THE DEVIL IS IN THE DETAILS

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 Joined: Tue Apr 24, 2018 9:33 pm
THE DEVIL IS IN THE DETAILS
Let's suppose you have a hand at the Dealer seat, the score is 7 to 7, (10D) is the upcard. Your hand consists of: . Three Passes, your decision.
Lets also suppose you know that if you make Diamonds trump, the euchre rate is 55 to 60% (estimate), and that you will make a point 40 to 45%. And if you pass the opponents will most likely score a point 70%, (basically loner 8%, sweep 20 & 1 point the remainder) and also will pass 30%.
Maybe you even know (estimates again) that the EV = .50 if you order and EV = .70 if you pass!
What will you do?
Will you take the chance of a euchre at ~60% and put the opponents to 9 points and their deal next hand? Or will you take a chance and pass hoping your partner gets that 30% chance to call?
If you order, what is your five card strategy of playing this hand?
What's your chance of winning the game going either direction?
Would you do the same if the score was 4 to 4?
This is the essence of Euchre.
IRISHWOLF
Lets also suppose you know that if you make Diamonds trump, the euchre rate is 55 to 60% (estimate), and that you will make a point 40 to 45%. And if you pass the opponents will most likely score a point 70%, (basically loner 8%, sweep 20 & 1 point the remainder) and also will pass 30%.
Maybe you even know (estimates again) that the EV = .50 if you order and EV = .70 if you pass!
What will you do?
Will you take the chance of a euchre at ~60% and put the opponents to 9 points and their deal next hand? Or will you take a chance and pass hoping your partner gets that 30% chance to call?
If you order, what is your five card strategy of playing this hand?
What's your chance of winning the game going either direction?
Would you do the same if the score was 4 to 4?
This is the essence of Euchre.
IRISHWOLF

 Posts: 238
 Joined: Thu Sep 16, 2021 6:56 pm
Irish, this is a straightforward math problem (not simple, but straightforward). You just need to define your probabilities very precisely.
I have worked on this sort of problem, and have a program which will calculate all the possible outcomes  of this and future games  and sum the probabilities of every outcome to determine the odds of either team winning the game to 10 pts.
As inputs I need to have the probability of each possible point outcome of each game (there are 6 of them: +4, +2 or +1 pt, for each of two teams). I already have a good estimate of this distribution for the future, random hands, and have simplified them to 4 situations:
1) both teams need at least 3 points to win [.067,.141,.459,.187,.121,.025]
2) dealing partnership is at 8 or 9 pts, and so won't bid alone [0, .223, .448, .187, .117, .025]
3) nondealing partnership is at 8 or 9 pts, and so won't bid alone [.067, .139, .459, .184, .151, 0]
4) both teams have 8+ points and so won't bid alone [0, .147, .182, .449, .222, 0]
What do these 6 numbers mean? The first 3 are the odds of the dealing team winning +4, +2 or +1 pts, respectively. The last 3 are the odds of the other team scoring +1, +2 or +4 pts, respectively. I calculated these probabilities based on several million random hands played by my program, so they may actually vary depending on the aggressiveness of the bidding, how well the hands are played, if other score considerations are taken into account in addition to "don't bid alone with 8+ pts, etc. But it's a good start, I think.
The final input is the point probability distribution for the initial hand, which I tried to create from the information you provided. I came up with the following:
[0, .15, .3, 0, .55, 0] if dealer bids; EV = 0.5
This means that dealer has a 55% chance of getting euchred, a 30% chance of scoring 1 pt., and a 15% chance of scoring 2 pts., for an overall EV of 0.5. I basically fixed the euchre rate at 55%, then played with odds of 1 or 2 pts to get an EV of 0.5. If you propose a different point probability distribution, let me know, but it will likely change the EV.
For the case where dealer passes, it was a bit more difficult to decipher the odds you wanted, but I came up with the following (trying to get an EV of 0.7):
[.02, .02, .26, .44, .2, .06]
So the dealer's team scores points 30% of the time (2% chance of 4 pts, 2% chance of 2 pts, 26% chance of 1pt), and the opposing team scores points 70% of the time (6% chance of scoring 4 pts  70% * 8%, 20% chance of scoring 2 pts, 44% chance of scoring 1pt). Again, if you want to propose a different point probability distribution, just let me know.
Using all the data above, here are the odds of winning the game to 10 pts:
bid: 40.2%
pass: 37.4%
It's just math, IF the input probabilities are correct. Let me know if you want to tweak those probabilities.
I have worked on this sort of problem, and have a program which will calculate all the possible outcomes  of this and future games  and sum the probabilities of every outcome to determine the odds of either team winning the game to 10 pts.
As inputs I need to have the probability of each possible point outcome of each game (there are 6 of them: +4, +2 or +1 pt, for each of two teams). I already have a good estimate of this distribution for the future, random hands, and have simplified them to 4 situations:
1) both teams need at least 3 points to win [.067,.141,.459,.187,.121,.025]
2) dealing partnership is at 8 or 9 pts, and so won't bid alone [0, .223, .448, .187, .117, .025]
3) nondealing partnership is at 8 or 9 pts, and so won't bid alone [.067, .139, .459, .184, .151, 0]
4) both teams have 8+ points and so won't bid alone [0, .147, .182, .449, .222, 0]
What do these 6 numbers mean? The first 3 are the odds of the dealing team winning +4, +2 or +1 pts, respectively. The last 3 are the odds of the other team scoring +1, +2 or +4 pts, respectively. I calculated these probabilities based on several million random hands played by my program, so they may actually vary depending on the aggressiveness of the bidding, how well the hands are played, if other score considerations are taken into account in addition to "don't bid alone with 8+ pts, etc. But it's a good start, I think.
The final input is the point probability distribution for the initial hand, which I tried to create from the information you provided. I came up with the following:
[0, .15, .3, 0, .55, 0] if dealer bids; EV = 0.5
This means that dealer has a 55% chance of getting euchred, a 30% chance of scoring 1 pt., and a 15% chance of scoring 2 pts., for an overall EV of 0.5. I basically fixed the euchre rate at 55%, then played with odds of 1 or 2 pts to get an EV of 0.5. If you propose a different point probability distribution, let me know, but it will likely change the EV.
For the case where dealer passes, it was a bit more difficult to decipher the odds you wanted, but I came up with the following (trying to get an EV of 0.7):
[.02, .02, .26, .44, .2, .06]
So the dealer's team scores points 30% of the time (2% chance of 4 pts, 2% chance of 2 pts, 26% chance of 1pt), and the opposing team scores points 70% of the time (6% chance of scoring 4 pts  70% * 8%, 20% chance of scoring 2 pts, 44% chance of scoring 1pt). Again, if you want to propose a different point probability distribution, just let me know.
Using all the data above, here are the odds of winning the game to 10 pts:
bid: 40.2%
pass: 37.4%
It's just math, IF the input probabilities are correct. Let me know if you want to tweak those probabilities.

 Posts: 1273
 Joined: Tue Apr 24, 2018 9:33 pm
RAY, (Thanks!)
Let's tweak a bit just for fun:
S2/S4  euchre rate 60%, let sweeps (sweeps 3% is okay, S2 passed.) & 1 pt adjust a accordingly. Let EV then as would result & not a set number. (ordering)
S4 Passes. What you have is realistic, keep the same (EV .70) but the EV could be higher so do .80, just for fun.
IRISH
Let's tweak a bit just for fun:
S2/S4  euchre rate 60%, let sweeps (sweeps 3% is okay, S2 passed.) & 1 pt adjust a accordingly. Let EV then as would result & not a set number. (ordering)
S4 Passes. What you have is realistic, keep the same (EV .70) but the EV could be higher so do .80, just for fun.
IRISH

 Posts: 1273
 Joined: Tue Apr 24, 2018 9:33 pm
I might add, that how this hand is played if Ordering by S4 MAY also be of interest.
For example, discarding the club vs heart is a variable.
If they lead to the off suit ace spade and it wins a trick, lead trump vs 9S.
Point being, I suggest it makes things worst by leading trump and taking your partner trump. Perhaps as much as 7 to 8%.
IRISH
For example, discarding the club vs heart is a variable.
If they lead to the off suit ace spade and it wins a trick, lead trump vs 9S.
Point being, I suggest it makes things worst by leading trump and taking your partner trump. Perhaps as much as 7 to 8%.
IRISH
 Dlan
 Site Admin
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 Location: Ohio
A diamond call here is a type of hand that I enjoy trying. While I've never tracked the outcome, a euchre rate of 5060% seems too high.irishwolf wrote: ↑Sun Jan 09, 2022 12:34 amLet's suppose you have a hand at the Dealer seat, the score is 7 to 7, (10D) is the upcard. Your hand consists of: . Three Passes, your decision.
Lets also suppose you know that if you make Diamonds trump, the euchre rate is 55 to 60% (estimate),
IRISHWOLF
You just need a little help from your partner. On those hands where your partner has nothing, yes, you will most likely be euchred.
How would I play it? The initial discard would be the 9H. I'd hope for a black lead by 1st. That should give me the first trick. Then I'd lead back the remaining black suit.
At 4 to 4, no question, I'd try it. At 7 to 7 I'd want a little more info on my partner and my opponents.

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With S4 bidding, R1, I compared discarding the 9H with discarding the 10C (100,000 hands; 6,100 make it S4 for decision):
discard 9H: EV = 0.80
discard 10: EV = 0.68
Seems reasonable, more likely to be able to trump C, the longer suit. [my program discards the 10C]
Then I looked at instances where S4 wins the first trick with the AS, and compared leading the 9S with leading the 10D (100,000 hands; 6,060 make it to S4 for decision; 10C discarded):
863 times this scenario arose
lead 9S, trick 2: EV = 0.702
lead 10D, trick 2: EV = 0.686
14 more sweeps and 28 fewer euchres when leading the 10D. The sample size may be a bit small to be definitive, but certainly not trending toward 78% worse.
Finally, I compared bidding with passing, discarding the 10C when bidding and playing the 10D after winning with the AS on trick 1 (10,000 hands; 6,097 make it to S4 for decision):
bid: [0, 0.02, 0.41, 0, 0.57, 0] EV = 0.68
pass: [0.01, 0.09, 0.13, 0.54, 0.15, 0.08] EV = 0.79
note the 57% euchre rate and 2% sweep rate when bidding, as Irish predicted; when passing, 8% 4 pt wins for the opponents and 15% 2pt wins, close to Irish's predictions as well
When I plug this point distribution into the "who wins the game" program:
bid: 37.0% chance of winning
pass: 36.3% chance of winning
So using realistic potential probabilities, too fine a difference to make a distinction. With a different score, the results would be different, and perhaps one strategy would be more clearly better. But then I'd need to rerun the scenarios if one team (or both) have 8+ points at the start of this hand and never bid alone, which I think I'll pass on.
But I did check what the results would be if the score were 44:
bid: 37.0% chance of winning
pass: 35.5% chance of winning
slightly clearer decision here, which makes sense since the better EV of bidding has more time to be relevant (less volatility around earning 6 pts than earning just 3).
discard 9H: EV = 0.80
discard 10: EV = 0.68
Seems reasonable, more likely to be able to trump C, the longer suit. [my program discards the 10C]
Then I looked at instances where S4 wins the first trick with the AS, and compared leading the 9S with leading the 10D (100,000 hands; 6,060 make it to S4 for decision; 10C discarded):
863 times this scenario arose
lead 9S, trick 2: EV = 0.702
lead 10D, trick 2: EV = 0.686
14 more sweeps and 28 fewer euchres when leading the 10D. The sample size may be a bit small to be definitive, but certainly not trending toward 78% worse.
Finally, I compared bidding with passing, discarding the 10C when bidding and playing the 10D after winning with the AS on trick 1 (10,000 hands; 6,097 make it to S4 for decision):
bid: [0, 0.02, 0.41, 0, 0.57, 0] EV = 0.68
pass: [0.01, 0.09, 0.13, 0.54, 0.15, 0.08] EV = 0.79
note the 57% euchre rate and 2% sweep rate when bidding, as Irish predicted; when passing, 8% 4 pt wins for the opponents and 15% 2pt wins, close to Irish's predictions as well
When I plug this point distribution into the "who wins the game" program:
bid: 37.0% chance of winning
pass: 36.3% chance of winning
So using realistic potential probabilities, too fine a difference to make a distinction. With a different score, the results would be different, and perhaps one strategy would be more clearly better. But then I'd need to rerun the scenarios if one team (or both) have 8+ points at the start of this hand and never bid alone, which I think I'll pass on.
But I did check what the results would be if the score were 44:
bid: 37.0% chance of winning
pass: 35.5% chance of winning
slightly clearer decision here, which makes sense since the better EV of bidding has more time to be relevant (less volatility around earning 6 pts than earning just 3).

 Posts: 1273
 Joined: Tue Apr 24, 2018 9:33 pm
Thanks again Ray,
Good stuff and here is my commentary on this hand, for what it is worth, and even reviewing your results:
Results is similar to what I expected except for leading 10D but I can see that. I was on on the fence but lean toward the 2nd spade lead. However, it is more important to pull trump by leading than the 2nd lead of spades as you still have that opportunity with QD and one void. Made a note of it!
But I will also say this about that hand, and considering this as a one game series that I want to win, competitively. Although the EV I know is negative both decisions, and greater if I Pass. However, I am Passing on this hand as this is the shortrun. Why, because, (ordering) I see it as the euchre rate is less (55 to 60%) than the opponent's probability of scoring just one point. Scoring 1 pt will be 65 to 70% of the time based on thousands of hands I have analyzed. Scoring a loner, I say so what, low probability. And that is just how the cookie crumbles, I can accept that! I will take my chances against a loner as I have one stopper. In addition, tipping the scales for me is that S2 will get to call about 30% of the time. Also, critical in my judgement that I do not want to put the opponent at 9 points with their deal next. I do not like euchre rates that exceed 40% and they have to be given careful consideration. And it does matter who is in Eldest seat as a consideration. All that factored together.  Irish
Ray said, "Then I looked at instances where S4 wins the first trick with the AS, and compared leading the 9S with leading the 10D (100,000 hands; 6,060 make it to S4 for decision; 10C discarded):
863 times this scenario arose
lead 9S, trick 2: EV = 0.702
lead 10D, trick 2: EV = 0.686
14 more sweeps and 28 fewer euchres when leading the 10D. The sample size may be a bit small to be definitive, but certainly not trending toward 78% worse."
Good stuff and here is my commentary on this hand, for what it is worth, and even reviewing your results:
Results is similar to what I expected except for leading 10D but I can see that. I was on on the fence but lean toward the 2nd spade lead. However, it is more important to pull trump by leading than the 2nd lead of spades as you still have that opportunity with QD and one void. Made a note of it!
But I will also say this about that hand, and considering this as a one game series that I want to win, competitively. Although the EV I know is negative both decisions, and greater if I Pass. However, I am Passing on this hand as this is the shortrun. Why, because, (ordering) I see it as the euchre rate is less (55 to 60%) than the opponent's probability of scoring just one point. Scoring 1 pt will be 65 to 70% of the time based on thousands of hands I have analyzed. Scoring a loner, I say so what, low probability. And that is just how the cookie crumbles, I can accept that! I will take my chances against a loner as I have one stopper. In addition, tipping the scales for me is that S2 will get to call about 30% of the time. Also, critical in my judgement that I do not want to put the opponent at 9 points with their deal next. I do not like euchre rates that exceed 40% and they have to be given careful consideration. And it does matter who is in Eldest seat as a consideration. All that factored together.  Irish
Ray said, "Then I looked at instances where S4 wins the first trick with the AS, and compared leading the 9S with leading the 10D (100,000 hands; 6,060 make it to S4 for decision; 10C discarded):
863 times this scenario arose
lead 9S, trick 2: EV = 0.702
lead 10D, trick 2: EV = 0.686
14 more sweeps and 28 fewer euchres when leading the 10D. The sample size may be a bit small to be definitive, but certainly not trending toward 78% worse."

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The question now is, after taking the first trick with our As, is leading the low trump best out of this configuration no matter what our 2 trump holding is.irishwolf wrote: ↑Sun Jan 09, 2022 9:42 pmResults is similar to what I expected except for leading 10D but I can see that. I was on on the fence but lean toward the 2nd spade lead. However, it is more important to pull trump by leading than the 2nd lead of spades as you still have that opportunity with QD and one void. Made a note of it!
IMO if we call with Td9d, Qd9d or QdTd we already have our answer. Lead trump. But what about when we have Kd9d, KdTd, KdQd? My guess would be still lead trump. But with Ad9d, AdTd, AdQd, AdKd I would guess that double leading the spade is best. The idea being having a high trump that can potentially overtrump the enemy later in the hand makes a meaningful difference. Plus a lot of times Ax in trump is effectively guarded. So that's my new hypothesis: If you have Ax in trump double lead spade, if you have Kx or lower in trump lead low trump. What do you think? Hopefully Ray can have time to test this. He would only have to test AdKd to potentially falsify my hypothesis. If leading trump is correct from that holding the rest follows.

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Don't get caught up in the math. Don't over analyze. At 7 to 7. This is a clear pass. The stats alone show it tho be a pass. Just because something is better ev while still being a negative ev. Doesn't necessarily necessitate making a call here. The euchre rate and the score should be your deciding factors. Being down 7 to 9 their deal with a high euchre rate is a recipe to losing the game.
Tbolt65
Edward
Tbolt65
Edward

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 Joined: Wed Jun 13, 2018 3:03 pm
I don't think it's a clear pass. The evidence on this spot so far doesn't suggest that at all. Unless I'm misreading something, even at 77 Rays numbers suggest that bidding slightly beats out passing. At worse it's a statistical tie. So nothing is really clear about this spot in my mind. In fact, IF it really is a statistical tie I'm definitely calling this whenever I have an amateur P. Having an amateur P means they will pass biddable hands from the 2SR1. Hands like R+1+0, L+1+A, 2 nonbower trump+2A, even marginal but profitable 2 nonbower trump +A hands. That changes the math. If it's a statistical tie at worse with a good P, it's gotta swing to a call with your typical amateur P which is an important conclusion since most of the time we will be playing with an amateur P.Tbolt65 wrote: ↑Tue Jan 25, 2022 7:25 pmDon't get caught up in the math. Don't over analyze. At 7 to 7. This is a clear pass. The stats alone show it tho be a pass. Just because something is better ev while still being a negative ev. Doesn't necessarily necessitate making a call here. The euchre rate and the score should be your deciding factors. Being down 7 to 9 their deal with a high euchre rate is a recipe to losing the game.
Tbolt65
Edward
What's more fascinating to me is just how much our equity gets punched in the face just by being dealt this hand. At 77 before the cards are out the dealer team has approx 53% equity. Just getting dealt this extremely marginal hand smashed our equity down to around 37% whether we call or pass. So at 77 you have this feeling that it's anyone's game and then the instant you get dealt this hand you're proverbially f***ed.
Also given how close the numbers are at 77 I think we can safely infer that at 88 this would now be a pass for sure with a good P in the 2S. But in my mind even this spot is a major mindf**k with an amateur P.

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Wes, Ray said below:
The euchre rate is 55%. More than half the time when played randomly you are giving the team 2pts to put your team at a deficit of 79. What happens to that rate when playing vs competent opponents that see their opponents are calling ligh? I suggest it will be even a higher euchre rate. Making it even clearer not to make this call at 77. Giving up 1pt on a pass on a possible s1 r2 order is better overall than picking and getting euchred. 78 is not what you would like but it is way better than 79. With that said. It can still get back to your partner in seat 2 a good amount of times.
At 4 to 4 I can understand why someone might call this in a normal game. I disagree with it but I can see the justifications. Me personally. I approach the game now for a long time. To not unnecessarily give points away. Don't get me wrong. I still take chances but I try to limit giving away points too much in all situations. If I can help it. As people will see as they play in tougher and tougher games the smallest of mistakes or over calling will be the difference maker.
Tbolt65
Edward
raydog wrote: ↑Sun Jan 09, 2022 12:32 pmIrish, this is a straightforward math problem (not simple, but straightforward). You just need to define your probabilities very precisely.
I have worked on this sort of problem, and have a program which will calculate all the possible outcomes  of this and future games  and sum the probabilities of every outcome to determine the odds of either team winning the game to 10 pts.
As inputs I need to have the probability of each possible point outcome of each game (there are 6 of them: +4, +2 or +1 pt, for each of two teams). I already have a good estimate of this distribution for the future, random hands, and have simplified them to 4 situations:
1) both teams need at least 3 points to win [.067,.141,.459,.187,.121,.025]
2) dealing partnership is at 8 or 9 pts, and so won't bid alone [0, .223, .448, .187, .117, .025]
3) nondealing partnership is at 8 or 9 pts, and so won't bid alone [.067, .139, .459, .184, .151, 0]
4) both teams have 8+ points and so won't bid alone [0, .147, .182, .449, .222, 0]
What do these 6 numbers mean? The first 3 are the odds of the dealing team winning +4, +2 or +1 pts, respectively. The last 3 are the odds of the other team scoring +1, +2 or +4 pts, respectively. I calculated these probabilities based on several million random hands played by my program, so they may actually vary depending on the aggressiveness of the bidding, how well the hands are played, if other score considerations are taken into account in addition to "don't bid alone with 8+ pts, etc. But it's a good start, I think.
The final input is the point probability distribution for the initial hand, which I tried to create from the information you provided. I came up with the following:
[0, .15, .3, 0, .55, 0] if dealer bids; EV = 0.5
This means that dealer has a 55% chance of getting euchred, a 30% chance of scoring 1 pt., and a 15% chance of scoring 2 pts., for an overall EV of 0.5. I basically fixed the euchre rate at 55%, then played with odds of 1 or 2 pts to get an EV of 0.5. If you propose a different point probability distribution, let me know, but it will likely change the EV.
For the case where dealer passes, it was a bit more difficult to decipher the odds you wanted, but I came up with the following (trying to get an EV of 0.7):
[.02, .02, .26, .44, .2, .06]
So the dealer's team scores points 30% of the time (2% chance of 4 pts, 2% chance of 2 pts, 26% chance of 1pt), and the opposing team scores points 70% of the time (6% chance of scoring 4 pts  70% * 8%, 20% chance of scoring 2 pts, 44% chance of scoring 1pt). Again, if you want to propose a different point probability distribution, just let me know.
Using all the data above, here are the odds of winning the game to 10 pts:
bid: 40.2%
pass: 37.4%
It's just math, IF the input probabilities are correct. Let me know if you want to tweak those probabilities.
The euchre rate is 55%. More than half the time when played randomly you are giving the team 2pts to put your team at a deficit of 79. What happens to that rate when playing vs competent opponents that see their opponents are calling ligh? I suggest it will be even a higher euchre rate. Making it even clearer not to make this call at 77. Giving up 1pt on a pass on a possible s1 r2 order is better overall than picking and getting euchred. 78 is not what you would like but it is way better than 79. With that said. It can still get back to your partner in seat 2 a good amount of times.
At 4 to 4 I can understand why someone might call this in a normal game. I disagree with it but I can see the justifications. Me personally. I approach the game now for a long time. To not unnecessarily give points away. Don't get me wrong. I still take chances but I try to limit giving away points too much in all situations. If I can help it. As people will see as they play in tougher and tougher games the smallest of mistakes or over calling will be the difference maker.
Tbolt65
Edward

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 Joined: Sun Jun 16, 2019 9:14 pm
 Location: Las Vegas
Tbolt65 wrote: ↑Wed Jan 26, 2022 12:05 pmWes, Ray said below:
raydog wrote: ↑Sun Jan 09, 2022 12:32 pmIrish, this is a straightforward math problem (not simple, but straightforward). You just need to define your probabilities very precisely.
I have worked on this sort of problem, and have a program which will calculate all the possible outcomes  of this and future games  and sum the probabilities of every outcome to determine the odds of either team winning the game to 10 pts.
As inputs I need to have the probability of each possible point outcome of each game (there are 6 of them: +4, +2 or +1 pt, for each of two teams). I already have a good estimate of this distribution for the future, random hands, and have simplified them to 4 situations:
1) both teams need at least 3 points to win [.067,.141,.459,.187,.121,.025]
2) dealing partnership is at 8 or 9 pts, and so won't bid alone [0, .223, .448, .187, .117, .025]
3) nondealing partnership is at 8 or 9 pts, and so won't bid alone [.067, .139, .459, .184, .151, 0]
4) both teams have 8+ points and so won't bid alone [0, .147, .182, .449, .222, 0]
What do these 6 numbers mean? The first 3 are the odds of the dealing team winning +4, +2 or +1 pts, respectively. The last 3 are the odds of the other team scoring +1, +2 or +4 pts, respectively. I calculated these probabilities based on several million random hands played by my program, so they may actually vary depending on the aggressiveness of the bidding, how well the hands are played, if other score considerations are taken into account in addition to "don't bid alone with 8+ pts, etc. But it's a good start, I think.
The final input is the point probability distribution for the initial hand, which I tried to create from the information you provided. I came up with the following:
[0, .15, .3, 0, .55, 0] if dealer bids; EV = 0.5
This means that dealer has a 55% chance of getting euchred, a 30% chance of scoring 1 pt., and a 15% chance of scoring 2 pts., for an overall EV of 0.5. I basically fixed the euchre rate at 55%, then played with odds of 1 or 2 pts to get an EV of 0.5. If you propose a different point probability distribution, let me know, but it will likely change the EV.
For the case where dealer passes, it was a bit more difficult to decipher the odds you wanted, but I came up with the following (trying to get an EV of 0.7):
[.02, .02, .26, .44, .2, .06]
So the dealer's team scores points 30% of the time (2% chance of 4 pts, 2% chance of 2 pts, 26% chance of 1pt), and the opposing team scores points 70% of the time (6% chance of scoring 4 pts  70% * 8%, 20% chance of scoring 2 pts, 44% chance of scoring 1pt). Again, if you want to propose a different point probability distribution, just let me know.
Using all the data above, here are the odds of winning the game to 10 pts:
bid: 40.2%
pass: 37.4%
It's just math, IF the input probabilities are correct. Let me know if you want to tweak those probabilities.
The euchre rate is 55%. More than half the time when played randomly you are giving the team 2pts to put your team at a deficit of 79. What happens to that rate when playing vs competent opponents that see their opponents are calling ligh? I suggest it will be even a higher euchre rate. Making it even clearer not to make this call at 77. Giving up 1pt on a pass on a possible s1 r2 order is better overall than picking and getting euchred. 78 is not what you would like but it is way better than 79. With that said. It can still get back to your partner in seat 2 a good amount of times.
At 4 to 4 I can understand why someone might call this in a normal game. I disagree with it but I can see the justifications. Me personally. I approach the game now for a long time. To not unnecessarily give points away. Don't get me wrong. I still take chances but I try to limit giving away points too much in all situations. If I can help it. As people will see as they play in tougher and tougher games the smallest of mistakes or over calling will be the difference maker.
Tbolt65
Edward
Irishwolf said in another post,
If this line of thinking or reasoning is valid for three trump two suited. Can't we apply the same to this threads hand of acequeen of diamonds? One of which I've already mentioned and pointed out.
Irishwolf I'm not using your words against you but to support and give credence to what I've already said. For us to think about said situations.
Tbolt65
Edward

 Posts: 1522
 Joined: Wed Jun 13, 2018 3:03 pm
You brought up an interesting idea. We know the crappy 3 trump, 3 suited, no ace hands from the dealer spot have a negative expected outcome but we call anyways because they are +EV, ie calling beats out passing. Part of the reason this is true is cuz we block some S1R2 4 pt loners. But at 88 S1 can't really get a 4pt loner as it's only worth 2 pts. So the math changes at that score. Passing now doesn't cost as much. The big question then is does the cost of passing drop enough to make passing with some of these hands better than calling. Idk, but I'm gonna default to calling until good evidence suggests otherwise.Tbolt65 wrote: ↑Wed Jan 26, 2022 4:43 pmTbolt65 wrote: ↑Wed Jan 26, 2022 12:05 pmWes, Ray said below:
raydog wrote: ↑Sun Jan 09, 2022 12:32 pmIrish, this is a straightforward math problem (not simple, but straightforward). You just need to define your probabilities very precisely.
I have worked on this sort of problem, and have a program which will calculate all the possible outcomes  of this and future games  and sum the probabilities of every outcome to determine the odds of either team winning the game to 10 pts.
As inputs I need to have the probability of each possible point outcome of each game (there are 6 of them: +4, +2 or +1 pt, for each of two teams). I already have a good estimate of this distribution for the future, random hands, and have simplified them to 4 situations:
1) both teams need at least 3 points to win [.067,.141,.459,.187,.121,.025]
2) dealing partnership is at 8 or 9 pts, and so won't bid alone [0, .223, .448, .187, .117, .025]
3) nondealing partnership is at 8 or 9 pts, and so won't bid alone [.067, .139, .459, .184, .151, 0]
4) both teams have 8+ points and so won't bid alone [0, .147, .182, .449, .222, 0]
What do these 6 numbers mean? The first 3 are the odds of the dealing team winning +4, +2 or +1 pts, respectively. The last 3 are the odds of the other team scoring +1, +2 or +4 pts, respectively. I calculated these probabilities based on several million random hands played by my program, so they may actually vary depending on the aggressiveness of the bidding, how well the hands are played, if other score considerations are taken into account in addition to "don't bid alone with 8+ pts, etc. But it's a good start, I think.
The final input is the point probability distribution for the initial hand, which I tried to create from the information you provided. I came up with the following:
[0, .15, .3, 0, .55, 0] if dealer bids; EV = 0.5
This means that dealer has a 55% chance of getting euchred, a 30% chance of scoring 1 pt., and a 15% chance of scoring 2 pts., for an overall EV of 0.5. I basically fixed the euchre rate at 55%, then played with odds of 1 or 2 pts to get an EV of 0.5. If you propose a different point probability distribution, let me know, but it will likely change the EV.
For the case where dealer passes, it was a bit more difficult to decipher the odds you wanted, but I came up with the following (trying to get an EV of 0.7):
[.02, .02, .26, .44, .2, .06]
So the dealer's team scores points 30% of the time (2% chance of 4 pts, 2% chance of 2 pts, 26% chance of 1pt), and the opposing team scores points 70% of the time (6% chance of scoring 4 pts  70% * 8%, 20% chance of scoring 2 pts, 44% chance of scoring 1pt). Again, if you want to propose a different point probability distribution, just let me know.
Using all the data above, here are the odds of winning the game to 10 pts:
bid: 40.2%
pass: 37.4%
It's just math, IF the input probabilities are correct. Let me know if you want to tweak those probabilities.
The euchre rate is 55%. More than half the time when played randomly you are giving the team 2pts to put your team at a deficit of 79. What happens to that rate when playing vs competent opponents that see their opponents are calling ligh? I suggest it will be even a higher euchre rate. Making it even clearer not to make this call at 77. Giving up 1pt on a pass on a possible s1 r2 order is better overall than picking and getting euchred. 78 is not what you would like but it is way better than 79. With that said. It can still get back to your partner in seat 2 a good amount of times.
At 4 to 4 I can understand why someone might call this in a normal game. I disagree with it but I can see the justifications. Me personally. I approach the game now for a long time. To not unnecessarily give points away. Don't get me wrong. I still take chances but I try to limit giving away points too much in all situations. If I can help it. As people will see as they play in tougher and tougher games the smallest of mistakes or over calling will be the difference maker.
Tbolt65
Edward
Irishwolf said in another post,
If this line of thinking or reasoning is valid for three trump two suited. Can't we apply the same to this threads hand of acequeen of diamonds? One of which I've already mentioned and pointed out.
Irishwolf I'm not using your words against you but to support and give credence to what I've already said. For us to think about said situations.
Tbolt65
Edward

 Posts: 1273
 Joined: Tue Apr 24, 2018 9:33 pm
Yes, Ed that is what I said, and still true, this hand or that (see below). I think we are in agreement in that 'reasoning' is what is important in the same in both in that we have to evaluate score, situation, resulting Expected Values of one vs the other. What different and also has to be consider is that EV relates to (Law of Large Numbers) Many Hands Played. But In any game or tournament it is all about the this hand and that is about the ShortTerm. In this particular hand, score, players, situation is what counts. Some might think that is double talk but it is not.
If I risk a hand that has a 57% euchre rate +/ vs a more negative EV. I see it a little different in that I have more confidence in holding them to 1 pt compared to taking a euchre and putting them at 9 pts and there deal. I consider that statistically 65% score 1 pt vs Sweeps 15%, and I will take the risk of a loner. So I just might Pass depending how good S1 & S2 players are. I have to know something about my partner (an unknown, or expert that would not pass with with 2 trumps, and all those things).
Score is so critical. 4 to 4, I have confidence I can make up some ground. But at score 8 to 8, I am not going to give the game away if that euchre rate is like 57%! I would have to take the risk of opponents scoring 1 point, even if it is their deal next hand. Both might be dismal! But which is worse? It's never over til its over. Sometimes just do it for the hell of it to push the boundaries to the extreme! What can I get away with. The other thing about euchre, it's not just my five cards, my partner has five as well!
The hand in this post,The Devil is in the Details was posted to make people think. It's also about Longterm a more negative EV vs Shortterm decision making. Perhaps, there is no right or wrong answer except what occurs in the present. There is also third factor governing situations that sometimes I go against what 'probability' prescribes based on Intuition. I have found my Intuition usually out weighs probability.
IRISH
Irishwolf said in another post,
irishwolf wrote: ↑
Tue Jan 11, 2022 9:46 pm
Bidding with 3 trumps of any size has been a RULE OF THUMB in Euchre. However, that has to be qualified based on the Score. (Such as 8 to 8.) Probability of taking a euchre vs opponents scoring 1 point needs to be carefully evaluated.
IRISH
If this line of thinking or reasoning is valid for three trump two suited. Can't we apply the same to this threads hand of acequeen of diamonds? One of which I've already mentioned and pointed out.
Irishwolf I'm not using your words against you but to support and give credence to what I've already said. For us to think about said situations.
Tbolt65
Edward
If I risk a hand that has a 57% euchre rate +/ vs a more negative EV. I see it a little different in that I have more confidence in holding them to 1 pt compared to taking a euchre and putting them at 9 pts and there deal. I consider that statistically 65% score 1 pt vs Sweeps 15%, and I will take the risk of a loner. So I just might Pass depending how good S1 & S2 players are. I have to know something about my partner (an unknown, or expert that would not pass with with 2 trumps, and all those things).
Score is so critical. 4 to 4, I have confidence I can make up some ground. But at score 8 to 8, I am not going to give the game away if that euchre rate is like 57%! I would have to take the risk of opponents scoring 1 point, even if it is their deal next hand. Both might be dismal! But which is worse? It's never over til its over. Sometimes just do it for the hell of it to push the boundaries to the extreme! What can I get away with. The other thing about euchre, it's not just my five cards, my partner has five as well!
The hand in this post,The Devil is in the Details was posted to make people think. It's also about Longterm a more negative EV vs Shortterm decision making. Perhaps, there is no right or wrong answer except what occurs in the present. There is also third factor governing situations that sometimes I go against what 'probability' prescribes based on Intuition. I have found my Intuition usually out weighs probability.
IRISH
Irishwolf said in another post,
irishwolf wrote: ↑
Tue Jan 11, 2022 9:46 pm
Bidding with 3 trumps of any size has been a RULE OF THUMB in Euchre. However, that has to be qualified based on the Score. (Such as 8 to 8.) Probability of taking a euchre vs opponents scoring 1 point needs to be carefully evaluated.
IRISH
If this line of thinking or reasoning is valid for three trump two suited. Can't we apply the same to this threads hand of acequeen of diamonds? One of which I've already mentioned and pointed out.
Irishwolf I'm not using your words against you but to support and give credence to what I've already said. For us to think about said situations.
Tbolt65
Edward

 Posts: 1273
 Joined: Tue Apr 24, 2018 9:33 pm
There appears to be an issue that no one has YET picked up on about the hand and the EV's posted for Ordering vs Passing (by the Dealer). Just to review the Euchre rate of Ordering for the Dealer is 57% +/3 vs Passing EV is actually more negative EV = .70.
So the concept is that those EV's posted by Ray are of many hands, LONGTERM (Probabilities). The hand in question was posted as a 'Single' occurrence (Shortterm). And the EVs of those Hand examples where Ray does 100,000 hands are all of the LAW OF LARGE NUMBERS.
Thus you have to be careful applying Longterm Probability to a particular Single Hand. Longterm frequencies cannot be used to determine whether an outcome will occur on a single occasion, but can be used to predict longterm proportion of times the outcome will occur.
I gave the hint several times, but no one seemed to acknowledge. A good example of that, is to flip a coin 6 times and see what the frequencies will be of Heads & Tails? If it is exactly 3 : 3, very doubtful. You might get 5 heads in a row, etc. etc (all in the shortterm). But do it 10,000 times with a fair coin will be within a fraction of 50:50.
That's the Devil in The Details.
The answer to the hand is  PASS! Go figure!
IRISH
Let's suppose you have a hand at the Dealer seat, the score is 7 to 7, (10D) is the upcard. Your hand consists of: . Three Passes, your decision.
Lets also suppose you know that if you make Diamonds trump, the euchre rate is 55 to 60% (estimate), and that you will make a point 40 to 45%. And if you pass the opponents will most likely score a point 70%, (basically loner 8%, sweep 20 & 1 point the remainder) and also will pass 30%.
Maybe you even know (estimates again) that the EV = .50 if you order and EV = .70 if you pass!
So the concept is that those EV's posted by Ray are of many hands, LONGTERM (Probabilities). The hand in question was posted as a 'Single' occurrence (Shortterm). And the EVs of those Hand examples where Ray does 100,000 hands are all of the LAW OF LARGE NUMBERS.
Thus you have to be careful applying Longterm Probability to a particular Single Hand. Longterm frequencies cannot be used to determine whether an outcome will occur on a single occasion, but can be used to predict longterm proportion of times the outcome will occur.
I gave the hint several times, but no one seemed to acknowledge. A good example of that, is to flip a coin 6 times and see what the frequencies will be of Heads & Tails? If it is exactly 3 : 3, very doubtful. You might get 5 heads in a row, etc. etc (all in the shortterm). But do it 10,000 times with a fair coin will be within a fraction of 50:50.
That's the Devil in The Details.
The answer to the hand is  PASS! Go figure!
IRISH
Let's suppose you have a hand at the Dealer seat, the score is 7 to 7, (10D) is the upcard. Your hand consists of: . Three Passes, your decision.
Lets also suppose you know that if you make Diamonds trump, the euchre rate is 55 to 60% (estimate), and that you will make a point 40 to 45%. And if you pass the opponents will most likely score a point 70%, (basically loner 8%, sweep 20 & 1 point the remainder) and also will pass 30%.
Maybe you even know (estimates again) that the EV = .50 if you order and EV = .70 if you pass!

 Posts: 795
 Joined: Sun Jun 16, 2019 9:14 pm
 Location: Las Vegas
Thats what I've been saying. The 2pts deficit is key here vs seat 1 calling more frequently and getting 1pt more often. You are saving 1 point which is critical and crucial. Your team is in a better spot to still win the game.irishwolf wrote: ↑Wed Jan 26, 2022 10:18 pmYes, Ed that is what I said, and still true, this hand or that (see below). I think we are in agreement in that 'reasoning' is what is important in the same in both in that we have to evaluate score, situation, resulting Expected Values of one vs the other. What different and also has to be consider is that EV relates to (Law of Large Numbers) Many Hands Played. But In any game or tournament it is all about the this hand and that is about the ShortTerm. In this particular hand, score, players, situation is what counts. Some might think that is double talk but it is not.
If I risk a hand that has a 57% euchre rate +/ vs a more negative EV. I see it a little different in that I have more confidence in holding them to 1 pt compared to taking a euchre and putting them at 9 pts and there deal. I consider that statistically 65% score 1 pt vs Sweeps 15%, and I will take the risk of a loner. So I just might Pass depending how good S1 & S2 players are. I have to know something about my partner (an unknown, or expert that would not pass with with 2 trumps, and all those things).
Score is so critical. 4 to 4, I have confidence I can make up some ground. But at score 8 to 8, I am not going to give the game away if that euchre rate is like 57%! I would have to take the risk of opponents scoring 1 point, even if it is their deal next hand. Both might be dismal! But which is worse? It's never over til its over. Sometimes just do it for the hell of it to push the boundaries to the extreme! What can I get away with. The other thing about euchre, it's not just my five cards, my partner has five as well!
The hand in this post,The Devil is in the Details was posted to make people think. It's also about Longterm a more negative EV vs Shortterm decision making. Perhaps, there is no right or wrong answer except what occurs in the present. There is also third factor governing situations that sometimes I go against what 'probability' prescribes based on Intuition. I have found my Intuition usually out weighs probability.
IRISH
Irishwolf said in another post,
irishwolf wrote: ↑
Tue Jan 11, 2022 9:46 pm
Bidding with 3 trumps of any size has been a RULE OF THUMB in Euchre. However, that has to be qualified based on the Score. (Such as 8 to 8.) Probability of taking a euchre vs opponents scoring 1 point needs to be carefully evaluated.
IRISH
If this line of thinking or reasoning is valid for three trump two suited. Can't we apply the same to this threads hand of acequeen of diamonds? One of which I've already mentioned and pointed out.
Irishwolf I'm not using your words against you but to support and give credence to what I've already said. For us to think about said situations.
Tbolt65
Edward
Tbolt65
Edward