QUIZ  HAND 342

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QUIZ  HAND 342
You are playing a game, all good players. The KC is the up card and you just ordered the dealer from S3. You hold JC JS QC AS QD. S1 played the KD, S2 the AD, S3 the QD and Dealer played the 9S.
S2 now leads the AH.
PART I: What will you play to this trick?
Why do you think this is the best card to play?
What is your chance of getting euchred even though you chose this card?
PART II: Suppose what you played won the trick! What will you play to trick 3?
Why did you play this card?
Did you reassess your chance of being euchre choosing this card?
S2 now leads the AH.
PART I: What will you play to this trick?
Why do you think this is the best card to play?
What is your chance of getting euchred even though you chose this card?
PART II: Suppose what you played won the trick! What will you play to trick 3?
Why did you play this card?
Did you reassess your chance of being euchre choosing this card?

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 Joined: Wed Jun 13, 2018 3:03 pm
Looks like nobody wants to do your quiz which is rather sad, becuz I believe I know what you're up to, and I am certain you are right!irishwolf wrote: ↑Fri Feb 14, 2020 10:56 amYou are playing a game, all good players. The KC is the up card and you just ordered the dealer from S3. You hold JC JS QC AS QD. S1 played the KD, S2 the AD, S3 the QD and Dealer played the 9S.
S2 now leads the AH.
PART I: What will you play to this trick?
Why do you think this is the best card to play?
What is your chance of getting euchred even though you chose this card?
PART II: Suppose what you played won the trick! What will you play to trick 3?
Why did you play this card?
Did you reassess your chance of being euchre choosing this card?
The answer to Question 9 on the "Rate your euchre skills" quiz for this site is indeed wrong!

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LOL, maybe just coincidence?
"Looks like nobody wants to do your quiz which is rather sad, becuz I believe I know what you're up to, and I am certain you are right!"
It too makes me wonder? Two "possible right answers" (all depends on assumptions). In reality, one is better than the other! And the right answer is the one that makes a point!
~Irishwolf
"Looks like nobody wants to do your quiz which is rather sad, becuz I believe I know what you're up to, and I am certain you are right!"
It too makes me wonder? Two "possible right answers" (all depends on assumptions). In reality, one is better than the other! And the right answer is the one that makes a point!
~Irishwolf

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 Joined: Wed Jun 13, 2018 3:03 pm
I don't think there's two possible right answers here becuz you effectively eliminated the "depends on assumptions" issue by saying "all good players", altho I'll admit I may be taking some liberties on the "good player" part. For instance, in doing my analysis of S2's range I assumed he can't have 2 trump in his hand becuz a good player is not passing 2 trump + 2 aces from 2S, 1st rd. I think that's the only possible point of contention. Either way, assuming that's the case, I think one strategy clearly beats the other albeit not by that much, but by enough where the answer is clear to me. S3 should trump the AH with the QC. That strategy has a higher EV than trumping with a bower and then sending a bower on the next lead. Hence, question 9 on the "Rate your euchre skills" quiz for this site is wrong.

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I don't feel like calculating the percentage of getting euchred. But here's a high level overview. You already said my partner is a good player. I called trump from third seat and they lead the KD. That's a pretty sure bet that they don't have any trump and probably don't have aces either. That puts me in a bind because my hand is probably not strong enough to take 3 tricks alone. I might get lucky with the AS but anecdotally that fails more than 50% of the time, especially if your partner has no trump.PART I: What will you play to this trick?
Why do you think this is the best card to play?
What is your chance of getting euchred even though you chose this card?
The only realistic way you're getting a point here is to trump it with the queen and pray that the dealer doesn't overtrump with the king.Your partner probably can't trump this. No good player will lead a green king on a third seat call unless he's void in trump.

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Ok, time to do some math.Wes (aka the legend) wrote: ↑Sun Feb 16, 2020 1:17 amS3 should trump the AH with the QC. That strategy has a higher EV than trumping with a bower and then sending a bower on the next lead. Hence, question 9 on the "Rate your euchre skills" quiz for this site is wrong.
First let's look at the "Trump the AH with a bower and then lead the other bower" strategy. We need to figure out when this strategy will get euchred and how often. This S3 strategy will get euchred when S4 holds a total of 3 trump (KC+2) and no more spades or when S4 has 4 trump.
S4 3trump + no more spades = 3C0 x 3C2 x 5C1 x 3C0 = 15 combos
Note: The first 3C0 represents the 3 diamonds left in the deck that S4 cannot have. 3C2 = 3 trump left besides the KC and we're choosing 2. 5C1 = 5 hearts left and we're choosing 1, and the second 3C0 = 3 spades left and we're choosing none of them.
S4 4trump = 3C0 x 3C3 x 5C0 x 3C0 = 1 combo
Now figure out the total range of possibilities for S4. Since we know after the first lead that S4 cannot have JdTd9d in his range, there are 11 cards left unseen. So S4's total range is 11C3 = 165 combos
This means S3 will be euchred (15 + 1)/165 = 9.70% of the time
Ok now let's look at the other strategy of trumping the AH with the QC and thus risking getting overtrumped. First how often will we get overtrumped? First figure out how many combos in S4's range that are void in hearts.
3C0 x 5C0 x 6C3 = 20 combos
Note: 3C0 = zero diamonds, 5C0 = zero hearts, 6C3 = there are 6 unseen nondiamond/nonhearts and we're choosing 3.
Note 2: This means S4's hand including the KC is made up of some combination of clubs and spades, and precisely one combo of all clubs no spades.
Ok we already know S4's total range is 165 combos. So if S3 trumps the AH with the QC he will get overtrumped 20/165 = 12.12% of the time.
HEY WAIT A SECOND, 12.12% is worse than 9.70%! Why the hell do you want me to trump low with the QC!!?? Becuz we don't always get euchred the 12.12% of the time our QC gets overtrumped and that makes all the difference!
Once we get overtrumped, we are euchred those times S4 leads a spade (which he should always do if he has one) and S2 is void in spades and trumps in. There's basically 4 scenarios we need to look at.
Scenario 1: After S4 overtrumps the QC with the KC, his remaining 3 cards are KsQsTs (1 combo):
And given that S4 has this hand, what is the probability that S2 will trump a Spade lead. This part gets a little tricky. First figure out S2's total range, and to do that you have to remember that S1, a good player led a nontrump card on the first lead, revealing that he has no trump in his hand which means his remaining 4 cards are effectively exposed as nontrump cards. So to figure out S2's total range we have minus those 4 cards out. At the moment that S4 overtrumps S3's QC with the KC there are 11 unseen cards left  S1's effectively exposed 4 card hand = 7 unseen cards left. So S2's total range is 7C3 = 35 combos, and since S2 can never have 3 trump or 2 trump in this spot, his range is really only 22 combos:
18 combos where he has 1 trump in his hand: 3C1 x 4C2 = 18
4 combos where he has no trump: 3C0 x 4C3 = 4
So S2 will have 1 trump 18/22 = 81.82% of the time given the 1/20 = 5% of the time S4 has precisely KsQsTs left.
Scenario 2: After S4 overtrumps the QC with the KC, his remaining 3 cards are 1 club and 2 spades (9 combos):
3C0 x 3C1 x 5C0 x 3C2 = 9 combos
Given that S4 has 1 trump and 2 spades, S2 will be able to trump a spade lead those times he does not hold the remaining spade AND has 1 trump. Ok I'm gonna skip the combo counting step and just cut to the chase to save some space: 55.56% of the time S1 does not have that spade, S2 will be void in Spades and be able to trump 12/30 = 40% of the time. When S1 does have that spade 44.44% of the time, S2 will have a trump 20/30 = 66.67% of the time.
(.5556 x .40) + (.4444 x .6667) = 51.85%
So S2 will have 1 trump AND a void in spades 51.85% of the time given the 9/20 = 45% of the time S4 has 1 trump and 2 spades left.
Scenario 3: After S4 overtrumps the QC with the KC, his remaining 3 cards are 2 clubs and 1 spade (9 combos):
3C0 x 3C2 x 5C0 x 3C1 = 9 combos
Given that S4 has 2 trump and 1 spade, S2 will be able to trump a spade lead 17.14% of the time given the 33.33% of the time S1 has no spades, 28.57% of the time given the 53.33% of the time S1 has 1 spade, and 42.86% of the time given the 13.33% of the time S1 has both remaining spades.
(.3333 x .1714) + (.5333 x .2857) + (.1333 x .4286) = 26.66%
Scenario 4: This one is easy. 1/20 = 5% of the time S4 will have 4 trump (Kc+3) and it's game over for this strategy. S3 will get euchred 100% of the time. This is true for the trump with the bower strategy too and this rare scenario is accounted for in that strategy also.
Ok now we gotta add up the conditional probabilities:
(1/20)(.8182) + (9/20)(.5185) + (9/20)(.2666) + (1/20 x 1) = 44.42%
So the 12.12% of the time our QC gets overtrumped we will get euchred 44.42% of the time which translates to a euchre rate of (.1212 x .4442) = 5.38%
So the trump high with a bower strategy gets euchred 9.70% of the time, and the trump low with a QC gets euchred 5.38% of the time. Trumping low is cleary the best strategy.
I was wrong when I said this:
That's a pretty big difference. To reiterate, Question 9 on the "Rate your euchre skills" quiz for this site is clearly wrong.Wes (aka the legend) wrote: ↑Sun Feb 16, 2020 1:17 amI think one strategy clearly beats the other albeit not by that much, but by enough where the answer is clear to me.
Last edited by Wes (aka the legend) on Sun Feb 16, 2020 3:55 pm, edited 7 times in total.

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This isn't true. The strategy of trumping the AH with a bower and then sending a bower gets a point 90.3% of the time. That's certainly a "realistic way" to get a point. The other "realistic way", trumping the AH with the QC at the risk of getting overtrumped, does a bit better by getting a point 94.62% of the time.

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To All:
There is one main element disclosed after five cards have been played. I have not observed much discussion around this fact that is critical to the situation?
There is one main element disclosed after five cards have been played. I have not observed much discussion around this fact that is critical to the situation?

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 Joined: Tue Apr 24, 2018 9:33 pm
I think there are some issues with your Scenarios 1 & 2:
Scenario 1: After S4 overtrumps the QC with the KC, his remaining 3 cards are KsQsTs (1 combo):
And given that S4 has this hand, what is the probability that S2 will trump a Spade lead.
It would make no sense for a skilled player to hold 3 spades and lead the AH to trick 2. You would logically lead the spade and hold the AH as S3 will have to play a heart some time if he has it. But that does assume and take a chance S4 has no spades.
Scenario 2: After S4 over trumps the QC with the KC, his remaining 3 cards are 1 club and 2 spades (9 combos): Given that S4 has 1 trump and 2 spades..
This also makes no sense as S2 would have to have a club and 2 spades and lead a singleton AH.
We know that S2 can have no diamonds as good player will lead a 2nd one to trick 2. So S2 has only clubs, hearts and/or spades 11 cards and the same for S4.
I do agree with Scenario 3 & 4.
Scenario 1: After S4 overtrumps the QC with the KC, his remaining 3 cards are KsQsTs (1 combo):
And given that S4 has this hand, what is the probability that S2 will trump a Spade lead.
It would make no sense for a skilled player to hold 3 spades and lead the AH to trick 2. You would logically lead the spade and hold the AH as S3 will have to play a heart some time if he has it. But that does assume and take a chance S4 has no spades.
Scenario 2: After S4 over trumps the QC with the KC, his remaining 3 cards are 1 club and 2 spades (9 combos): Given that S4 has 1 trump and 2 spades..
This also makes no sense as S2 would have to have a club and 2 spades and lead a singleton AH.
We know that S2 can have no diamonds as good player will lead a 2nd one to trick 2. So S2 has only clubs, hearts and/or spades 11 cards and the same for S4.
I do agree with Scenario 3 & 4.

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I misread your Scenario 1, thought you had S2 with 3 spades. But I what stated is true that S2 cannot have 3 spades.
Statement I made on Scenario 2 is also true about leading lead the AH.
The hands of issue for S3 trumping with a bower, then lead the other bower then the AS is when S4/S2 holds these combos at the same time (I did not include the obvious 3 & 4 club hands.):
S4 (vertical column) & S2 (horizontal line) "C" means S2 trumped resulting in a euchre.
JS/JC AS 1H  2C
3S  KC JCJS S/"C"
2S  2C JCJS S/"C"
1S  3C JCJS S/"C"
3H  KC JCJS S/"C"
1S2HKC JCJS S/"C"
The hands off issue for S4 over trumping the QC are below. I did not include those for 3 & 4 clubs.
S4 (vertical column) & S2 (horizontal line) "C" means S2 trumped resulting in a euchre.
PLAY QC 2H  1C 1H  2C
3S  KC QC S "C" QC S "C"
2S  2C QC S "C" QC S "C"
1S  3C QC S "C" N/A
Note some similar results with some holdings regardless of KC or Bower.
Statement I made on Scenario 2 is also true about leading lead the AH.
The hands of issue for S3 trumping with a bower, then lead the other bower then the AS is when S4/S2 holds these combos at the same time (I did not include the obvious 3 & 4 club hands.):
S4 (vertical column) & S2 (horizontal line) "C" means S2 trumped resulting in a euchre.
JS/JC AS 1H  2C
3S  KC JCJS S/"C"
2S  2C JCJS S/"C"
1S  3C JCJS S/"C"
3H  KC JCJS S/"C"
1S2HKC JCJS S/"C"
The hands off issue for S4 over trumping the QC are below. I did not include those for 3 & 4 clubs.
S4 (vertical column) & S2 (horizontal line) "C" means S2 trumped resulting in a euchre.
PLAY QC 2H  1C 1H  2C
3S  KC QC S "C" QC S "C"
2S  2C QC S "C" QC S "C"
1S  3C QC S "C" N/A
Note some similar results with some holdings regardless of KC or Bower.

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Irishwolf,
Before I start correcting my mistakes, would you agree with this statement:
Given that S2 cannot have diamonds, and can only have 1 trump at most, his range boils down to 3 hand types:
1) Xh, Xh, Xs
Note: It's logical for S2 to still lead the AH from this configuration since the probability of S4 being void in hearts or spades is exactly the same (As far as doing the combos from S2's perspective, there are 9 cards exposed: S2's 4 remaining cards before he chooses what to lead on the 2nd trick, the KC upcard and the 4 cards played on the first lead).
Assuming S2 has that hand type:
Probability S4 is void in hearts = (3C0 x 3C0 x 9C3)/12C3 = 84/220 = 38.18%
Probability S4 is void in spades = (3C0 x 3C0 x 9C3)/12C3 = 84/220 = 38.18%
Note: The 12C3 represents S4's total range. From S2's perspective there are 15 unseen cards but we found out S4 has no diamonds on the first lead, so his range is really 12C3.
The other two hand types:
2) Xh, Xh, Xc
3) Xh, Xh, Xh
Do you agree with this so far IW. Just making sure I get this right before I start the work.
Before I start correcting my mistakes, would you agree with this statement:
Given that S2 cannot have diamonds, and can only have 1 trump at most, his range boils down to 3 hand types:
1) Xh, Xh, Xs
Note: It's logical for S2 to still lead the AH from this configuration since the probability of S4 being void in hearts or spades is exactly the same (As far as doing the combos from S2's perspective, there are 9 cards exposed: S2's 4 remaining cards before he chooses what to lead on the 2nd trick, the KC upcard and the 4 cards played on the first lead).
Assuming S2 has that hand type:
Probability S4 is void in hearts = (3C0 x 3C0 x 9C3)/12C3 = 84/220 = 38.18%
Probability S4 is void in spades = (3C0 x 3C0 x 9C3)/12C3 = 84/220 = 38.18%
Note: The 12C3 represents S4's total range. From S2's perspective there are 15 unseen cards but we found out S4 has no diamonds on the first lead, so his range is really 12C3.
The other two hand types:
2) Xh, Xh, Xc
3) Xh, Xh, Xh
Do you agree with this so far IW. Just making sure I get this right before I start the work.

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Ok here's my new post with the new math based on the fact that S2 cannot have any diamonds.
First let's look at the "Trump the AH with a bower and then lead the other bower" strategy. We need to figure out when this strategy will get euchred and how often. This S3 strategy will get euchred when S4 holds a total of 3 trump (KC+2) and no more spades or when S4 has 4 trump.
S4 3trump + no more spades = 3C0 x 3C2 x 5C1 x 3C0 = 15 combos
Note: The first 3C0 represents the 3 diamonds left in the deck that S4 cannot have. 3C2 = 3 trump left besides the KC and we're choosing 2. 5C1 = 5 hearts left and we're choosing 1, and the second 3C0 = 3 spades left and we're choosing none of them.
S4 4trump = 3C0 x 3C3 x 5C0 x 3C0 = 1 combo
Now figure out the total range of possibilities for S4. Since we know after the first lead that S4 cannot have JdTd9d in his range, there are 11 cards left unseen. So S4's total range is 11C3 = 165 combos
This means S3 will be euchred (15 + 1)/165 = 9.70% of the time
Ok now let's look at the other strategy of trumping the AH with the QC and thus risking getting overtrumped. First how often will we get overtrumped? First figure out how many combos in S4's range that are void in hearts.
3C0 x 5C0 x 6C3 = 20 combos
Note: 3C0 = zero diamonds, 5C0 = zero hearts, 6C3 = there are 6 unseen nondiamond/nonhearts and we're choosing 3.
Note 2: This means S4's hand including the KC is made up of some combination of clubs and spades, and precisely one combo of all clubs no spades.
Ok we already know S4's total range is 165 combos. So if S3 trumps the AH with the QC he will get overtrumped 20/165 = 12.12% of the time.
HEY WAIT A SECOND, 12.12% is worse than 9.70%! Why the hell do you want me to trump low with the QC!!?? Becuz we don't always get euchred the 12.12% of the time our QC gets overtrumped and that makes all the difference!
Once we get overtrumped, we are euchred those times S4 leads a spade (which he should always do if he has one) and S2 is void in spades and trumps in. There's basically 4 scenarios we need to look at.
Scenario 1: After S4 overtrumps the QC with the KC, his remaining 3 cards are KsQsTs (1 combo):
And given that S4 has this hand, what is the probability that S2 will trump a Spade lead. First we gotta figure out S2's total range. Ok given that S2 cannot have a diamond, and can only have 1 trump at most, and S4 has all the remaining spades, S2's 3 remaining cards are either all hearts or 2 hearts and 1 club which makes up 40 combos.
30 combos where S2 has 1 trump in his hand: 5C2 x 3C1 = 30
10 combos where S2 has no trump: 5C3 x 3C0 = 10
So S2 will have 1 trump 30/40 = 75% of the time given the 1/20 = 5% of the time S4 has precisely KsQsTs left.
Scenario 2: After S4 overtrumps the QC with the KC, his remaining 3 cards are 1 club and 2 spades (9 combos):
3C0 x 3C1 x 5C0 x 3C2 = 9 combos
Given that S4 has 1 trump and 2 spades, S2 will be able to trump a spade lead those times he does not hold the remaining spade AND has 1 trump. Ok I'm gonna skip the combo counting step and just cut to the chase to save some space: 55.56% of the time S1 does not have that spade, S2 will be void in Spades and be able to trump 30/50 = 60% of the time.
Note: In this scenario S2's remaining 3 cards are gonna be 3 different hand types:
1) HHS = 5C2 x 1C1 x 3C0 = 10 combos
2) HHH = 5C3 x 1C0 x 3C0 = 10 combos
3) HHC = 5C2 x 1C0 x 3C1 = 30 combos
Note 2: S2 can never have a club and a spade at the same time becuz that would mean he would've led a spade on the 2nd trick instead of the AH.
When S1 does have that spade 44.44% of the time, S2 will have a trump 30/40 = 75% of the time.
(.5556 x .60) + (.4444 x .75) = 66.67%
So S2 will have 1 trump AND a void in spades 66.67% of the time given the 9/20 = 45% of the time S4 has 1 trump and 2 spades left.
Scenario 3: After S4 overtrumps the QC with the KC, his remaining 3 cards are 2 clubs and 1 spade (9 combos):
3C0 x 3C2 x 5C0 x 3C1 = 9 combos
Given that S4 has 2 trump and 1 spade, S2 will be able to trump a spade lead 40% of the time given the 33.33% of the time S1 has no spades, 50% of the time given the 53.33% of the time S1 has 1 spade, and 66.67% of the time given the 13.33% of the time S1 has both remaining spades.
(.3333 x .40) + (.5333 x .50) + (.1333 x .6667) = 48.88%
Scenario 4: This one is easy. 1/20 = 5% of the time S4 will have 4 trump (Kc+3) and it's game over for this strategy. S3 will get euchred 100% of the time. This is true for the trump with the bower strategy too and this rare scenario is accounted for in that strategy also.
Ok now we gotta add up the conditional probabilities:
(1/20)(.75) + (9/20)(.6667) + (9/20)(.4888) + (1/20 x 1) = 60.75%
So the 12.12% of the time our QC gets overtrumped we will get euchred 60.75% of the time which translates to a euchre rate of (.1212 x .6075) = 7.36%
So the trump high with a bower strategy gets euchred 9.70% of the time, and the trump low with a QC gets euchred 7.36% of the time. Trumping low is cleary the best strategy.
So question 9 on the "Rate your euchre skills" quiz for this site is still wrong.
First let's look at the "Trump the AH with a bower and then lead the other bower" strategy. We need to figure out when this strategy will get euchred and how often. This S3 strategy will get euchred when S4 holds a total of 3 trump (KC+2) and no more spades or when S4 has 4 trump.
S4 3trump + no more spades = 3C0 x 3C2 x 5C1 x 3C0 = 15 combos
Note: The first 3C0 represents the 3 diamonds left in the deck that S4 cannot have. 3C2 = 3 trump left besides the KC and we're choosing 2. 5C1 = 5 hearts left and we're choosing 1, and the second 3C0 = 3 spades left and we're choosing none of them.
S4 4trump = 3C0 x 3C3 x 5C0 x 3C0 = 1 combo
Now figure out the total range of possibilities for S4. Since we know after the first lead that S4 cannot have JdTd9d in his range, there are 11 cards left unseen. So S4's total range is 11C3 = 165 combos
This means S3 will be euchred (15 + 1)/165 = 9.70% of the time
Ok now let's look at the other strategy of trumping the AH with the QC and thus risking getting overtrumped. First how often will we get overtrumped? First figure out how many combos in S4's range that are void in hearts.
3C0 x 5C0 x 6C3 = 20 combos
Note: 3C0 = zero diamonds, 5C0 = zero hearts, 6C3 = there are 6 unseen nondiamond/nonhearts and we're choosing 3.
Note 2: This means S4's hand including the KC is made up of some combination of clubs and spades, and precisely one combo of all clubs no spades.
Ok we already know S4's total range is 165 combos. So if S3 trumps the AH with the QC he will get overtrumped 20/165 = 12.12% of the time.
HEY WAIT A SECOND, 12.12% is worse than 9.70%! Why the hell do you want me to trump low with the QC!!?? Becuz we don't always get euchred the 12.12% of the time our QC gets overtrumped and that makes all the difference!
Once we get overtrumped, we are euchred those times S4 leads a spade (which he should always do if he has one) and S2 is void in spades and trumps in. There's basically 4 scenarios we need to look at.
Scenario 1: After S4 overtrumps the QC with the KC, his remaining 3 cards are KsQsTs (1 combo):
And given that S4 has this hand, what is the probability that S2 will trump a Spade lead. First we gotta figure out S2's total range. Ok given that S2 cannot have a diamond, and can only have 1 trump at most, and S4 has all the remaining spades, S2's 3 remaining cards are either all hearts or 2 hearts and 1 club which makes up 40 combos.
30 combos where S2 has 1 trump in his hand: 5C2 x 3C1 = 30
10 combos where S2 has no trump: 5C3 x 3C0 = 10
So S2 will have 1 trump 30/40 = 75% of the time given the 1/20 = 5% of the time S4 has precisely KsQsTs left.
Scenario 2: After S4 overtrumps the QC with the KC, his remaining 3 cards are 1 club and 2 spades (9 combos):
3C0 x 3C1 x 5C0 x 3C2 = 9 combos
Given that S4 has 1 trump and 2 spades, S2 will be able to trump a spade lead those times he does not hold the remaining spade AND has 1 trump. Ok I'm gonna skip the combo counting step and just cut to the chase to save some space: 55.56% of the time S1 does not have that spade, S2 will be void in Spades and be able to trump 30/50 = 60% of the time.
Note: In this scenario S2's remaining 3 cards are gonna be 3 different hand types:
1) HHS = 5C2 x 1C1 x 3C0 = 10 combos
2) HHH = 5C3 x 1C0 x 3C0 = 10 combos
3) HHC = 5C2 x 1C0 x 3C1 = 30 combos
Note 2: S2 can never have a club and a spade at the same time becuz that would mean he would've led a spade on the 2nd trick instead of the AH.
When S1 does have that spade 44.44% of the time, S2 will have a trump 30/40 = 75% of the time.
(.5556 x .60) + (.4444 x .75) = 66.67%
So S2 will have 1 trump AND a void in spades 66.67% of the time given the 9/20 = 45% of the time S4 has 1 trump and 2 spades left.
Scenario 3: After S4 overtrumps the QC with the KC, his remaining 3 cards are 2 clubs and 1 spade (9 combos):
3C0 x 3C2 x 5C0 x 3C1 = 9 combos
Given that S4 has 2 trump and 1 spade, S2 will be able to trump a spade lead 40% of the time given the 33.33% of the time S1 has no spades, 50% of the time given the 53.33% of the time S1 has 1 spade, and 66.67% of the time given the 13.33% of the time S1 has both remaining spades.
(.3333 x .40) + (.5333 x .50) + (.1333 x .6667) = 48.88%
Scenario 4: This one is easy. 1/20 = 5% of the time S4 will have 4 trump (Kc+3) and it's game over for this strategy. S3 will get euchred 100% of the time. This is true for the trump with the bower strategy too and this rare scenario is accounted for in that strategy also.
Ok now we gotta add up the conditional probabilities:
(1/20)(.75) + (9/20)(.6667) + (9/20)(.4888) + (1/20 x 1) = 60.75%
So the 12.12% of the time our QC gets overtrumped we will get euchred 60.75% of the time which translates to a euchre rate of (.1212 x .6075) = 7.36%
So the trump high with a bower strategy gets euchred 9.70% of the time, and the trump low with a QC gets euchred 7.36% of the time. Trumping low is cleary the best strategy.
So question 9 on the "Rate your euchre skills" quiz for this site is still wrong.

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S2 hand, 3 unknown cards. There are 11 cards available 5 H's, 3 S's & 3 C's
S2 hands possible (6): 3 H's // 3 S's // 3 C's // 1 H & 2C's // 2 H's & 1 C // 2 H's & 1 S // doubtful he has 1 H & 2 S's as S2 should hold the AH.
Wes, do you agree these are the most likely hands for S2?
S4 hand, 4 unknown cards. (and the same as those above available)
S4 hands possible in addition to the KC (9): 1 S & 2 C's// 2 S's & 1 C// 3 S's // 2 S's & 1 H // 3 C's // 2 H's & 1 S // 3 H's // 2 H's 1 C // 2 C's & 1 H
Of course these have to mesh to stay within those possible or S2 & S4.
Wes, do you agree for the hands possible for S4?
I would have to review your math above to see if it included the combinations for those possible.
~Irishwolf
S2 hands possible (6): 3 H's // 3 S's // 3 C's // 1 H & 2C's // 2 H's & 1 C // 2 H's & 1 S // doubtful he has 1 H & 2 S's as S2 should hold the AH.
Wes, do you agree these are the most likely hands for S2?
S4 hand, 4 unknown cards. (and the same as those above available)
S4 hands possible in addition to the KC (9): 1 S & 2 C's// 2 S's & 1 C// 3 S's // 2 S's & 1 H // 3 C's // 2 H's & 1 S // 3 H's // 2 H's 1 C // 2 C's & 1 H
Of course these have to mesh to stay within those possible or S2 & S4.
Wes, do you agree for the hands possible for S4?
I would have to review your math above to see if it included the combinations for those possible.
~Irishwolf

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I agree.
I eliminated 3 spades from S2's range since if he had this hand he should've led a spade on trick 2 instead of the AH. I eliminated 1H & 2S's for the same reason. I eliminated 3 clubs from S2's range becuz he would've called with that hand. I also eliminated all 2 club hands since if he has 2 clubs he would've had two trump + two aces, a hand he should call with also (I'll admit this is debatable, but that's what I did). You're missing those times S2 can have 1H,1S, & 1C, but I eliminated that hand from S2's range becuz if he had that hand, he should lead the 1S over the doubleton AH. When S2 holds this hand, his P is still more likely to be void in spades than hearts. I'll admit it's a bit of a stretch to eliminate this hand as I am now assuming S2 plays perfectly and good players don't play perfectly. Nevertheless that's what I did. So after S2 leads the AH, there are only 3 configurations for his remaining 3 cards:
HHH
HHC
HHS
There are only 3 unknown cards for S4. He picked up the KC and he played the 9S on the first lead, so there's 3 cards left unaccounted for.
Since there are 11 unseen cards, and we have 3 slots to fill for S4's hand, we can say his total possible range is 11C3 = 165 combos.
Right.
I mean, as long as the hands possible add up to 165 combos than yeah we're on the same page I think.
Of course. No problem.

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Irishwolf, I just realized a pretty big mistake I made in assessing S4's range. Remember I said S4's range = 11C3 = 165 combos and that those times he sets us when we utilize the "Trump with a bower, send a bower" strategy depends on if he has 3 trump (KC + 2) and no spade or 4 trump (KC+3) and this happens 16/165 = 9.70% of the time. The problem is we know something about S1's range given that he didn't lead trump and we need to factor this into S4's range.
We know S1's 4 other cards are nontrump cards. This dramatically affects S4's range when we're trying to figure out how often he'll have 3 trump and 4 trump. S1's hand is basically 4 more exposed cards as we know they are all nontrump cards. So S4's real range is not 11C3 but instead 7C3 = 35 combos. And the real chance this strategy get's euchred is not 16/165 but rather 16/35 = 45.71%
So It appears that despite S3's relatively strong holding he is actually in a rather precarious spot. This also changes how often S3 gets euchred on the "Trump low with the QC at the risk of getting overtrumped" strategy. Instead of S3 getting overtrumped 20/165 = 12.12% of the time, this really happens 20/35 = 57.14% of the time. Once this happens, we're still euchred 60.75% of the time, so our total euchre rate of the trump low strategy = (.5714 x .6075) = 34.71%
34.71% is still considerably less than 45.71%, so again, we're still better off trumping low.
We know S1's 4 other cards are nontrump cards. This dramatically affects S4's range when we're trying to figure out how often he'll have 3 trump and 4 trump. S1's hand is basically 4 more exposed cards as we know they are all nontrump cards. So S4's real range is not 11C3 but instead 7C3 = 35 combos. And the real chance this strategy get's euchred is not 16/165 but rather 16/35 = 45.71%
So It appears that despite S3's relatively strong holding he is actually in a rather precarious spot. This also changes how often S3 gets euchred on the "Trump low with the QC at the risk of getting overtrumped" strategy. Instead of S3 getting overtrumped 20/165 = 12.12% of the time, this really happens 20/35 = 57.14% of the time. Once this happens, we're still euchred 60.75% of the time, so our total euchre rate of the trump low strategy = (.5714 x .6075) = 34.71%
34.71% is still considerably less than 45.71%, so again, we're still better off trumping low.

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Ugh, found another mistake. I said S3 'trumping the AH with a bower and sending a bower' strategy will get euchred when S4 holds 3 trump (KC+2) and no more spades, implying that IF S4 has KC+2 and a spade, S3 is safe, which is false becuz when S4 has KC+2 and a spade he is gonna get rid of his spade when S3 trumps high on the AH. This means that ANY time S4 has 3 trump S3 is euchred when deploying this strategy:Wes (aka the legend) wrote: ↑Mon Feb 17, 2020 7:32 pmOk here's my new post with the new math based on the fact that S2 cannot have any diamonds.
First let's look at the "Trump the AH with a bower and then lead the other bower" strategy. We need to figure out when this strategy will get euchred and how often. This S3 strategy will get euchred when S4 holds a total of 3 trump (KC+2) and no more spades or when S4 has 4 trump.
S4 3 trump + no more spades = 3C0 x 3C2 x 5C1 x 3C0 = 15 combos
S4 has 3 trump = 3C0 x 3C2 x 8C1 = 24 combos
S4 has 4 trump = 3C0 x 3C3 x 8C0 = 1 combo
And S4's effect range is still 7C3 = 35 combos becuz this range takes into account that the real probability of S4 having additional trump is much higher in reality given that we KNOW S1's 4 remaining cards are all nontrump cards.
Ok so now if S3 deploys the "trump high, send high" strategy he will get euchred 25/35 = 71.43% of the time. Man this is unreal. I started out saying 9.70% and now I'm up to 71.43%. JFC. The trump low strategy still gets euchred 34.71% of the time AFAICT. My god what a HUGE gap. So trumping with the QC at the risk of getting overtrumped is massively better than trumping high as 34.71% euchre rate is WAY better than 71.43%. You literally more than double your euchre rate with the "trump high send high" strategy. Sounds unbelievable. I gotta run some kitchen table samples to see if this plays out.

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Ok, I did 60 kitchen table hands comparing both strategies each hand. I took out all the cards that had been played, made sure S1 can never have a trump, and made sure S4 or S2 could never have a diamond (any time S2 got 2 trump I reshuffled the hand but I honestly don't think it makes much of a difference if we include this hand in S2's range). Here's what I got:Wes (aka the legend) wrote: ↑Wed Feb 19, 2020 9:31 pmI gotta run some kitchen table samples to see if this plays out.
The "trump with a bower, send a bower" strategy got euchred 19/60 = 31.67% of the time.
The "trump low with the QC and risk getting overtrumped" strategy got euchred 2/60 = 3.33%
Ok this sample doesn't come close to comporting with the math I've done. I mean I know it's a small sample but something is off here. And I'm too exhausted to figure out what's wrong. That said, altho the numbers have changed as mistakes have been corrected, the theme has always stayed the same. Trumping low with the QC has always achieved better results than trumping high.

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I know where I messed up with my kitchen table sample. I shouldn't deal S4 three cards. I should actually deal him 4 cards and let him discard 1 into the kitty. The dealer always gets to look at 6 cards and keep 5. Forgot to factor that in. Will do another sample when I get the energy.Wes (aka the legend) wrote: ↑Wed Feb 19, 2020 11:31 pmOk, I did 60 kitchen table hands comparing both strategies each hand. I took out all the cards that had been played, made sure S1 can never have a trump, and made sure S4 or S2 could never have a diamond (any time S2 got 2 trump I reshuffled the hand but I honestly don't think it makes much of a difference if we include this hand in S2's range). Here's what I got:Wes (aka the legend) wrote: ↑Wed Feb 19, 2020 9:31 pmI gotta run some kitchen table samples to see if this plays out.
The "trump with a bower, send a bower" strategy got euchred 19/60 = 31.67% of the time.
The "trump low with the QC and risk getting overtrumped" strategy got euchred 2/60 = 3.33%
Ok this sample doesn't come close to comporting with the math I've done. I mean I know it's a small sample but something is off here. And I'm too exhausted to figure out what's wrong. That said, altho the numbers have changed as mistakes have been corrected, the theme has always stayed the same. Trumping low with the QC has always achieved better results than trumping high.

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Ok this time I did it right, giving S4 one extra card and allowing him to discard. First I need to explain how I set up my kitchen table sample. I gave S1 eight different hand configurations for his remaining 4 nontrump cards, which would directly impact what's in the kitty. E.G. if S1 has HHSS then you would know that the kitty contained DDDX, X being the extra card S4 got that he chose to discard.Wes (aka the legend) wrote: ↑Wed Feb 19, 2020 11:49 pmI know where I messed up with my kitchen table sample. I shouldn't deal S4 three cards. I should actually deal him 4 cards and let him discard 1 into the kitty. The dealer always gets to look at 6 cards and keep 5. Forgot to factor that in. Will do another sample when I get the energy.
BTW, The 8 different hand configurations is not by accident. Based on my S1 leading strategy, if I have no trump, no acesI try to lead the longest suit possible veering towards singleton green to minimize my P getting overtrumpedand when I lead a Diamond, my remaining 4 cards can only be one of the 8 below configurations:
DHHS (90 combos)
DHHH (30 combos)
DSSS (3 combos)
DDSS (9 combos)
HSSS (5 combos)
HHSS (30 combos)
HHHS (30 combos)
HHHH (5 combos)
If S2 got dealt 2 or more clubs, I reshuffled and did it again. I really don't think much would change if we included all S2 combos with 2 trump, in fact based on the mental notes I made doing these samples, I suspect including this hand hurts the "Trump high, send high" strategy more relative to the trump low strategy.
I did a 10 hand sample for each S1 hand configuration. Here's what I got:
The "Trump high, send high" strategy got euchred 68/80 = 85%
The "Trump low with the Qc" strategy got euchred 19/80 = 23.75%
The problem with this analysis is that S1 combos that barely happen (DSSS) get the same weighting as S1 combos that happen very often (DHHS). To fix this issue I simply did a weighted average euchre rate for each strategy, and here's the adjusted numbers:
The "Trump high, send high" strategy got euchred 72.52%
The "Trump Low with the QC" strategy got euchred 23.76%
Remember, based on the math, the "Trump high, send high" strategy got euchred 71.43% of the time, while the "Trump low with the QC" got euchred 34.71% of the time. BTW even if I did a trillion hand sample it would never line up with the actual math becuz the math calcs include S1 hand configurations that are not in my sample, I.E. hands I could never have. For example, If I had DDDSH in S1 I would never lead a diamond as a diamond lead would more likely put my P in a potential squeeze, I would always lead a heart. Either way this sample lines up with those numbers a lot better now. Anybody can set this hand up on their kitchen table and they'll quickly reach the inevitable conclusion: Trumping low with the QC is a VASTLY superior strategy to trumping high with a bower and sending another bower in this situation.

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S1 hand is important in that he would only lead KD as a singleton or doubleton diamond. His other cards have to be combinations of hearts & spades. Probably low ones. It would be very difficult to to don kitchen table dealing an analysis, IMO. This can actually be quite complex.
I have been out of town, just got back but will respond soon!
I have been out of town, just got back but will respond soon!

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Try this on for size:
Excluding with the dealer has 4 clubs as a euchre results regardless of what is done.
There are only two situation that results in a euchre when the QC trumps the AH, and they are:
S4 has 3S/KC S2 has 2H/1C QC lead Euchred, JC lead gets a point.
S4 has 3S/KC S2 has 1H/2C QC lead Euchred, JC lead gets a point.
Note also having all three remaining spades is a very low probability.
There are five hand situations that results in a euchre when the JC trumps the AH, and they are:
S4 has 1S/2H/KC S2 has 1H/2C QC lead gets a point, JC lead is euchred but must have AC 9c/10c.
S4 has 2H/2C S2 has 1H/2C QC lead gets a point, JC lead is Euchred (MUST HAVE AC IN COMBO)
S4 has 1H/3C S2 has 2H/1C QC lead gets a point, JC lead is euchred (AC & 9c/10c COMBO).
S4 has 1H/3C S2 has 2H/1S QC lead gets a point, JC lead is euchred (AC & 9c/10c COMBO).
S4 has 1H/3C S2 has 3H QC lead gets a point, JC lead is euchred (AC & 9c/10c COMBO).
Thus trumping with the QC has a higher EV than trumping with a bower. Some of those situations it may be better after trumping with a bower to lead the AS instead of leading the other bower. This forces the dealer to lead back to S3 who has JC/QC combo.
The AC generally has to be one of the two trumps in the opponents holdings as 9C/10C combo by itself by S2 does not result in a euchre.
~Irishwolf
x
Excluding with the dealer has 4 clubs as a euchre results regardless of what is done.
There are only two situation that results in a euchre when the QC trumps the AH, and they are:
S4 has 3S/KC S2 has 2H/1C QC lead Euchred, JC lead gets a point.
S4 has 3S/KC S2 has 1H/2C QC lead Euchred, JC lead gets a point.
Note also having all three remaining spades is a very low probability.
There are five hand situations that results in a euchre when the JC trumps the AH, and they are:
S4 has 1S/2H/KC S2 has 1H/2C QC lead gets a point, JC lead is euchred but must have AC 9c/10c.
S4 has 2H/2C S2 has 1H/2C QC lead gets a point, JC lead is Euchred (MUST HAVE AC IN COMBO)
S4 has 1H/3C S2 has 2H/1C QC lead gets a point, JC lead is euchred (AC & 9c/10c COMBO).
S4 has 1H/3C S2 has 2H/1S QC lead gets a point, JC lead is euchred (AC & 9c/10c COMBO).
S4 has 1H/3C S2 has 3H QC lead gets a point, JC lead is euchred (AC & 9c/10c COMBO).
Thus trumping with the QC has a higher EV than trumping with a bower. Some of those situations it may be better after trumping with a bower to lead the AS instead of leading the other bower. This forces the dealer to lead back to S3 who has JC/QC combo.
The AC generally has to be one of the two trumps in the opponents holdings as 9C/10C combo by itself by S2 does not result in a euchre.
~Irishwolf
x

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Wes
I don't think your euchre rate is correct for either. Remember I said S4's range = 11C3 = 165 combos and this statement is true. However, consider that S2's range is 8c3 = 56 (minus the 3 cards S4 hands for each 165 combos). For every hand that S4 has, there are 56 hands combinations for each that S2 has.
Thus, there are a total of 165 x 56 = 9240 combinations possible. So the euchre rate is a percentage of the 9240 total hands. Of course there are some hands not likely of the 9240 possible and it will still favor trumping with the QC, not the bower. The percentage for each are in line of 4 to 1 favoring the QC. I don't know the exact calculations at this point.
~Irishwolf
I don't think your euchre rate is correct for either. Remember I said S4's range = 11C3 = 165 combos and this statement is true. However, consider that S2's range is 8c3 = 56 (minus the 3 cards S4 hands for each 165 combos). For every hand that S4 has, there are 56 hands combinations for each that S2 has.
Thus, there are a total of 165 x 56 = 9240 combinations possible. So the euchre rate is a percentage of the 9240 total hands. Of course there are some hands not likely of the 9240 possible and it will still favor trumping with the QC, not the bower. The percentage for each are in line of 4 to 1 favoring the QC. I don't know the exact calculations at this point.
~Irishwolf

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My euchre rates will be off a tad becuz my numbers do not account for the fact that S1 will not be leading out of certain hand configurations like DDDHS for example but I don't think this oversight is important enough to correct. We don't need to be THAT precise on this. Same goes for the possibility S2 can have two trump. I eliminated this possibility, but even if S2 can have 2 trump in this spot, it hardly changes anything as far as figuring out which strategy is best. My euchre rates of 71.43% for "trump high, send high" and 34.71% for "trump low" are close enough to reality and I feel very confident in that now after doing my kitchen table 80 hand sample.
This is incorrect. I initially thought the same thing but I was wrong. In order to figure out S4's range we HAVE to account for the fact that S1 has no trump in his hand. Since S1 has 4 nontrump cards, S4's effective range is really 114 = 7C3 = 35 combos. If we don't make this adjustment we dramatically underestimate how often S4 will have 3 trump and 4 trump which is the most important probability in this hand.
S2's range is almost not even relevant. Every time S4 has 3 trump or 4 trump, the "trump high, send high" strategy is euchred regardless of what S2 has. S2's range becomes somewhat relevant when we trump low and get overtrumped, as now we are concerned with those hand combos S2 can have that have both 1 trump and a void in Spades which I've accounted for earlier in this thread (if our QC does not get overtrumped, S2's range is again irrelevant as we then have a sure point no matter what he has).
We don't even need to know the exact calculations at this point. There is no corrective tweak we can make that can change the theme here (which we both agree on). The "trump high, send high" strategy massively underperforms the "trump low" strategy. Again, I encourage anybody to set this hand up on their kitchen table, run some samples and see for themselves. They will quickly see what I'm talking about. The difference is large enough to reveal itself in very small samples.

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The euchre rate I get is 13% for QC trumping the AH and ~39% for trumping with a bower and leading the other bower. I am confident in the 13% as it closely follows the 5c0 6c3 11c3 = 12% for hearts. The 39% +/ 3% is my best estimate. My original 4 to 1 was too high, 3 to 1 is much closer to reality.
~Irishwolf
~Irishwolf

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The problem is the 12% for hearts part is wrong (I.E. the chance our trump low strategy gets overtrumped = 20/165 = 12.12% is wrong). Again, S4's range is not 11C3 = 165 combos. We have to account for the fact that S1 never has trump in his hand. If we don't then we dramatically underestimate how often S4 has 3 or 4 trump in his hand which corrupts everything.irishwolf wrote: ↑Fri Feb 21, 2020 9:01 pmThe euchre rate I get is 13% for QC trumping the AH and ~39% for trumping with a bower and leading the other bower. I am confident in the 13% as it closely follows the 5c0 6c3 11c3 = 12% for hearts. The 39% +/ 3% is my best estimate. My original 4 to 1 was too high, 3 to 1 is much closer to reality.
~Irishwolf
That said, I now just realized I was technically wrong when I said that S4's effective range was 114 = 7C3. S4's effective range is actually dynamic. It depends on S1's exact 4 remaining nontrump cards. For example, if S1 has no more diamonds left in his hand (E.G. HHSS) then I am right. S4's effective range is indeed 7C3 = 35 combos. But if S1 has another diamond, say he has DHHSsomething completely logical for him to have imothen S4's range is 8C3 = 56 combos. Either way, whether we're talking about 35 combos or 56 combos, those are much smaller numbers than the original incorrect assumption of 165 combos. If we don't make this adjustment, our euchre rates will be hopelessly amiss, dramatically underestimating the actual euchre rate for S3. S3 is actually in a much more precarious position than any of us initially thought.
So your euchre rate for trumping low, 13%, and your euchre rate for the trump high strategy, 39%, will be too low. All you'd have to do is deal this out on a kitchen table and you'll quickly see that. It doesn't take much of a sample size for the trend to show. Working from the erroneous 165 combo number is corrupting your data.

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I have to disagree:
"So your euchre rate for trumping low, 13%, and your euchre rate for the trump high strategy, 39%, will be too low. All you'd have to do is deal this out on a kitchen table and you'll quickly see that. It doesn't take much of a sample size for the trend to show. Working from the erroneous 165 combo number is corrupting your data."
I completed 150 hands and that is what I got. 150 / 165 hands for 11c4 is a very high percentage of the entire population.
"So your euchre rate for trumping low, 13%, and your euchre rate for the trump high strategy, 39%, will be too low. All you'd have to do is deal this out on a kitchen table and you'll quickly see that. It doesn't take much of a sample size for the trend to show. Working from the erroneous 165 combo number is corrupting your data."
I completed 150 hands and that is what I got. 150 / 165 hands for 11c4 is a very high percentage of the entire population.

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That's crazy how our samples can be so far off. Well at least both our math and our samples point to the same provocative conclusionand it's not even closeQuestion 9 on the Main quiz for this site is wrong. Trumping low with the QC is clearly the best play. Your sample makes me think there's something I may have done wrong. I've made so many mistakes already in this thread it wouldn't surprise me. I have to think about this some more.irishwolf wrote: ↑Fri Feb 21, 2020 11:22 pmI have to disagree:
"So your euchre rate for trumping low, 13%, and your euchre rate for the trump high strategy, 39%, will be too low. All you'd have to do is deal this out on a kitchen table and you'll quickly see that. It doesn't take much of a sample size for the trend to show. Working from the erroneous 165 combo number is corrupting your data."
I completed 150 hands and that is what I got. 150 / 165 hands for 11c4 is a very high percentage of the entire population.
BTW do you agree that we can't use 11C3 for S4's range? That we must account for the fact that S1 has no trump or we dramatically underestimate the probability S4 has 3 trump or 4 trump?

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Wes (aka the legend) wrote: ↑Thu Feb 20, 2020 7:35 amI know where I messed up with my kitchen table sample. I shouldn't deal S4 three cards. I should actually deal him 4 cards and let him discard 1 into the kitty. The dealer always gets to look at 6 cards and keep 5. Forgot to factor that in. Will do another sample when I get the energy.
Ok, I found a mistake in my kitchen table methodology. I tried to account for the fact that S4 is the dealer and has the privilege to see 6 cards and discard 1, hence dealing S4 four cards and letting him get rid of one after 2 cards of S4 had already been exposed (the 9S and the KC) to simulate what S4's final 3 cards could be, but when I did this I forgot the fact that S4 could've discarded a diamond. That changes things. Correctly putting a diamond into S4's range that he could discard weakens his overall range somewhat. So I don't like my sample anymore. Son of a *****. I simultaneously hate this thread and love this thread.Wes (aka the legend) wrote: ↑Thu Feb 20, 2020 7:35 amthis time I did it right, giving S4 one extra card and allowing him to discard.

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Wes,
I suspect our methodology of dealing the cards was different. I found a way to ignore all but 11 cards. So yes, 11c3, and just ignore S1 during the dealing. The 11 cards 9H to KH, 10S QS KS, 9C 10C AC are pulled out of the deck of 24. Shuffle the cards and deal the cards by one each to three piles, S2, S4 and Stock with two left over. You record S2 & S4. Just ignore those five left over card as any trump would go to the stock and any hearts would go to S1. Now just play out what you determined was S2 hand. Only 11 cards so it can be done very quickly. Ignore any discard, it's irrelevant for this purpose. You know S2 & S4 can only have 3 each of the 11 cards randomly dealt.
Just try it!
~Irishwolf
"BTW do you agree that we can't use 11C3 for S4's range? That we must account for the fact that S1 has no trump or we dramatically underestimate the probability S4 has 3 trump or 4 trump..."
I suspect our methodology of dealing the cards was different. I found a way to ignore all but 11 cards. So yes, 11c3, and just ignore S1 during the dealing. The 11 cards 9H to KH, 10S QS KS, 9C 10C AC are pulled out of the deck of 24. Shuffle the cards and deal the cards by one each to three piles, S2, S4 and Stock with two left over. You record S2 & S4. Just ignore those five left over card as any trump would go to the stock and any hearts would go to S1. Now just play out what you determined was S2 hand. Only 11 cards so it can be done very quickly. Ignore any discard, it's irrelevant for this purpose. You know S2 & S4 can only have 3 each of the 11 cards randomly dealt.
Just try it!
~Irishwolf
"BTW do you agree that we can't use 11C3 for S4's range? That we must account for the fact that S1 has no trump or we dramatically underestimate the probability S4 has 3 trump or 4 trump..."

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Wes,
I should have also said, that if S2 is dealt any hands with 2 or 3 clubs, just switch that hand with the 3rd hand you dealt (or ignore that hand, your choice) because he would have assisted.
~Irishwolf
I should have also said, that if S2 is dealt any hands with 2 or 3 clubs, just switch that hand with the 3rd hand you dealt (or ignore that hand, your choice) because he would have assisted.
~Irishwolf

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Wes,
One more thing on shuffling as you might notice you cannot riffle shuffle 11 cards very well. Just turn all 11 face down and stir them to get a random shuffle, Then just pick them up and deal one at a time.
You can also check to verify that you have a random deal. Count the hands that have one or more hearts as you know statistically it should be 88% and 12% void of hearts. You also know that each hand should have one spade (same with clubs) 50.9%. So it should be close (+/ 2%) to confirm you do not have some lopsided shuffling. I did all that to confirm my random dealing.
~Irishwolf
One more thing on shuffling as you might notice you cannot riffle shuffle 11 cards very well. Just turn all 11 face down and stir them to get a random shuffle, Then just pick them up and deal one at a time.
You can also check to verify that you have a random deal. Count the hands that have one or more hearts as you know statistically it should be 88% and 12% void of hearts. You also know that each hand should have one spade (same with clubs) 50.9%. So it should be close (+/ 2%) to confirm you do not have some lopsided shuffling. I did all that to confirm my random dealing.
~Irishwolf

 Posts: 1314
 Joined: Tue Apr 24, 2018 9:33 pm
Okay, here is my final final on the S3 play the QC vs JC. The correct answer is still the same, EV of the QC will forever exceed the JC on the AH. However, I have to eat some humble pie in that I made a mistake with the euchre rate for the JC. It is not 39% as I suggested and I corrected my error(s).
The rate of euchre is 16 to 17% in that range. And so obvious as well in that S4 must have 2 clubs + the KC, (and a void in HEARTS (NOT SPADES), total of 3 trumps in order to euchre the maker. S4 will lose one trump to the bower lead, then must have a void to the AS lead, then leads back to pick up the QC from S3. The maximum possible, statistically, is ~16%.
This also makes sense statistically as the 9c 10 AC, 3c2, 8c1 11c3 = 14.4%. But can also euchre the maker when S4 3c3 all clubs, another percentage point.
The QC trumping the AH, statistically is is about 12% and requires most of the time for S2 to have a void spades and one trump of any size. It also requires S4 to have a Heart void to the AH lead to over trump the QC with his KC, then leads a spades which S3 must follow suit and S2 trumps the spade lead.
Thus the euchre rate will be almost double when a bower is used on the AH However, still making a point 6 out of 7 times on the average. Actual hands were used for the results.
~Irishwolf
The rate of euchre is 16 to 17% in that range. And so obvious as well in that S4 must have 2 clubs + the KC, (and a void in HEARTS (NOT SPADES), total of 3 trumps in order to euchre the maker. S4 will lose one trump to the bower lead, then must have a void to the AS lead, then leads back to pick up the QC from S3. The maximum possible, statistically, is ~16%.
This also makes sense statistically as the 9c 10 AC, 3c2, 8c1 11c3 = 14.4%. But can also euchre the maker when S4 3c3 all clubs, another percentage point.
The QC trumping the AH, statistically is is about 12% and requires most of the time for S2 to have a void spades and one trump of any size. It also requires S4 to have a Heart void to the AH lead to over trump the QC with his KC, then leads a spades which S3 must follow suit and S2 trumps the spade lead.
Thus the euchre rate will be almost double when a bower is used on the AH However, still making a point 6 out of 7 times on the average. Actual hands were used for the results.
~Irishwolf

 Posts: 1314
 Joined: Tue Apr 24, 2018 9:33 pm
I can't convince you or anyone else, that's not my job.
Only you can convince yourself.
It's up to you to deal out those trial hands to
see how things work out but you can't do 25 hands to see it.
~Irishwolf
Only you can convince yourself.
It's up to you to deal out those trial hands to
see how things work out but you can't do 25 hands to see it.
~Irishwolf

 Posts: 1538
 Joined: Wed Jun 13, 2018 3:03 pm
Been distracted the last few days, but I'm gonna take a closer look at everything soon.

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 Joined: Wed Jun 13, 2018 3:03 pm
As long as you have S4's range = 11C3 you're not doing it right in my estimation. We have to account for the fact that we know S1 has no trump in his hand, and thus the probability S4 has 3 trump and 4 trump goes up dramatically. When trying to figure out the probability of S4 having 3 trump or 4 trump, S1's hand is effectively exposed. Those exposed cards need to be accounted for. After acknowledging that fact, then we then have to realize that S4's effective range is dependent on S1's exact 4 remaining cards. Specifically, if S1 does not have a diamond left, say S1 has HHSS, then S4's range is 7C3 = 35 combos. If S1 has a diamond still in his hand, say he has a hand like HHHD, then S4's range is 8C3 = 56 combos. If S1 has two diamonds left in his hand, say DDSS, then S4's range is 9C3 = 84 combos.irishwolf wrote: ↑Sat Feb 22, 2020 10:38 amWes,
I suspect our methodology of dealing the cards was different. I found a way to ignore all but 11 cards. So yes, 11c3, and just ignore S1 during the dealing. The 11 cards 9H to KH, 10S QS KS, 9C 10C AC are pulled out of the deck of 24. Shuffle the cards and deal the cards by one each to three piles, S2, S4 and Stock with two left over. You record S2 & S4. Just ignore those five left over card as any trump would go to the stock and any hearts would go to S1. Now just play out what you determined was S2 hand. Only 11 cards so it can be done very quickly. Ignore any discard, it's irrelevant for this purpose. You know S2 & S4 can only have 3 each of the 11 cards randomly dealt.
Just try it!
~Irishwolf
One could even go further and make another reasonable assumption: If S1 had KDXD he would always lead under the KD in hopes to promote it to a boss card that may be useful later. If that's the case then when S1 leads the KD he cannot have another Diamond. If we agree with this assumption, then S1's remaining 4 cards never have a diamond in it and thus S4's effective range is always 7C3 = 35 combos.
Now the next step is where I think I messed up:
I think the 25 combo part may be wrong. If we know S1 has no trump and has no diamonds left, then we can simply assign him his 4 remaining cards, say HHSS, and then figure out how many combos S4 can have where he has 3 trump and 4 trump:Ugh, found another mistake. I said S3 'trumping the AH with a bower and sending a bower' strategy will get euchred when S4 holds 3 trump (KC+2) and no more spades, implying that IF S4 has KC+2 and a spade, S3 is safe, which is false becuz when S4 has KC+2 and a spade he is gonna get rid of his spade when S3 trumps high on the AH. This means that ANY time S4 has 3 trump S3 is euchred when deploying this strategy:
S4 has 3 trump = 3C0 x 3C2 x 8C1 = 24 combos
S4 has 4 trump = 3C0 x 3C3 x 8C0 = 1 combo
And S4's effect range is still 7C3 = 35 combos becuz this range takes into account that the real probability of S4 having additional trump is much higher in reality given that we KNOW S1's 4 remaining cards are all nontrump cards.
Ok so now if S3 deploys the "trump high, send high" strategy he will get euchred 25/35 = 71.43% of the time.
S4 has 3 trump: 3C2 x 4C1 = 12 combos
S4 has 4 trump: 3C3 x 4C0 = 1 combos
Every time S4 has 3 trump or 4 trump S3 is euchred when deploying the "trump high, send high" strategy. Thus the probability of S3 getting euchred if he does that is (12+1)/7C3 = 13/35 = 37.14%
This isn't true. S4 does not need a void in spades + 3 trump to euchre S3's trump high, send high strategy. S4 just needs 3 trump becuz he will get rid of that spade when S3 trumps high on the AH.irishwolf wrote: ↑Sun Feb 23, 2020 12:47 amOkay, here is my final final on the S3 play the QC vs JC. The correct answer is still the same, EV of the QC will forever exceed the JC on the AH. However, I have to eat some humble pie in that I made a mistake with the euchre rate for the JC. It is not 39% as I suggested and I corrected my error(s).
The rate of euchre is 16 to 17% in that range. And so obvious as well in that S4 must have 2 clubs + the KC, (and a void in spades), total of 3 trumps in order to euchre the maker. S4 will lose one trump to the bower lead, then must have a void to the AS lead, then leads back to pick up the QC from S3. The maximum possible, statistically, is ~16%.

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Thinking about this some more, I think our samples and math will always be off if we don't incorporate the fact that S4 had the luxury of seeing an extra card before he discarded to short suit himself. The fact that S4 had the ability to create a void and exposed the 9S on the first trick suggests he probably has another spade (or is at least more likely to have a spade than our combination functions are indicating) which also suggests he is less likely to be void in Hearts than our combination functions are indicating. IDK man. This sh** is getting too complicated for me. I honestly don't think I have the ability to solve this problem anymore. The samples and the math DO strongly point in one directionthat trumping low is best, and I would bet alot of money that is the case, but I've given up hope of reaching any kind of convincing precision on this.

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More than likely S4 discarded a diamond (3 unkn) it could have been the 9D and the 9S could be S4 only spade. However, I agree more likely he has a another spade but could also have 2 H's.
Regardless, if he has a spade(s) and over trumps the QC, he will have to lead a spade back to S3 who has the AS. S3 can only be euchred if S2 (or S4 all clubs about 1%) has a spade void and can trump it (either way, a low probability as indicated above).
If S3 uses a bower and assumed S4 did have a void in hearts, then S3 can only be euchred if S4 has 3 Clubs (2 clubs along with his KC). If S4 has all clubs it is the same regardless of which ever S3 does. S4 has to have 3 clubs for a euchre, one club is lost with a bower lead, one to trump the AS leaving KC to lead back  15 to 17%). Double euchre rate of using the QC to trump the AH.
As simple as that. The discard is irrelevant and the above covers all situations known to man.
Regardless, if he has a spade(s) and over trumps the QC, he will have to lead a spade back to S3 who has the AS. S3 can only be euchred if S2 (or S4 all clubs about 1%) has a spade void and can trump it (either way, a low probability as indicated above).
If S3 uses a bower and assumed S4 did have a void in hearts, then S3 can only be euchred if S4 has 3 Clubs (2 clubs along with his KC). If S4 has all clubs it is the same regardless of which ever S3 does. S4 has to have 3 clubs for a euchre, one club is lost with a bower lead, one to trump the AS leaving KC to lead back  15 to 17%). Double euchre rate of using the QC to trump the AH.
As simple as that. The discard is irrelevant and the above covers all situations known to man.

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I meant to say he is more likely to be void in Hearts than our combination function indicates.Wes (aka the legend) wrote: ↑Wed Feb 26, 2020 3:44 amThinking about this some more, I think our samples and math will always be off if we don't incorporate the fact that S4 had the luxury of seeing an extra card before he discarded to short suit himself. The fact that S4 had the ability to create a void and exposed the 9S on the first trick suggests he probably has another spade (or is at least more likely to have a spade than our combination functions are indicating) which also suggests he is less likely to be void in Hearts than our combination functions are indicating. IDK man. This sh** is getting too complicated for me. I honestly don't think I have the ability to solve this problem anymore. The samples and the math DO strongly point in one directionthat trumping low is best, and I would bet alot of money that is the case, but I've given up hope of reaching any kind of convincing precision on this.

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irishwolf wrote: ↑Wed Feb 26, 2020 11:00 amMore than likely S4 discarded a diamond (3 unkn) it could have been the 9D and the 9S could be S4 only spade. However, I agree more likely he has a another spade but could also have 2 H's.
Regardless, if he has a spade(s) and over trumps the QC, he will have to lead a spade back to S3 who has the AS. S3 can only be euchred if S2 (or S4 all clubs about 1%) has a spade void and can trump it (either way, a low probability as indicated above).
If S3 uses a bower and assumed S4 did have a void in hearts, then S3 can only be euchred if S4 has 3 Clubs (2 clubs along with his KC). If S4 has all clubs it is the same regardless of which ever S3 does. S4 has to have 3 clubs for a euchre, one club is lost with a bower lead, one to trump the AS leaving KC to lead back  15 to 17%). Double euchre rate of using the QC to trump the AH.
Your euchre rate for the "Trump high send high" strategy is still WAY too low. Think about it man. Once S2 leads the AH, 10 cards have been exposed (AdKdQd9sKcAh and S3's remaining hand JcJsQcAs). We know S1 has no trump in his hand and we're making the reasonable assumption S2 has zero trump or 1 trump in his hand. We also know that S4 and S2 have no diamonds in their hand, and there's a very good chance S1 has no diamonds either. So play out a hypothetical:
Give S1:
Give S2:
Now there's only 4 cards Left:
And S4 has 3 unknown slots to fill.
There is a 50% chance S4 will have 3 trump and thus there is a 50% chance S3 will be euchred with a "trump high, send high" approach. And this is not some rigged setup to prove a point. This is actually close to a best case scenario for S3's prospects. Imagine S3's dire prospects when S2 doesn't have a trump! Actually no, don't imagine. Just play it out:
S1:
S2:
There are 4 cards left unaccounted for:
S4 has 3 unknown slots to fill.
When S2 has no trump, there is now a 75% chance S4 has 3 trump and a 25% chance S4 has 4 trump. IOW in this scenario S4 will have at least 3 trump 100% of the time, and thus S3 is euchred 100% of the time!
Somehow your math/samples are not capturing this realitythe reality that S3 is in a really bad situation if he chooses to trump high, send highbut nevertheless, this is the dire situation S3 is in. All one has to do is lay the cards out on the table to see it.
If we ignore the fact that S4 had the privilege to discard we end up putting hands in his range that cannot be there. For example, after S4 plays the 9S on the first trick his final 3 unexposed cards can never be SHC, SSH, HCC*. If we ignore the fact that S4 discarded earlier in the hand we will erroneously put those hands in his range.
*: S4 shouldn't have HCC becuz if he had that then he should've thrown off the Heart, I.E. the longer suit, on the first trick instead of the 9S.

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Not true as he could have discarded the 9D!
"S4 shouldn't have HCC becuz if he had that then he should've thrown off the Heart, I.E. the longer suit, on the first trick instead of the 9S."
S1 could have one more diamond and leads his highest one. S4 could have discarded the 9D. So your assumption he has none is not correct. He could have DDHHH, DDSSS OR DHHSS I think are reasonable holdings. S2 could have one spade. Based on that you hand holdings are not all inclusive of the combinations, IMO. Regardless, we know trumping with the QC has fewer euchres.
Holdings for S2/S4 should be based on 11 cards.
I corrected the S4 void in spades, I mention to say Hearts, QC hand over.
Did you do more hands with the 11 cards?
"S4 shouldn't have HCC becuz if he had that then he should've thrown off the Heart, I.E. the longer suit, on the first trick instead of the 9S."
S1 could have one more diamond and leads his highest one. S4 could have discarded the 9D. So your assumption he has none is not correct. He could have DDHHH, DDSSS OR DHHSS I think are reasonable holdings. S2 could have one spade. Based on that you hand holdings are not all inclusive of the combinations, IMO. Regardless, we know trumping with the QC has fewer euchres.
Holdings for S2/S4 should be based on 11 cards.
I corrected the S4 void in spades, I mention to say Hearts, QC hand over.
Did you do more hands with the 11 cards?

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He can't have HCCor I should say he shouldn't have HCC leftbecuz if he had Kc + 9s + HCC he would've thrown off with the Heart instead of the 9s given that the heart is the longer suit.
Ok, so you don't like my assumption that if S1 had KdXd he would lead the lower one to promote the Kd for later. That's fine. No biggie for me. Moving on, ok he can still have one more diamond. Now lets see what happens if we give S1 one more diamond:
Give S1:
Give S2:
Now there are 5 cards left:
S4 has 3 slots to fill.
When S2 has 1 trump, there is now a 30% chance S4 has 3 trump, and therefore S3's "trump high, send high" strategy will have a 30% euchre rate. What happens when S2 has no trump in his hand:
S1:
S2:
Remaining cards:
S4 now has a 70% chance of having 3 trump or more, therefore S3 is getting euchred 70% of the time he trumps high and sends high.
Just playing those scenarios out should make it clear, that giving the S3 strategy of "trump high, send high" a euchre rate in the teens is glaringly wrong.
I know but nevertheless these scenarios should be illuminating. It should make someone realize they are doing something wrong if their estimated euchre rate on a "trump high, send high" strategy is in the teens
That's where you're making a mistake. We CANNOT ignore the fact that S1 has no trump in his hand. When trying to figure out how often S4 has 3 trump or more, and thus how often S3 gets euchred, S1's hand is effectively exposed. If we don't account for this we dramatically underestimate S3's euchre rate.
You can't do the hands with 11 cards. That's the problem. You have to assign S1 4 nontrump cards and deal from there, and you'd have to deal from every logical 4 nontrump card holding S1 could have (I think there's 8 logical hand configurations S1 can have if we reject the assumption S1 can't have another diamond). So we would need an adequate sample starting from each of the possible eight S1 configurations. And then do a weighted average euchre rate to encompass the fact that some S1 hand configurations are more common than others. It's too bad my assumption that "S1 can't have another diamond becuz if he had KdXd he would lead from the lower diamond to promote the King" was rejected becuz it would make solving this logical puzzle easier as we would only have to work with 4 different S1 hand configurations consisting of 70 combos:
HSSS (5 combos)
HHSS (30 combos)
HHHS (30 combos)
HHHH (5 combos)

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Not so, he can have one diamond! Then three other cards consisting of H's & S's.

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There's no dispute. I dropped the assumption that "S1 cannot have another diamond since if he had KdXd he would lead the lower diamond to potentially promote his Kd to boss status for possible use later".
That's why I did those scenarios where S1 has 1 diamond. S3's euchre rate is still very high if he "trumps high, sends high".