Imagine you are sitting in 1st seat, with your team trailing 4 to 9. This is not looking good. In fact, your odds of winning the game (to 10 pts.) are about 4.2%.
You look at your cards and, WOW!, you suddenly feel like you've been given a new lease on life. You have been dealt JS + JC + AD + KD + QD, with the KS turned! You try hard not to salivate. Now it's decision time: do you bid S alone, R1 [choice A], or wait and bid C alone, R2 [choice B]?
Being a maven of statistics and probabilities, you are able to estimate the following distribution of outcomes should you make either choice (first number is for choice A; the 2nd is for choice B):
opponents score 4 pts: 0 / 0
opponents score 2 pts: 0.238 / 0.081 (by euchring you)
opponents score 1 pt: 0 / 0.237 (they will actually bid and make a pt in S almost a quarter of the time when you pass, R1)
your team scores 1 pt: 0.103 / 0.061
your team scores 2 pts: 0 / 0.111 (by euchring opponents)
your team scores 4 pts.: 0.659 / 0.510 (your euchre success rate is higher if you bid C, R2, but you get more actual 4 pt wins if you bid S, R1)
What does all this mean? If you bid S alone, R1, you expect a success rate of 65.9% and you expect to get euchred 23.8% of the time. EV = 2.26
If you pass R1 you expect your opponents to bid a little more than third of the time, with a euchre rate of slightly under a third. If it does get around to you again, R2, you can expect a loner success rate of almost 80% and a euchre rate of about 12% on your clubs call. EV = 1.93
Assume for the moment that these probabilities are correct (it's important to delve into that, but it's not the point of this exercise). Remember, your team is losing 4 - 9. Do you choose to bid R1, or pass and bid R2? The advantage of the R1 bid is a higher EV and a greater chance of scoring 4 pts. The downside is that you get euchred almost twice as often, ending the game there and then. Which call gives you the best chance of winning the game?
Note: it's difficult to present this data in a very clear format, so I encourage you to write it down in whatever way you find easiest to digest, then try to intuitively weigh the odds of each bid. There is a clear correct answer...
Odds of winning a game
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I think you have a lot of erroneous data and/or assumptions on this. Score is 9 to 4 therefore good players, this being close out, S2 will order most times he has 2 trumps (25.4%) with or with out an off suit ace. For sure will will order with 3 or 4 and that number, add another ~5%. So from S2 seat that basically 30%.
From Dealer's seat, will have 2 + upcard - 25.4%, and 3 + upcard 4.3%. And many times just 1 trump with upcard with one ace and that all depends. Does he have Next blocked? Good players are not going to let R2 come up ahead 9 to 4. That is how I would play, anyway. You can't avoid looking at the probabilities but don't get buried with all what could happen on R1, too variable, not just the cards but what Players do or don't do.
If I am the dealer, I will pass with 3 clubs with that score! There goes you Next loner.
So if you go alone, it will have to be R1 because I am going to block you R2. So if you go alone, you will have to do it R1, and will make a loner about 67% of the time and get euchred 30 to 33%.
All that just destroys your thesis! IMO So it is intuitively obvious S1 must call and go alone R1, and scoring the most points because against good players your opportunity of going alone in R2 is so much diminished.
I believe I have posted what is relevant, if not shoot it down with Facts! Opinions don't count.
IRISH
From Dealer's seat, will have 2 + upcard - 25.4%, and 3 + upcard 4.3%. And many times just 1 trump with upcard with one ace and that all depends. Does he have Next blocked? Good players are not going to let R2 come up ahead 9 to 4. That is how I would play, anyway. You can't avoid looking at the probabilities but don't get buried with all what could happen on R1, too variable, not just the cards but what Players do or don't do.
If I am the dealer, I will pass with 3 clubs with that score! There goes you Next loner.
So if you go alone, it will have to be R1 because I am going to block you R2. So if you go alone, you will have to do it R1, and will make a loner about 67% of the time and get euchred 30 to 33%.
All that just destroys your thesis! IMO So it is intuitively obvious S1 must call and go alone R1, and scoring the most points because against good players your opportunity of going alone in R2 is so much diminished.
I believe I have posted what is relevant, if not shoot it down with Facts! Opinions don't count.
IRISH
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Irish, I always appreciate your feedback, ever insightful.
I like your idea of S2 bidding more aggressively at this score (winning 9-5) in order to close out the game. I tested that decision with 100,000 games (completely random hands, except for KS turned), forcing S2 to bid with any 2 or more trump (spades) and found the following:
1) S2 bids 36,200 hands instead of 17,200 (so 19,000 more)
2) S4 bids about 8,000 less hands (because would have bid had S2 not bid)
3) S2/S4 earn at least a point on 68,100 hands (as compared to 65,600 hands if S2 doesn't take this aggressive bidding strategy)
4) Overall, S2/S4 only loses about 1,300 pts (over 100,000 games) by using this strategy, partially off-set by the fact that S1/S3 lose 800 pts (by bidding R1, S2 takes away opportunities for S1 especially to successfully bid R2)
It's almost worth doing this all the time, except that I find the overall EV to be about 0.07 lower (calculated on hands bid) - and every small edge counts.
The odds of S2 or S4 having 2+ or 3+ spades is just a probability calculation, so I find the same numbers as you. But S1 is not euchred as often as you suggest if they call, R1, because there is a good chance that whichever opponent stopped the loner with a 3rd trump has a diamond they will have to lead on the 4th or 5th trick - it works out to about 1/3 of the time. Which explains the euchre rate I found of 23.7%, with 1 pt being scored 10.3% of the time.
So ultimately I agree with you - my data is no good, and doesn't show anything. S1 needs to bid R1 [which, by the way, was the result I found], as S2 will really sabotage any effort to bid next R2. S1/S3 increase their odds of winning the overall game from about 21% to 25% by bidding R1, and that effect will be even more pronounced by S2 bidding more often.
I will come up with a better example to make the point I was trying to make, which is that assessing the tradeoff between more 4 pt wins coupled with more euchres is not always intuitive.
BTW, have a look at question 16 in the Euchre Quiz. My problem with that question is that it is a nonsensical scenario, since it should never arise (S1 will bid R1, as you have eloquently explained).
I like your idea of S2 bidding more aggressively at this score (winning 9-5) in order to close out the game. I tested that decision with 100,000 games (completely random hands, except for KS turned), forcing S2 to bid with any 2 or more trump (spades) and found the following:
1) S2 bids 36,200 hands instead of 17,200 (so 19,000 more)
2) S4 bids about 8,000 less hands (because would have bid had S2 not bid)
3) S2/S4 earn at least a point on 68,100 hands (as compared to 65,600 hands if S2 doesn't take this aggressive bidding strategy)
4) Overall, S2/S4 only loses about 1,300 pts (over 100,000 games) by using this strategy, partially off-set by the fact that S1/S3 lose 800 pts (by bidding R1, S2 takes away opportunities for S1 especially to successfully bid R2)
It's almost worth doing this all the time, except that I find the overall EV to be about 0.07 lower (calculated on hands bid) - and every small edge counts.
The odds of S2 or S4 having 2+ or 3+ spades is just a probability calculation, so I find the same numbers as you. But S1 is not euchred as often as you suggest if they call, R1, because there is a good chance that whichever opponent stopped the loner with a 3rd trump has a diamond they will have to lead on the 4th or 5th trick - it works out to about 1/3 of the time. Which explains the euchre rate I found of 23.7%, with 1 pt being scored 10.3% of the time.
So ultimately I agree with you - my data is no good, and doesn't show anything. S1 needs to bid R1 [which, by the way, was the result I found], as S2 will really sabotage any effort to bid next R2. S1/S3 increase their odds of winning the overall game from about 21% to 25% by bidding R1, and that effect will be even more pronounced by S2 bidding more often.
I will come up with a better example to make the point I was trying to make, which is that assessing the tradeoff between more 4 pt wins coupled with more euchres is not always intuitive.
BTW, have a look at question 16 in the Euchre Quiz. My problem with that question is that it is a nonsensical scenario, since it should never arise (S1 will bid R1, as you have eloquently explained).
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The only thing I would add to your last post is, If I am playing the Dealer seat and S2 does not assist, unless I have Next blocked, I am picking up the KS. It's playing Defense on Offense and I will take a hit rather than passing at 9 to 54. It is not always about scoring a point - what the EV of calling trump vs Passing? Tis the the question.
I don't know how your simulator is set up BUT S2 or S4 does not lead trump at any time unless the Right bower is led or you have two off suit aces, or one has three trumps. [I suspect your program is leading trump at first opportunity, in particular S4.] This is close out, with thin orders - trump leads can destroy you. It's a Myth in euchre that you must lead trump to make a point. It can have a big impact on the Euchre rate.
BTW, thanks for do extra work on this. It is important, IMO.
IRISH
P.S. I hope you go back and finish the hand JD DOWN, 16,920 HANDS JC 10C 9D KD QH - using your simulator and run it with 9D being led to the first trick. I think you shall be surprised!
I don't know how your simulator is set up BUT S2 or S4 does not lead trump at any time unless the Right bower is led or you have two off suit aces, or one has three trumps. [I suspect your program is leading trump at first opportunity, in particular S4.] This is close out, with thin orders - trump leads can destroy you. It's a Myth in euchre that you must lead trump to make a point. It can have a big impact on the Euchre rate.
BTW, thanks for do extra work on this. It is important, IMO.
IRISH
P.S. I hope you go back and finish the hand JD DOWN, 16,920 HANDS JC 10C 9D KD QH - using your simulator and run it with 9D being led to the first trick. I think you shall be surprised!
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Ray,
You can still use your scenario, just change the score to 6 to 6. Now at that score, S2/S4 has to be more careful of ordering aggressive not to get euchred with their deal next. IMO Now it's valid.
IRISH
You can still use your scenario, just change the score to 6 to 6. Now at that score, S2/S4 has to be more careful of ordering aggressive not to get euchred with their deal next. IMO Now it's valid.
IRISH
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I checked a few hands, and my program never had S2 or S4 leading trump at first possibility, on a thin call. The mistake I've seen lots of people make, in live games, is having 3 trump to the Left and being afraid to lead the Left. One opponent trumps a trick, another opponent trumps a different trick, opponents win with an off-suit Ace (because they overtrumped declarer earlier), or perhaps with the Right they still hold, it's euchre. Not every time, not even most times, but more often than necessary.