leading the JD. No need to spell those out.
So why get cute and concede a GIVING away a small trump? Look at all the hands that now exposes S1/S3 to getting euchred? You now increase your vulnerability by leading the 9D or any other card than the JD. If S2 has QD with AC and S4 has 9D JH AH DOUBLETON to S1 ace lead you are euchred. Or if S4 has two clubs. So what is risk that S4 has 2 of JH AD QD - 16%. The euchre will be much much higher than 1.2%.
What if S2 & S4 have two trumps each, ~8%. ? Basically, you increase your chance here too of being euchred to a greater extent than leading the JD. I say even the worse comginations you do not increase your chance of being euchred as leading the JD to trick 1. If one disagrees, lay out those hands where this is so?
Yes, leading low you will win your point the majority of the time but that is not optimizing your chances of winning the game.
Compare the two scenarios. I conclude, lead the JD then either ace, I like the AS. This forces the opponent with two trumps.
IRISH
P.S. (after giving some more thought to this)
Ray indicated a euchre rate of ..." euchre rate was 4.2%. I haven't explored the euchre scenarios, but am confident..." with many hands randomly dealt & played. That's a good number.
I calculated 4 trumps at S4 as 1.2%. x 2 = 2.4% as S2 could also have the 3 unknown trumps at the same rate. So where are the rest of those 'euchres' coming from. Well, of course when S2/S4 each have two trumps (about 8% - but also the voids and leads have to be perfect and reduce the euchres down from 8% to less than 2%) to complement their play.
I think that adds some validity to the simulator.
-
irishwolf
- Posts: 1307
- Joined: Tue Apr 24, 2018 9:33 pm
Ray, if you are listening.
It would be interesting (and of value in other situations) to do your 10K or 100K simulator with leading the 9D and compare that to leading the JD. I predict a 10 to 14 euchre rate leading the 9D. You already did the JD lead, (4.2%).
IRISH
It would be interesting (and of value in other situations) to do your 10K or 100K simulator with leading the 9D and compare that to leading the JD. I predict a 10 to 14 euchre rate leading the 9D. You already did the JD lead, (4.2%).
IRISH
-
raydog
- Posts: 260
- Joined: Thu Sep 16, 2021 6:56 pm
I simulated this hand, thought it would be simple but it wasn't.
I first imagined that the score was irrelevant - say it's 0-0. S1 has AS + AH + K-J-10D, with 9D turned. S1 calls alone. 100,000 hands.
JD led: (37,880 / 57,789 / 4,331) [sweep for 4 pts. / 1 pt. / euchred]
10D led: (6,805 / 89,166 / 4,029)
If the goal is to score the most points, best to lead the JD. If the goal is to just score at least a point (not get euchred), then best to lead the 10D.
But the nuance here is that, since S1/S3 already have 9 pts., the general strategy is to not call alone - take your partner along to ensure a better chance of not getting euchred.
So I did the same simulation, but with S1 calling with partner:
JD led: (33,660 / 61,545 / 4,795)
10D led: (31,053 / 64,113 / 4,834)
In this case, it's better to lead the JD. But very close. In fact, too close to be definitive. While I am confident my program plays correctly when S1 calls alone (there just isn't much choice in how to play), I am less sure it's optimal when all 4 players are involved. So here it's a toss-up.
This result leads to a larger question: when do you call alone, despite having 8 or 9 pts, because your chances of scoring a point (or a sweep) are actually GREATER calling alone than with your partner? I know some expert players have ideas on this topic, but I have not come across any guidelines, so for the moment I am simply collecting evidence of hands (like this one) which break the general rule, to hopefully eventually come up with such guidance.
Suffice to say that for this hand, if S1 calls alone it's best to lead the JD, but if S1 calls with partner it's unclear which lead is best.
...................................................
One thing I did note is that my program leads high trump on the second trick (so S1 leads the KD in this instance), rather than the off-suit Ace, as Irish suggested. So I compared the two scenarios.
100,000 totally random hands, looking at the ones where S1 calls alone (about 7%).
S1 calls alone:
lead high trump, 2nd trick:
(7,105 / 15,133 / 2,727) R1 calls
(14,211 / 31,137 / 3,552) R2 calls
lead off-suit A, 2nd trick (because don't have boss trump):
(6,251 / 16,109 / 2,605) R1 calls
(12,570 / 32,892 / 3,438) R2 calls
* be sure to compare R1 vs R1, R2 vs R2!
In both cases (R1 and R2 calls), the euchre rate is higher when high trump is led on the second trick, but this is more than offset by the fact that the loner success rate is also higher. So, in general, it's best to lead the highest trump on the 2nd trick if you are in S1 and going alone, even if you have an off-suit Ace.
I also tested this for this specific hand for S1. I found an EV of 1.61 when the highest trump was led on the second trick vs. an EV of 1.52 when the off-suit Ace was led.
For what it's worth...
I first imagined that the score was irrelevant - say it's 0-0. S1 has AS + AH + K-J-10D, with 9D turned. S1 calls alone. 100,000 hands.
JD led: (37,880 / 57,789 / 4,331) [sweep for 4 pts. / 1 pt. / euchred]
10D led: (6,805 / 89,166 / 4,029)
If the goal is to score the most points, best to lead the JD. If the goal is to just score at least a point (not get euchred), then best to lead the 10D.
But the nuance here is that, since S1/S3 already have 9 pts., the general strategy is to not call alone - take your partner along to ensure a better chance of not getting euchred.
So I did the same simulation, but with S1 calling with partner:
JD led: (33,660 / 61,545 / 4,795)
10D led: (31,053 / 64,113 / 4,834)
In this case, it's better to lead the JD. But very close. In fact, too close to be definitive. While I am confident my program plays correctly when S1 calls alone (there just isn't much choice in how to play), I am less sure it's optimal when all 4 players are involved. So here it's a toss-up.
This result leads to a larger question: when do you call alone, despite having 8 or 9 pts, because your chances of scoring a point (or a sweep) are actually GREATER calling alone than with your partner? I know some expert players have ideas on this topic, but I have not come across any guidelines, so for the moment I am simply collecting evidence of hands (like this one) which break the general rule, to hopefully eventually come up with such guidance.
Suffice to say that for this hand, if S1 calls alone it's best to lead the JD, but if S1 calls with partner it's unclear which lead is best.
...................................................
One thing I did note is that my program leads high trump on the second trick (so S1 leads the KD in this instance), rather than the off-suit Ace, as Irish suggested. So I compared the two scenarios.
100,000 totally random hands, looking at the ones where S1 calls alone (about 7%).
S1 calls alone:
lead high trump, 2nd trick:
(7,105 / 15,133 / 2,727) R1 calls
(14,211 / 31,137 / 3,552) R2 calls
lead off-suit A, 2nd trick (because don't have boss trump):
(6,251 / 16,109 / 2,605) R1 calls
(12,570 / 32,892 / 3,438) R2 calls
* be sure to compare R1 vs R1, R2 vs R2!
In both cases (R1 and R2 calls), the euchre rate is higher when high trump is led on the second trick, but this is more than offset by the fact that the loner success rate is also higher. So, in general, it's best to lead the highest trump on the 2nd trick if you are in S1 and going alone, even if you have an off-suit Ace.
I also tested this for this specific hand for S1. I found an EV of 1.61 when the highest trump was led on the second trick vs. an EV of 1.52 when the off-suit Ace was led.
For what it's worth...
-
irishwolf
- Posts: 1307
- Joined: Tue Apr 24, 2018 9:33 pm
Here are my comments:
I totally agree if going alone, lead the KD to the 2nd trick.
I cannot buy that euchre rate of leading the JD followed by KD is close to leading AS. This is why:
1) Leading JD, the AS is that the only potentially euchre(s) will result when Dealer has 3/4 trumps and has to win the with a Club lead. Having 3/4 is 4.2%. But also has to have that club to (S2/S4 - has boss club). It will not be 4.2% euchres but less than 2%. But you have to add to this when S2/S4 each have two trumps but consider they have to work together to not lead Hearts or Spades, only clubs. So what might that be, maybe a another 1%?
2) However, leading JD followed by KD - and Dealer has AD & QD or JH - just two plus the up card with Clubs doubleton S1 gets euchred. S4 only needs 2 of 3 unknown, 15% has two + upcard. but also Club doubleton or his partner has boss clubs. This is now added to number 1_. above. So S4 having 9D AD QD, 9D JH QD, 9D JH AD COMBOS with boss clubs results in a euchre when S1 leads KD. S4 wins 2nd trick and leads clubs S1 has to use his last trump but S4 still has a trump and a club. HE'S Dead! What that is statistically should be about 33% of 15% plus #1 above. Thus, I think closer to 6 to 8% euchre rate when leading KD to 2nd trick. So is S2/S4 playing correctly? If KD is led and S4 has AD/JH or JH/QD is he leading back to take S1 last trump then leading CLubs. S1 can't get back in. Euchre rate will be higher do that! Some issues, I think. Simulator is not OPTIMIZING by leading the KD, IMO.
3) Now for leading the 10D taking going alone but for 1 point. There is no need to lead low - result is the same leading JD followed by the AS. The Euchre rate is going to be less than 2%. It is not rocket science. Can't be 4.2% as S4 has to have Clubs. So 1/3 or slightly more. There are more clubs out than Hearts or Spades. So more than 33% of 4.2%
I am not sure taking your partner along would be better if going for 1 point. I know for FACT the euchre rate is less than 2%. And the euchre rate will be higher leading the 10D.
IRISH
I totally agree if going alone, lead the KD to the 2nd trick.
I cannot buy that euchre rate of leading the JD followed by KD is close to leading AS. This is why:
1) Leading JD, the AS is that the only potentially euchre(s) will result when Dealer has 3/4 trumps and has to win the with a Club lead. Having 3/4 is 4.2%. But also has to have that club to (S2/S4 - has boss club). It will not be 4.2% euchres but less than 2%. But you have to add to this when S2/S4 each have two trumps but consider they have to work together to not lead Hearts or Spades, only clubs. So what might that be, maybe a another 1%?
2) However, leading JD followed by KD - and Dealer has AD & QD or JH - just two plus the up card with Clubs doubleton S1 gets euchred. S4 only needs 2 of 3 unknown, 15% has two + upcard. but also Club doubleton or his partner has boss clubs. This is now added to number 1_. above. So S4 having 9D AD QD, 9D JH QD, 9D JH AD COMBOS with boss clubs results in a euchre when S1 leads KD. S4 wins 2nd trick and leads clubs S1 has to use his last trump but S4 still has a trump and a club. HE'S Dead! What that is statistically should be about 33% of 15% plus #1 above. Thus, I think closer to 6 to 8% euchre rate when leading KD to 2nd trick. So is S2/S4 playing correctly? If KD is led and S4 has AD/JH or JH/QD is he leading back to take S1 last trump then leading CLubs. S1 can't get back in. Euchre rate will be higher do that! Some issues, I think. Simulator is not OPTIMIZING by leading the KD, IMO.
3) Now for leading the 10D taking going alone but for 1 point. There is no need to lead low - result is the same leading JD followed by the AS. The Euchre rate is going to be less than 2%. It is not rocket science. Can't be 4.2% as S4 has to have Clubs. So 1/3 or slightly more. There are more clubs out than Hearts or Spades. So more than 33% of 4.2%
I am not sure taking your partner along would be better if going for 1 point. I know for FACT the euchre rate is less than 2%. And the euchre rate will be higher leading the 10D.
IRISH
Last edited by irishwolf on Wed Sep 07, 2022 9:04 pm, edited 2 times in total.
-
RedDuke
- Posts: 473
- Joined: Sun Dec 02, 2018 2:22 am
First of all, you made a huge mistake not going alone. There is absolutely nothing that your partner can possibly have that will change the outcome of this hand:
Dealer has all 4 trumps that you don't - you get euchred.
Either opponent has two trumps, one of which is either the left or the ace - it's a one-point hand unless that opponent is a total moron.
Partner has either the left, ace or queen - you take it or the partner takes it, resulting in the hand having exactly the same outcome as if you removed these cards from play.
There is nothing your partner can do to change any of these outcomes. Having him in it might actually hurt your chances depending on exactly what he has. So there is literally no point in bringing him along.
Playing the ace first instead of the right was your other big mistake. The scenario you don't want is for your opponents to have voids to trump your aces. This is especially true because you don't know where the three cards that can stop you are - the queen, ace, and left. What you want to do then is try to reduce the trumps that the opponents have before you play your aces. The best way to do this is by playing the right. This sort of gives you an idea of what they have too. Ideally, the dealer's partner will play the ace or the left, then you don't have to worry about that anymore. In that case, I'd probably play the king and see if it gets taken by the remaining trouble card. Use either your aces or the 10-trump to retake control and finish off the hand.
Playing your ace first is just asking the opponents to take a trick that you could have taken later in the hand if you took out their trump first. The best way to take out their trump is to lead the right.
I can't think of too many cases in which I won't lead the right in first seat. Maybe if the opponent called and I'm trying to sandbag him... if I call, the only time I won't lead it is if it's my only trump.
Dealer has all 4 trumps that you don't - you get euchred.
Either opponent has two trumps, one of which is either the left or the ace - it's a one-point hand unless that opponent is a total moron.
Partner has either the left, ace or queen - you take it or the partner takes it, resulting in the hand having exactly the same outcome as if you removed these cards from play.
There is nothing your partner can do to change any of these outcomes. Having him in it might actually hurt your chances depending on exactly what he has. So there is literally no point in bringing him along.
Playing the ace first instead of the right was your other big mistake. The scenario you don't want is for your opponents to have voids to trump your aces. This is especially true because you don't know where the three cards that can stop you are - the queen, ace, and left. What you want to do then is try to reduce the trumps that the opponents have before you play your aces. The best way to do this is by playing the right. This sort of gives you an idea of what they have too. Ideally, the dealer's partner will play the ace or the left, then you don't have to worry about that anymore. In that case, I'd probably play the king and see if it gets taken by the remaining trouble card. Use either your aces or the 10-trump to retake control and finish off the hand.
Playing your ace first is just asking the opponents to take a trick that you could have taken later in the hand if you took out their trump first. The best way to take out their trump is to lead the right.
I can't think of too many cases in which I won't lead the right in first seat. Maybe if the opponent called and I'm trying to sandbag him... if I call, the only time I won't lead it is if it's my only trump.
-
raydog
- Posts: 260
- Joined: Thu Sep 16, 2021 6:56 pm
Irish, you are correct. My program was playing incorrectly when S4 had 3 trump. After S4 wins the 2nd trick with the AD (or JH), they should lead their remaining trump card (which is the boss trump) on the 3rd trick, which gives a great chance to euchre if they then lead a C on the 4th trick.
I tested this as a general rule when defending against a lone, and it is the better play from all seats.
I incorporated this new rule, then re-ran the scenarios:
JD led, 1st trick, then:
KD led, trick 2: (37/971 / 54,148 / 7,881) EV = +1.90
AS led, trick 2: (28,655 / 70,072 / 1,273) EV = +1.82
Over 6 times as many euchres when the KD is lead on the 2nd trick. BUT, the surplus of sweeps for 4 pts. still makes leading the KD the better play.
I also looked again at leading the JD vs. leading the 10D (when S1 bids alone and is looking for just 1 pt.)
JD led (then KD): (37,920 / 54,111 / 7,969)
10D led: (6,933 / 89,026 / 4,041)
Euchre rate is still just over 4% when leading the 10D. I had a look at when S1 was euchred when leading the 10D, and it boiled down to 2 scenarios:
1) S4 has 4 trump
2) S4 has 3 trump + 2 clubs
How often do these scenarios occur? I calculated S4 as having 4 trump 1.23% of the time and having 3 trump 15.93% of the time. To check this, I ran 100,000 random hands [I only fixed S1's hand and the turn card] and found S4 had 4 trump 1.26% of the time and 3 trump 15.75% of the time. So my calculation seems correct.
But what about S4 having 3 trump AND 2 clubs? Thats very hard to calculate, so I simply ran the 100,000 random hands again and counted the number of times S4 had 3 trump + 2 clubs. It turned out to be 2.88% of the time.
Notice that the odds of scenario 1) or 2) occurring is 1.26% + 2.88% = a bit over 4%, which matches the euchre rate I found when S1 leads the 10D. I'm not sure why you are so sure the euchre rate should be less than 2%.
One final thought. If the goal is to gain that final point needed for victory, the best play is for S1 to call ALONE, then lead the JD followed by the AS. This has the lowest euchre rate of all, a bit under 1.3%, so offers the best odds of winning the game. The reason the AS lead is better is that it forces S4 to use a 2nd trump to win the trick (when S4 has 3 trump + 2 clubs), thus avoiding letting S4 take control of the hand. If S4 is allowed to draw out all the trump and lead clubs, S1's AS is discarded worthlessly on a club lead.
I tested this as a general rule when defending against a lone, and it is the better play from all seats.
I incorporated this new rule, then re-ran the scenarios:
JD led, 1st trick, then:
KD led, trick 2: (37/971 / 54,148 / 7,881) EV = +1.90
AS led, trick 2: (28,655 / 70,072 / 1,273) EV = +1.82
Over 6 times as many euchres when the KD is lead on the 2nd trick. BUT, the surplus of sweeps for 4 pts. still makes leading the KD the better play.
I also looked again at leading the JD vs. leading the 10D (when S1 bids alone and is looking for just 1 pt.)
JD led (then KD): (37,920 / 54,111 / 7,969)
10D led: (6,933 / 89,026 / 4,041)
Euchre rate is still just over 4% when leading the 10D. I had a look at when S1 was euchred when leading the 10D, and it boiled down to 2 scenarios:
1) S4 has 4 trump
2) S4 has 3 trump + 2 clubs
How often do these scenarios occur? I calculated S4 as having 4 trump 1.23% of the time and having 3 trump 15.93% of the time. To check this, I ran 100,000 random hands [I only fixed S1's hand and the turn card] and found S4 had 4 trump 1.26% of the time and 3 trump 15.75% of the time. So my calculation seems correct.
But what about S4 having 3 trump AND 2 clubs? Thats very hard to calculate, so I simply ran the 100,000 random hands again and counted the number of times S4 had 3 trump + 2 clubs. It turned out to be 2.88% of the time.
Notice that the odds of scenario 1) or 2) occurring is 1.26% + 2.88% = a bit over 4%, which matches the euchre rate I found when S1 leads the 10D. I'm not sure why you are so sure the euchre rate should be less than 2%.
One final thought. If the goal is to gain that final point needed for victory, the best play is for S1 to call ALONE, then lead the JD followed by the AS. This has the lowest euchre rate of all, a bit under 1.3%, so offers the best odds of winning the game. The reason the AS lead is better is that it forces S4 to use a 2nd trump to win the trick (when S4 has 3 trump + 2 clubs), thus avoiding letting S4 take control of the hand. If S4 is allowed to draw out all the trump and lead clubs, S1's AS is discarded worthlessly on a club lead.
-
irishwolf
- Posts: 1307
- Joined: Tue Apr 24, 2018 9:33 pm
Ray,
I am in 100% agreement with your latest Post. (P.S. - I know for FACT the euchre rate is less than 2%. i did not check my calculation, yours is correct. And the euchre rate will be higher leading the 10D.) BUT YOU NAILED IT! GOOD JOB!
Of course, if you are down and need a loner, lead the KD to the 2nd trick. AGREED!
Taking your partner along, he can only - not an option as S1 will score a point all except when the Dealer has 9D QD JH AD - 1.23%. I do see anomaly when S1 has two trumps, void in Spades with a Club to lead 3rd trick while S4 has two trumps with 2 Boss Clubs a void in Hearts that results in a euchre. So small, who cares.
And I agree about leading JD vs 10D.
GOOD JOB!
IRISH
I am in 100% agreement with your latest Post. (P.S. - I know for FACT the euchre rate is less than 2%. i did not check my calculation, yours is correct. And the euchre rate will be higher leading the 10D.) BUT YOU NAILED IT! GOOD JOB!
Of course, if you are down and need a loner, lead the KD to the 2nd trick. AGREED!
Taking your partner along, he can only - not an option as S1 will score a point all except when the Dealer has 9D QD JH AD - 1.23%. I do see anomaly when S1 has two trumps, void in Spades with a Club to lead 3rd trick while S4 has two trumps with 2 Boss Clubs a void in Hearts that results in a euchre. So small, who cares.
And I agree about leading JD vs 10D.
GOOD JOB!
IRISH