WHAT WOULD YOU DO?

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WHAT WOULD YOU DO?
You are going alone QS was the upcard and you now hold JS QS 10S AD KD
A club is led, S3 has a void and trumps with AS. What will you now do, and justify your decision?
A club is led, S3 has a void and trumps with AS. What will you now do, and justify your decision?

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This is probably the one scenario that you really don't want to see in this spot.
You could over trump S3 and pray that the ace was his big trump. That's probably what I'd do in this spot. But it's risky  if either opponent has the left or the king, you're now in trouble. It might be best to throw off here and play one of your diamonds, giving up on getting the four points. I can't prove which option would be better. My gut is telling me to overtrump the ace and lead back one of the diamonds, but I can't prove I'm right.
You could over trump S3 and pray that the ace was his big trump. That's probably what I'd do in this spot. But it's risky  if either opponent has the left or the king, you're now in trouble. It might be best to throw off here and play one of your diamonds, giving up on getting the four points. I can't prove which option would be better. My gut is telling me to overtrump the ace and lead back one of the diamonds, but I can't prove I'm right.

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I am overtrumping. I am not ready to give up so quickly. Yes you could get euchured but there are three trump left with a decent chance 12 are buried. I would then play the AD.

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So Wolfhills, since you over trumped the AS, and not ready to give up on your loner attempt  having led the AD. Suppose it went through on trick 2. What will you lead to trick 3? What have you assumed now about the 9S, KS and JC still unknown? What are your odds of a loner?

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I agree with RedDuke, this situation sucks.
It is still a power hand and I would play with power.
But that power is defined as near certainty of taking 3 tricks.
The probability of taking 5 tricks is rather weak.
There are 3 trump that if protected or available because of a void (Irish’s scenario) then my short cut to “guesstimate” probability is:
3 trump, 2 parlays, each parlay equivalent to 1/2 of a stopper.
So 3 stopper equivalents.
Each stopper has a 55.5% probability of being in an opponents hand.
That leaves a 45% chance of Loner success for 1 stopper equivalent.
45% x 45% = 20% for 2 stoppers and about
20% x 45% or 10% for 3.
The point is that compared to my average 30% Loner 4 point success rate this is a weak Loner attempt. (But with 3 near certain tricks, there has to be a good reason not to try, IMO & IMX).
Because this is a weak Loner hand, I switch quickly to preventing euchre, when confronted with adverse card distribution like this hand. This is especially important with this near 3 certain trick hand, there is no trick to spare.
The power in this hand is 3 trump with a Right. If I over trump with the Right, I only have 2 weak trump vs. 3 in the wild. I still have power in the number of trump. So I will lead the Qs expecting to draw out the remaining trump and preserving my slim opportunity at 4 points. I would estimate the probability of an opponent still having 2 trump at less than 15%. That is acceptable IMO with the adverse card distribution of this post.
If I lead the Ad, I would still lose if trump breaks against me (roughly the same as leading Qs). In addition I have an additional added loser now since the Ad may be trumped. So my analysis is the Ad lead is weaker than the Qs lead.
I have practical reasons for consistently leading from strength with a strong hand. I don’t get caught in overthinking, vacillating and making a random decision. IMO, a good decision is better than a random decision. I make less mistakes. We don’t have Wes’ “Mythical Euchre Tester”, so I rely on card counting and evaluating adverse card distributions. If one of my opponents has 3 trump, I Immediately move on, I don’t worry about being euchred. The bumper sticker says it all, “it happens.” I can readily observe adverse card distributions. I cannot properly evaluate probabilities, especially in real time. Simplifying allows me to move on to the next hand with a clear mind.
I also have an emotional reason for leading strength. I like it! I would rather go down swinging than taking a called 3rd strike.
It is still a power hand and I would play with power.
But that power is defined as near certainty of taking 3 tricks.
The probability of taking 5 tricks is rather weak.
There are 3 trump that if protected or available because of a void (Irish’s scenario) then my short cut to “guesstimate” probability is:
3 trump, 2 parlays, each parlay equivalent to 1/2 of a stopper.
So 3 stopper equivalents.
Each stopper has a 55.5% probability of being in an opponents hand.
That leaves a 45% chance of Loner success for 1 stopper equivalent.
45% x 45% = 20% for 2 stoppers and about
20% x 45% or 10% for 3.
The point is that compared to my average 30% Loner 4 point success rate this is a weak Loner attempt. (But with 3 near certain tricks, there has to be a good reason not to try, IMO & IMX).
Because this is a weak Loner hand, I switch quickly to preventing euchre, when confronted with adverse card distribution like this hand. This is especially important with this near 3 certain trick hand, there is no trick to spare.
The power in this hand is 3 trump with a Right. If I over trump with the Right, I only have 2 weak trump vs. 3 in the wild. I still have power in the number of trump. So I will lead the Qs expecting to draw out the remaining trump and preserving my slim opportunity at 4 points. I would estimate the probability of an opponent still having 2 trump at less than 15%. That is acceptable IMO with the adverse card distribution of this post.
If I lead the Ad, I would still lose if trump breaks against me (roughly the same as leading Qs). In addition I have an additional added loser now since the Ad may be trumped. So my analysis is the Ad lead is weaker than the Qs lead.
I have practical reasons for consistently leading from strength with a strong hand. I don’t get caught in overthinking, vacillating and making a random decision. IMO, a good decision is better than a random decision. I make less mistakes. We don’t have Wes’ “Mythical Euchre Tester”, so I rely on card counting and evaluating adverse card distributions. If one of my opponents has 3 trump, I Immediately move on, I don’t worry about being euchred. The bumper sticker says it all, “it happens.” I can readily observe adverse card distributions. I cannot properly evaluate probabilities, especially in real time. Simplifying allows me to move on to the next hand with a clear mind.
I also have an emotional reason for leading strength. I like it! I would rather go down swinging than taking a called 3rd strike.

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This is what I would do, maybe I am more conservative player and would play it safe.
As soon as the the AS is played a decision has to be made do I want to give up going for a loner, play conservative. Or to be stubborn and still hope for a loner that has low odds of success? Can you afford being euchred?
If you over trump you have now weakened your chance for getting out of this hand without being euchred, especially by leading QS. One opponent could have both KS/JC which is double with two opponents. And we know shit happens in euchre. Your chance of you winning all five tricks is very low if you over trump and zero if you don't.
GIVE IT UP would be my advice! You don't have to over trump just because you can. Bowers are for pulling trump from opponents, not for over trumping.
If you decide to be happy with one point, then slough the AD and you still hold JS QS 10S KD. The opponent gets fooled to thinking you do not hold diamonds and will lead a heart which is your void. Even if you get over trumped S1 has to lead into your strong hand and you can over trump the remaining KS or JC with your 10S as best man. S1 will also be reluctant to lead a diamond. You will get out with one point.
"Because this is a weak Loner hand, I switch quickly to preventing euchre, when confronted with adverse card distribution like this hand. This is especially important with this near 3 certain trick hand, there is no trick to spare.
The power in this hand is 3 trump with a Right. If I over trump with the Right, I only have 2 weak trump vs. 3 in the wild. I still have power in the number of trump. So I will lead the Qs expecting to draw out the remaining trump and preserving my slim opportunity at 4 points."
~Irishwolf
As soon as the the AS is played a decision has to be made do I want to give up going for a loner, play conservative. Or to be stubborn and still hope for a loner that has low odds of success? Can you afford being euchred?
If you over trump you have now weakened your chance for getting out of this hand without being euchred, especially by leading QS. One opponent could have both KS/JC which is double with two opponents. And we know shit happens in euchre. Your chance of you winning all five tricks is very low if you over trump and zero if you don't.
GIVE IT UP would be my advice! You don't have to over trump just because you can. Bowers are for pulling trump from opponents, not for over trumping.
If you decide to be happy with one point, then slough the AD and you still hold JS QS 10S KD. The opponent gets fooled to thinking you do not hold diamonds and will lead a heart which is your void. Even if you get over trumped S1 has to lead into your strong hand and you can over trump the remaining KS or JC with your 10S as best man. S1 will also be reluctant to lead a diamond. You will get out with one point.
"Because this is a weak Loner hand, I switch quickly to preventing euchre, when confronted with adverse card distribution like this hand. This is especially important with this near 3 certain trick hand, there is no trick to spare.
The power in this hand is 3 trump with a Right. If I over trump with the Right, I only have 2 weak trump vs. 3 in the wild. I still have power in the number of trump. So I will lead the Qs expecting to draw out the remaining trump and preserving my slim opportunity at 4 points."
~Irishwolf

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I'm also giving up here except when super desperate, say down 96/97 or down big.
For those who are curious, after S1 leads a club and S3 trumps in with the AS and the hero presumably overtrumps with the Right, what is the probability that the enemy will have at least 1 higher trump left?
First figure out how many cards have been exposed from the dealer's perspective:
The dealer has seen 6 cards, the 5 in his hand plus the one he discarded. We've also seen the Club lead by seat 1 and the AS from S3. That's 8 cards exposed out of 24 possible. So 16 cards left unseen.
Ok, now figure out how many combos the enemy has when they have neither the KS or the JC:
2C0 x 14C8 = 3003 combos.
Then figure out total possible combos:
16C8 = 12,870 total possible combos.
Now we have all the info we need. The probability that the enemy team will have at least one higher trump if we overtrump with the Right:
(12,870  3003)/12870 = 76.67%
Note: This number is not precisely accurate. It doesn't factor in how S2's range changes after he passes in the first round. For example, if S2 had JcKs9sXX he's obviously not passing, but this analysis erroneously assumes this hand IS in his passing range. Same could be said for a lot of Left+1+an off ace combos S2 could have or other 2 trump + two ace hands, etc. Becuz of this we should expect the actual probability that the enemy has at least 1 of those higher trump cards to be slightly higher than 76.67%, becuz in real life, S2 passing means he can't have those hands in his range which increases the chance the Ks or Jc is in enemy hands. Either way, roughly 77% is a great approximation.
Note 2: But aren't there some calling combos falsely assumed in S1 and S3's passing range too? Yes, but not nearly as many as S2, so the above dynamic still holds.

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Let’s continue this discussion. I am not aggressively pursuing 4 points. As I stated, this is a weak 4 point hand. I am playing my cards aggressively because it still has an 85% chance of success and thanks to Wes’ analysis an improved 23% chance of 4 points (compared to my analysis before the Ac was played of 10%).
Wes’ 77% probability of my Qc being overtrumped does not mean that I have been euchred. Opponents still need another trick. So more than likely, I will trump L3 (Lead 3) and then my AKd will run the table. I estimated that 15% of the time Opponents will have another trump.
So the EV at this point is:
23% 4 point Loner, based on Wes’ calcs
62% 1 point
15% euchred, based on Richard’s calcs
23%x4+62%x115%x2= .92+.62.3= 1.24
EV of tossing Ad or Kd, even at 100% success= 1.00
So where did I go wrong?
Wes’ 77% probability of my Qc being overtrumped does not mean that I have been euchred. Opponents still need another trick. So more than likely, I will trump L3 (Lead 3) and then my AKd will run the table. I estimated that 15% of the time Opponents will have another trump.
So the EV at this point is:
23% 4 point Loner, based on Wes’ calcs
62% 1 point
15% euchred, based on Richard’s calcs
23%x4+62%x115%x2= .92+.62.3= 1.24
EV of tossing Ad or Kd, even at 100% success= 1.00
So where did I go wrong?

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Bottom line is that you double your chance of getting euchred by over trumping the AS and leading the QS. The opponent who has the KS/JC combo (although low percentage) then gets your 10S and a lead back of hearts or clubs you are euchred. Let it go!
You do not have to expose yourself to worst odds of making a point by over trumping.
You do not have to expose yourself to worst odds of making a point by over trumping.

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Granted, I double my odds of getting euchred. But avoiding an euchre is not the name of the game. Getting to 10 points first is the goal of the game. An EV of 1.24 gets me there more quickly than an EV of 1.00.irishwolf wrote: ↑Sat Feb 08, 2020 9:36 pmBottom line is that you double your chance of getting euchred by over trumping the AS and leading the QS. The opponent who has the KS/JC combo (although low percentage) then gets your 10S and a lead back of hearts or clubs you are euchred. Let it go!
You do not have to expose yourself to worst odds of making a point by over trumping.
Do you disagree with my EV evaluation?

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The math looks good. When the math looks good there is only one way to challenge an EV argument, and that's to attack the assumptions. Obviously your weakest assumption is the 15% euchre rate becuz it's just a guesstimate. Intuitively I'm not sure what the euchre rate is after we overtrump the AS and lead the QS. Is it 15%? 20%? 25%? IDK. And this is a very difficult math problem becuz there are multiple ways to get euchred depending how the 9SKSJC are spread out, and depending on who's void in diamonds or not. This is the kinda spot I would want to simulate damnit!!Richardb02 wrote: ↑Sat Feb 08, 2020 8:03 pmLet’s continue this discussion. I am not aggressively pursuing 4 points. As I stated, this is a weak 4 point hand. I am playing my cards aggressively because it still has an 85% chance of success and thanks to Wes’ analysis an improved 23% chance of 4 points (compared to my analysis before the Ac was played of 10%).
Wes’ 77% probability of my Qc being overtrumped does not mean that I have been euchred. Opponents still need another trick. So more than likely, I will trump L3 (Lead 3) and then my AKd will run the table. I estimated that 15% of the time Opponents will have another trump.
So the EV at this point is:
23% 4 point Loner, based on Wes’ calcs
62% 1 point
15% euchred, based on Richard’s calcs
23%x4+62%x115%x2= .92+.62.3= 1.24
EV of tossing Ad or Kd, even at 100% success= 1.00
So where did I go wrong?
The other assumption that is slightly weak is the 23% 4 pt sweep number as that is not a conservative assumption. This number is obviously based on my math of the enemy team having a higher trump approx 77% of the time. But like I said, that number is actually an underestimate due to the range distortion problem of S2.
For instance, Let's say S2 is calling in the first round if he has 3 trump, 2 trump + 2 off aces or Left + 1 + an off ace when he only blocks 1 out of 3 second round suits.
3 trump: 3C3 x 13C2 = 78 combos
2 trump + 2 Aces = 3C2 x 2C2 x 11C1 = 33 combos  2 euchre hand combos = 31 combos
L+1+off ace = 1C1 x 2C1 x 2C1 x 2C0 x 9C2 = 144 combos (2C0 represents the exclusion of both red bowers to eliminate most of the combos S2 could have that block 2+ out of 3 suits)
Total S2 calling combos = 253
Total possible combos = 16C5 = 4,368
So S2's passing range is distorted by 253/4368 = 5.79%
BTW S1 and S3's passing range are barely distorted at all.
S1 is only calling with JcKs9sXX if he has no 2nd round hand, which means if he has 1 more club he's not calling and if he has good defense with all suits blocked he's aslo not calling. I count only 27 calling combos that are erroneously in S1's passing range. That's a distortion rate of 27/4368 = .62%. And S3 is even less than that. S3 is only calling if he has 4 trump (JcAsK9sX) or let's also say 3 trump + 2 aces (even tho I often bag most of those hands, let's say this S3 doesn't for the sake of argument ). Well that's only 17 total combos so S3's range is only distorted by 17/4368 = .39%. I suppose we could net that out. 253  27  17 = 209, meaning the enemy's range is 209/4368 = 4.78% stronger than indicated by my initial combo math since my initial math is based on a random distribution that isn't quite random.
So the enemy will have a higher trump somewhat greater than 77%. Maybe it's 80% instead and thus your 4 pt sweep rate is 20%. IDK, I'm just guesstimating.
That said, let's say we assume a 20% euchre rate, and only a 20% 4 pt sweep rate, making this new EO model significantly more conservative than your model:
EO of overtrumping Ace and leading the Qs: (.20 x 4) + (.6 x 1) + (.20 x 2) = 1
EO of playing it safe: "EV of tossing Ad or Kd, even at 100% success= 1.00"
Since we both know that playing it safe will not actually be 100% successful (btw that was a good example of appropriately using a conservative assumption!), these new more conservative numbers still suggest that NOT playing it safethat going for the gusto and overtrumping the AS and leading the QS is the higher EV play.
I think there's still room for rational skepticism here, but man Richard, I think you may be on to something. It all boils down to what is the true euchre rate of this aggressive strategy (wish we had that simulator). If one believes it is 15% then certainly they should go for the gusto. If I had to guess I would say there is a greater than 50% chance Richard is right about this.

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I also suspect the EV for over trumping and leading the QS is higher than being conservative, not trumping. Any time the loners made is greater than the euchre rate, of course the EV will always be greater. However, statistical calculations and EV values are for the Longrun, many, many hands played. For the short run, it depends on the situation, how have things been going, the opponents, where are you scorewise in the game, etc. etc. Can you afford a euchre? How bad do you need the 4 points? Each point in euchre is worth, roughly, 10% to the goal of winning. A euchre is a 30 point (%) swing and at score of 6 to 6 or 7 to 7, the game is too short to make up the difference. At low score or way down, of course go for it!
I see players over trump just because they can with the right bower when they should not. I put the situation with the AS because most players will over trump without thinking about what it does and how the situation changes.
Great discussion!
I see players over trump just because they can with the right bower when they should not. I put the situation with the AS because most players will over trump without thinking about what it does and how the situation changes.
Great discussion!

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Thank you Wes and Irish for revisiting this topic and not just for this post. I have noticed a shift on this forum away from aggressive play and towards finesse play. So this was an strategic discussion.
I am not nitpicking but let’s review my 15% expected euchre rate. I have confidence in the 15%. I extracted this discussion from one of my previous posts:
LOL. I just reread the last line of my quote, “ I also have an emotional reason for leading strength. I like it! I would rather go down swinging than taking a called 3rd strike.”
I would like to add, “Ordering Alone psyches out Opponent, earning 4 totally deflates them. Euchre favors the aggressive player.” “Euchre favors the aggressive player also paraphrases OE lessons.
Quoting from my post, “ The power in this hand is 3 trump with a Right. If I over trump with the Right, I only have 2 weak trump vs. 3 in the wild. I still have power in the number of trump.”
Statistically there are 3 trump in the wild and the wild consists of 16 cards. One of my opponents has to have 2 of the 3 trump to euchre me, if I lead the Qs. There are other remote possibilities, but they are remote, not worthy of calculating. I my simplified approach, each opponent has A 3/16 chance of having a trump with his 2nd card and then a 2/15 chance with cards 3 through 5. So for quick math I used:
3/16 x 2/15x3 = 3/16 x 6/15 = .1875 x .4 = 7.5%, 15% since there are 2 Opps
More accurate is 3/16 x (113/15^3) = .1875 x .349 = 5.9%, 12% for 2 Opps
So I suggest a 15% euchre rate is a conservative estimate.
Secondly I expected a 10% 4 point success rate before the first round of cards. Once the As was played, the card distribution changed in our favor for an euchre. I instinctively doubled my expectation to 20%. Wes’ calcs matched a 23% expectation. So I have a high confidence in EV:
65% x 1 + 20% x 4 15% x 2 = .65 +.8  .3 = 1.15
I am well aware that my aggressive style influences my conclusions. So I value any feedback that challenges my estimates and conclusions.
I am not nitpicking but let’s review my 15% expected euchre rate. I have confidence in the 15%. I extracted this discussion from one of my previous posts:
Richardb02 wrote: ↑Sat Feb 08, 2020 11:36 amI agree with RedDuke, this situation sucks.
It is still a power hand and I would play with power.
But that power is defined as near certainty of taking 3 tricks.
The power in this hand is 3 trump with a Right. If I over trump with the Right, I only have 2 weak trump vs. 3 in the wild. I still have power in the number of trump. So I will lead the Qs expecting to draw out the remaining trump and preserving my slim opportunity at 4 points. I would estimate the probability of an opponent still having 2 trump at less than 15%. That is acceptable IMO with the adverse card distribution of this post.
If I lead the Ad, I would still lose if trump breaks against me (roughly the same as leading Qs). In addition I have an additional added loser now since the Ad may be trumped. So my analysis is the Ad lead is weaker than the Qs lead.
I have practical reasons for consistently leading from strength with a strong hand. I don’t get caught in overthinking, vacillating and making a random decision. IMO, a good decision is better than a random decision. I make less mistakes. We don’t have Wes’ “Mythical Euchre Tester”, so I rely on card counting and evaluating adverse card distributions. If one of my opponents has 3 trump, I Immediately move on, I don’t worry about being euchred. The bumper sticker says it all, “it happens.” I can readily observe adverse card distributions. I cannot properly evaluate probabilities, especially in real time. Simplifying allows me to move on to the next hand with a clear mind.
I also have an emotional reason for leading strength. I like it! I would rather go down swinging than taking a called 3rd strike.
LOL. I just reread the last line of my quote, “ I also have an emotional reason for leading strength. I like it! I would rather go down swinging than taking a called 3rd strike.”
I would like to add, “Ordering Alone psyches out Opponent, earning 4 totally deflates them. Euchre favors the aggressive player.” “Euchre favors the aggressive player also paraphrases OE lessons.
Quoting from my post, “ The power in this hand is 3 trump with a Right. If I over trump with the Right, I only have 2 weak trump vs. 3 in the wild. I still have power in the number of trump.”
Statistically there are 3 trump in the wild and the wild consists of 16 cards. One of my opponents has to have 2 of the 3 trump to euchre me, if I lead the Qs. There are other remote possibilities, but they are remote, not worthy of calculating. I my simplified approach, each opponent has A 3/16 chance of having a trump with his 2nd card and then a 2/15 chance with cards 3 through 5. So for quick math I used:
3/16 x 2/15x3 = 3/16 x 6/15 = .1875 x .4 = 7.5%, 15% since there are 2 Opps
More accurate is 3/16 x (113/15^3) = .1875 x .349 = 5.9%, 12% for 2 Opps
So I suggest a 15% euchre rate is a conservative estimate.
Secondly I expected a 10% 4 point success rate before the first round of cards. Once the As was played, the card distribution changed in our favor for an euchre. I instinctively doubled my expectation to 20%. Wes’ calcs matched a 23% expectation. So I have a high confidence in EV:
65% x 1 + 20% x 4 15% x 2 = .65 +.8  .3 = 1.15
I am well aware that my aggressive style influences my conclusions. So I value any feedback that challenges my estimates and conclusions.