Unread post
by raydog » Wed Dec 07, 2022 10:12 pm
For what it's worth, I tried simulating this situation. But what struck me is that it is somehow unsatisfying to simply know what the better play is. Myriad situations arise, and it is impossible (at least for me) to consistently make the right choice; my memory is just too flawed. What is more important is to understand WHY the correct play is what it is, something which I recognize that Irish and a few others try to explain in many of these posts. Being able to do a quick and accurate analysis, on the fly (in a real game setting, where you don't have 5 minutes to make a decision), is crucial. So what's important is to develop a good sense of how to analyze these scenarios.
I have developed a habit of trying to EXPLAIN the results of my simulator, partially because it can help me uncover errors in the program (or in how I programmed a specific scenario), but also because it helps make me a better player.
Philosophy aside, here is what I found.
S4 has AS + K-10H + K-10C (JH turned). What to discard?
discard 10C (what my program currently does)
bid alone: (15,787 / 72,337 / 7,067) [sweep / 1 pt. / euchred] EV = +1.27
bid w/p: (22,564 / 66,391 / 6,236) EV = 1.04
discard AS:
bid alone: (11,578 / 70,605 / 12,862) EV = +0.96
bid w/p: (18,551 / 67,341 / 9,153) EV = +0.91
As was mentioned in the thread, this is definitely a loner hand, and it is clearly better to discard the 10C. The KC is only ever going to win if the AC is buried or with S2, no matter what is discarded. So it is simply a question of whether the singleton green AS is more likely to win a trick than the doubleton green 10C. Note that having to waste a trump on winning a S lead (because you discarded your AS), and very likely the K trump at that, doesn't seem likely to help your chances of getting a sweep, and the 4pts. that go with it. Also, as was previously mentioned, you may be able to dump your KC to create a void (assuming you discarded the 10C initially) if a D is lead and your partner is winning the trick. So this result makes sense.
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But what if we replace that AS with the KS?
discard KS (what my program currently does - creating a void):
bid alone: (11,932 / 69,986 / 12,557) EV = +0.98
bid w/p: (18,350 / 66,315 / 9,810) EV = +0.88
discard 10C:
bid alone: EV = +0.97
bid w/p: EV = +0.67
Here there doesn't seem to be any difference. Why? If the opponents have the AS, it would seem better to discard the KS, but that will cost you a trump if S are led, first trick, and still require that an opponent doesn't have a doubleton C - two ways of still getting your sweep thwarted. I guess the odds cancel out, making the choice moot.
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Wes suggested the hand where S4 has AS + KH + JD + K-QC (JH) [Wes: you have a typo in your post - you gave S4 the JH instead of the JD!]
discard QC, call alone: (32,487 / 62,081 / 689) EV = +2.00
discard AS, call alone: (33,200 / 61,399 / 577) EV = +2.03
I was surprised to see a virtual tie here - perhaps slightly better to discard the AS. What's going on? I think the crucial point here is that you need the AC to be buried or with your partner, otherwise you don't get your sweep. If the AC IS buried, you can still be thwarted if S1 leads a S AND S3 is void. If you kept your AS, you lose that trick. If you jettisoned your AS, you get to over-trump, win the next two trump leads, then (most likely) win with the KC and QC (unless S3 started w/ 4 trump or S1 started w/ 3 trump).
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Wes also suggested this hand: S4 has AS + KH + JD + A-QC
discard QC, call alone: (76,581 / 18,007 / 91) EV = +3.41
discard AS, call alone: (64,764 / 30,393 / 124) EV = +3.04
Better to discard the QC, in case an opponent has Kx of C.
But what if S4 has the A-KC:
discard KC, call alone: (75,652 / 19,492 / 68) EV = +3.38
discard AS, call alone: (78,969 / 16,241 / 95) EV = +3.48
In this case it's a bit better to discard the AS. Which is what Irish predicted. Here you have almost all boss cards, so you just need to avoid getting trumped on that first trick. So ditch the AS and get 2-suited.
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Concerning this last hand, Irish posited that the correct might be different if S1 or S3 called. So I tested this hand:
S4 has AS + 10-9H + A-KC (QH turned).
S1 orders up the QH, R1 [just 2,500 of 100,000 otherwise random hands tested]
discard KC: EV = -0.96
discard AS: EV = -1.24
S3 orders up the QH, R1 [just 3,200 of 100,000 otherwise random hands tested]
discard KC: EV = -0.87
discard AS: EV = -0.88
Given how infrequently this situation arises, it is VERY dependent on how my program selects that S1 and S3 call. So I unfortunately cannot offer any good explanation on why these results are what they are.