UPDATE: In the comments, Jonathan demonstrates some errors in my below-the-fold analysis. I'll try to summarize in a subsequent post, but briefly the chance that a suit splits evenly just because both opponents follow once or twice to a suit where they have several cards goes up much less than an empty-spaces approach would suggest. --Franco
I used to participate regularly on rec.games.bridge. Apparently I posted this hand back in 1995:
AJxx
KQ7x
9x
AK9
T8x
Axx
AKJT8x
x
Declare 6D on the SK lead (North opened 1N, the opponents did not bid).
You win SA, RHO follows. Do you:
a) Run the D9,
b) Play DA, CA, then hook DJ,
c) Play DA, CA, then run D9, or
d) Play DA, DK, then set about trying to pitch your spades?
There are a couple of potentially useful odds-related links in the links section to the upper right should you care to try for a very precise analysis.
Some further discussion is below the fold...
For some reason, when I first read _Winning Declarer Play_ many years ago, a hand around this trump suit tripped me up: I thought it was right to cash an honor then hook, but of course (as I'm sure all my readers know) that only picks up stiff Q offside and an immediate finesse gains against any other singleton (since you can repeat the finesse).
That's true in isolation, but here you have decent chances at an endplay: DA, CA, D9 as LHO shows out. Then ruff a club, HA, HK, CK (pitching a heart) HQ. If the CK or HQ is ruffed: overruff, draw trumps, and set up a spade. If they all survive, pitch a spade, ruff a heart, and exit a spade to score the last 2 in trumps. It's very close, but by my math it comes out slightly ahead (note that the immediate finesse might pick up 5-0 onside with a similar endplay which makes it even closer). Perhaps I'll work it out precisely in a subsequent post.
So, line B is easily dominated by C, and C appears to be very slightly better than A (which, in turn, is very slightly worse than 50% -- it picks up all onside Qs except Qxxxx with <2 hearts which happens about .5% of the time).
How about trying to drop the Q or get fast spade pitches? Roughly 1/3 chance of dropping the Q. The rest of the time, your only hope is 3-2 trumps (about 42% given the Q hasn't dropped) and 3-3 hearts (about 40% given all the known cards, so the parlay is 16.8%). This rough estimate comes in about 50%, so seems better than the immediate finesse. Whether it's better than C requires a more careful calculation.
In the small-world dept, Jonathan happened to comment on this thread back in '95. His analysis of the playing for the drop concluded it was inferior, but that analysis had a small but important error (I implicitly made a similar error at the time): as cards become known, the chance of even splits go up (where I said in the last paragraph 1/3 + 42% x 40%, he said 32% + 41% x 36%). These difference appear to be enough to swing in favor of the drop over the finesse.
Finally, to get this right note that you have to make some assumptions, such as whether 8-1 clubs is no longer possible. To me, the interesting one is what are the "known cards" after the SK lead and RHO following? Naively LHO is known to have started with at least 2 spades, while RHO only 1. But LHO had a chance to make a revealing lead that RHO didn't.
My head hurts.
Captions only
4 years ago
Hi,
ReplyDeleteWhen you restrict to the people interested in conditional bridge odds calculations, it seems to be a *very* small world indeed :-).
How many cards are you assuming to be known to get 40% for the conditional chance of 3-3 hearts? Just taking diamonds 3-2 gives 36.48% which would justify my math from 1995, when I may have chosen not to comment on small adjustments. There are some inferences about the other suits, but as you obviously know, you have to be careful about treating cards people choose to play as known for these calculations. Treating SK,SQ of spades as known is fine on the assumption he would lead SK iff he had KQ. Treating the low spade righty followed with as known, or the clubs people would follow with to two rounds, is wrong. Yes, you can condition on "spades not 6-0" but this is different from conditioning on "righty has S2" and has much less impact.
I actually think the lead makes it a smidge more likely the trump finesse is on, and a smidge less likely to get even breaks. (It's as if the "suit" consisting of the spade honors broke 2-0.) Definitely the nicest part of the hand is your observation that because of trump coup possibilities, c *might* be better than a. My calculations say (c) beats (a) by a *tiny* bit, under half a percent, partly because (c) sometimes survives even 5-0 trumps while (a) is down instantly against stiff Q. Fortunately for aesthetics, I do think it also beats (d).
Huh, I overlooked surviving 0=5 trumps after DA at trick 2. Seems like you can do it if RHO is exactly 2=3=5=3. Not nearly as probable as in line (a), though.
ReplyDeleteI got my 40% 3-3 estimate after I plugged in a few different things under different assumptions. For example, 6 known in each hand is 40.8%. I don't see where I got 6, but if we assume an 8 card suit would bid, we know each hand has at least 1 spade, 2 diamonds, and 2 clubs, which would give 39.2%.
Obviously which line is best here is a very close call and not important, but I'm curious if you think I have the general principle wrong as you don't count any spades or clubs as known cards in your calculation.
And I'll try to have a more exciting hand tomorrow.
This is long, I should really post it on my own blog :-).
ReplyDeleteHere is a good illustration of why you can't necessarily treat played cards as known: suppose you have AKQ2 opposite 543 and cash two rounds, everyone following. With no info about other suits, what are the odds the suit breaks? It could be tempting to say that each player has 2 known cards and we need the odds the remaining cards break 1-1, which would be 11/21, or 52%. This probably "feels wrong" because you know 4-2 is more likely than 3-3, and it *is* wrong -- since the evidence is equally consistent with 4-2 or 3-3, the relative odds haven't changed, and a 3-3 break is about 35/84. What was the flaw in the original logic? There is restricted choice on the spot cards; namely, the opponents have more ways to choose their spots on 3-3 breaks than 4-2 breaks, making 3-3 breaks less likely than they appear on the first calculation.
Moral: Any played card that would have been more "restricted" on some layouts and less on others cannot just be treated as a played card. When you say "clubs not 8-1" it's the same as saying "all follow to 2 rounds" and the same thing applies.
Note: the answer is very different if you have AKT9 opposite 432 and no one ever drops an honor unless forced...now when all follow small to 2 rounds it *is* 52% the suit breaks!
You're correct within a suit, though note even then that while the relative odds haven't changed, the probability of an even division has clearly gone up.
ReplyDeleteBut across suits the restricted choice is irrelevant. The odds of a 3-3 heart split conditioned on spades not 6-0 is the same as the odds of a 3-3 heart split given 2 hands with 12 spaces.
You said: 'Yes, you can condition on "spades not 6-0" but this is different from conditioning on "righty has S2" and has much less impact.'
What's the difference? I think both calculation lead to: 6-choose-3 * 18-choose-9 / 24-choose-12 = 35.959%.
No, the same thing applies across suits -- I just did within a suit for simplicity. Let's take event A = "spades not 6-0" vs. event B = "lho has S2, rho has S3." We agree these look different within the spade suit: Given A, the relative odds of 3-3,4-2,5-1 breaks stay the same, they just scale up, from 35.5,48.4,14.6 to 36.0,49.1,14.8. Given B, the odds shift very differently, to 41,49.6,9.4; basically we are now looking at the distribution of a 4-card suit, and even breaks have gotten much more likely. This actually has a semi-practical use: if your opponents *always* follow with their lowest spots, and you see them play the 2 and 3, the latter odds are actually correct. If, as they should to conceal information, they follow with random spots, the first set of odds are correct.
ReplyDeleteOK, what about other suits? The formula you gave, 6c3*18c9/24c12, is correct for P(3-3 hearts|B). It's useful to note we could get the identical answer a more complicated way: calculate P(3-3 hearts) given spades 3-3,4-2, or 5-1, and take a weighted average according to the second list of weights above, 41, 49.6, 9.4. This should suggest that P(3-3 hearts|A) must be different, assuming you agree with the first paragraph; you have to instead use the weights 36.0,49.1,14.8. There is no simpler way to calculate this that I know. It comes out to 35.67%, closer to the original odds, intuitively because A is less informative than B. If you accept the different implications within the suit, the different implications outside the suit follow.