Monday, July 27, 2009
Consider the following combinations:
A: AQ983 in hand opposite 542
B: AJ82 in hand opposite K543
and assume you are maximizing average tricks, with no entry restrictions.
Alan Applebaum was asking a bunch of people on Friday, just before the LM pairs started, about A. Almost everyone (including some top players) played low to the 8, then if that lost planned to hook the Q the second time. This is actually wrong, because it only gains when lho has JT doubleton and loses when he has stiff “jen,” compared to hooking the 9 the second round – it’s a restricted choice position, but one that seems to have gone unnoticed by almost everyone.
Then later that day at the table, position B came up, which happens to be a very close cousin. Our opponent cashed the K, and when the T appeared on the left hooked the J, commenting that she wasn’t sure that was right. Actually it’s right to hook the 8 (or cover righty’s 9 cheaply) on the second round, for similar reasons. You win against stiff T ( or 9), lose to doubleton T9; by restricted choice the stiff “tine” is more likely. At the table, only the third-best approach of going up ace worked; lefty had QT tight. I bet many good players would briefly consider going up ace, then hook the J because they know going up is too susceptible to falsecarding…but wouldn’t even consider the best option of hooking the 8! All I can tell from the recap sheet is that about 25% of the field dropped the Q; no idea how many hooked the J versus the 8.
By the way, there was another consideration on this hand: About 25%** of the field got to the optimal 6nt while the rest, as did my opponents, bid to 6h with this as the trump suit and no other losers anywhere. I’m ignoring the very few who went minus in a grand. Those in the inferior 6H would be justified in taking an anti-field approach in the play, which rates to gain about 70% of the mps when right but lose only about 30% when wrong. I can’t say for sure if the J or 8 is anti-field, but I do know the A is! Disappointingly, a glance at the recap seems to show no correlation between the contract and the number of tricks taken -- those who went +1020 should be a bit ashamed, as 990 was already a good score!
** Update, 7/30: I actually counted frequencies of the scores, see comments.
Saturday, July 18, 2009
It was with a bit of guilt that I wrote my first comment, which I knew only gave the solution assuming average-trick maximization when BAM play might be different. I knew the BAM analysis would be more complex and wasn’t sure anyone would care, but I am very glad FMB did care, because now that he got me to think about it more, this is a great hand from a theory point of view! As he points out, this is really a 4-player game: let’s call the players S1,E1 (team 1) and S2,E2 (team 2). Each South can plan to go up (U) or duck (D) when East keeps a club, and each East can bare (B) or not bare (N) the K when he has it. There are 3 distinct ways to look at equilibrium in this game:
I) Correlated team equilibrium
Each team picks an overall strategy which can be any mixture of the 4 pairs of South and East strategies, UB, UN, DB and DN. It’s an equilibrium if the other team has no better-than-even response.
This becomes in effect a two-“player” symmetric zero-sum game, thinking of each team as a player. I had to turn to Matlab to invert the 4x4 payoff matrix, and eventually found that the unique equilibrium was .5 UN, .25 DB, .25 DN. Notice that:
a) The percentage of B by the Easts (.25) and U by the Souths (.5) is the same as in the average-trick maximization analysis that looked at just one table.
b) S1 and E1, also S2 and E2, must correlate their strategies to never play UB. This may or may not be possible, but leave that aside for the moment.
c) No one ever wins the board by two tricks, because this would require both teams to play B with one D and one U, not possible if no team ever plays UB. This is why the average-trick equilibrium translates into a BAM equilibrium here – when the trick difference is always -1, 0 or 1 it gets converted linearly to the BAM scale.
d) Comment c provides some intuition for why UB is avoided; a team that plays UB would be in “danger” of winning the board by two tricks, which wastes some of their average trick total. Even more intuitively, if your teammate South is playing Up, there is some chance as East that you have already won the board by not baring, so baring could be pointlessly risking a valuable trick for a valueless one.
II) Independent team equilibrium. Each team announces their randomization which must consist of *independent* randomizations by S and E. It's an equilibrium if the other team has no better-than-even response.
Fascinating to me is that there is no type II equilibrium! If you believe me that the equilibrium in I is unique for its type, this is actually pretty immediate. A team has a better-than-even response to any randomization that isn’t the correlated one in I, which implies that one of the four pure strategies must be a winning response, and pure strategies of course satisfy independence.
How is this possible? Doesn’t it violate Nash’s Existence Theorem? No. Why? When Nash’s Theorem is applied to games of more than two players, you have to be careful about the definition of equilibrium. A 4-tuple of strategies is a Nash equilibrium if no *single* player can improve his outcome by deviating. Nash doesn’t recognize teams, so his theorem doesn’t apply to the equilibrium concept defined above (II), which says equilibrium can be broken by two players deviating together. Nevertheless concept II is fairly natural, and there is something disturbing about the fact that there is no such equilibrium. Think about it: If S1 and E1 huddle and decide on a randomization for each, then *whatever* they decided, team 2 has a winning response if they just know the frequencies. This is totally opposed to the usual intuition for mixed-strategy equilibrium.
III) 4-player independent Nash equilibrium. Each player announces a randomization – again, teams cannot correlate. It's an equilibrium if no player can improve his lot by changing his plan, with both his opponents *and* his teammate held fixed.
Nash tells us that such a thing definitely exists. FMB must have been referring to this type when he said East should bare 1-sqrt(.5) of the time. I haven’t been able to verify his calculations yet; I get some equations that are a bit hairy when I try to solve for the type III equilibrium.
All of this brings up the issues: Can teams correlate? Is it legal for them to correlate? I’ll comment on that sometime later, but for now will note that being able to correlate seems to provide some edge, although it is difficult to define how much: the two-team game where one team can correlate and the other must be independent has no equilibrium! Frankly, I find the conditions in II more natural bridgewise than I or III, but that's the one that leads to non-existence!
Saturday, July 11, 2009
Update: I decided to try out the BBO handviewer program -- thanks to Memphis Mojo for recommendation and advice, and to Fred Gitelman for the program. So, anyway, you can see the whole deal in much prettier format at the bottom of the post -- but may want to read the story first.
On Thursday, the A/X Swiss played 7 8-board matches, while the lower brackets played 7 X 7. The thing was, the directors put us all on the same clock (I think 40 minutes to start the last board.) Yikes, this wasn’t billed as the Speedball Swiss! Anyway, the constant time pressure figures a bit into this story, also from the 5th match:
I was dealer, neither vul, and the competitive auction went
3D showed a limit raise+ in spades.
Lefty led the HA (I played the T) and shifted to a club – J, A, ruff. By isolating the club menace against himself, he had given me a chance. I considered the possibilities:
A simple squeeze? This would require ducking a diamond to rectify the count, and the defense wouldn’t have to be especially brilliant to continue diamonds, leaving no entries – the end.
Ok, it would have to be a squeeze without the count. The conditions weren’t right for an ordinary one (control of both suits) – what about a ruffing squeeze without the count? Cash all the trumps but one, then hearts ending in dummy. That leaves Ax Qx in dummy and three diamonds and a trump in hand, and if lefty keeps Kx clubs he has only 2 diamonds, and you can establish the last one in hand…that works! Nice – the kind of hand that makes your day.
Dear Reader, I did work all this out before trick 3, and I did make the contract…I wish I could leave it at that, but journalistic ethics forbid. Recall that we were under time pressure; having taken several minutes to plan the play, I started to play fast once I knew the solution. I pulled 3 trumps (lefty had one), and then, somehow…a small heart came out of my hand. As it hit the table, I realized with horror I had wrecked my perfect ruffing-squeeze-without-the-count, because I could no longer cash hearts ending in dummy as was required. Bob may have seen me shake my head at this point and wondered why. Was there still hope? Yes! Go back to Plan A: If, as wasn’t extremely unlikely, righty had any stiff diamond honor, I could rectify the count for a simple squeeze, and it couldn’t be broken up. I finished the hearts (optional). These cards remained:
--- --- A8x Qx
xx --- T9x ---
I played a diamond to the 8 and K. Righty perforce played a club, ruffed, and I knew a simple squeeze was inevitable. Actually lefty had bared his CK earlier so it appeared now, but that didn’t matter. A less-than-completely-triumphant making 5. At least shifting to the simple squeeze had meant no count guess in the ending! Lefty started with x A QJxxxx Kxxxx.
The whole deal in bridgeviewer, some low spots approximate:
At the other table, Bobby Levin for once fell from grace and duplicated the defense at the first two tricks (HA, club) – hard to envision this layout, I guess. RJ did say he signaled for diamonds. But the opposing declarer fell at the first small hurdle and didn’t play a club honor from dummy! RJ could play the T and there was no chance of a squeeze – 11 to the good guys.
Friday, July 10, 2009
Between the trials in June and nationals in July, and the fact that I’m only living in Princeton for two more months until I have to go back to
I arrived in 3NT with lefty overcalling in spades: 1H-(1S)-2D-2NT-3NT
ST (rusinow) was led to my Q. I could make if diamonds came in, with maybe some additional chances. I cashed the DK and led the HK. Lefty won and continued spades; I won (ducking would have been fatal) and righty pitched a club. I tested diamonds, keeping all my hearts, but righty turned up with Jxxx. Time to try hearts; I hooked the T which held and played two more rounds, righty turning up with Jxxx in that suit also. Righty tried a low club and I had no guess; I had to try the CK and it held, so my last heart made nine tricks. Lefty’s hand was KJTxxx Ax xx Qxx. It always feels good to make a thin game with less-than-perfect breaks, but it didn’t feel as if I had done anything special or the opponents anything silly. It’s a hand where you hope for a swing but aren’t sure. There was a swing, because the defense *had* slipped, as Levin demonstrated at the other table. Do you see where?
After the first two tricks were the same (spade to Q, DK), Levin ducked the HK! This completely wrecks the entries, and declarer cannot make it even with mirrors. In fact, he went down 3. I think that for many players the duck would not be instinctive, but on a bit of thought, it almost can’t be wrong. The DK really looks like a singleton. If declarer has Qxx KQJxx Kx Jxx (so you need to win and cash clubs), he has made an implausibly subtle double-cross play by cashing the DK. With diamonds running, almost any declarer would either try to slip a heart by without cashing the DK, or run the whole diamond suit hoping for discard trouble. I am sure that Bobby’s instincts told him that diamonds weren’t running, (or at least that he should play for that), and he found the duck without much effort.
I’ll try to post the other hand, a cute squeeze hand, tomorrow.
Thursday, July 2, 2009
After the auction begins as in the previous post, you arrive in 4S with:
Dummy: A75 J93 K843 KT8
Declarer: KJ643 A62 AJ2 93
This is not one of the ideal combinations you were hoping for, but you still have play. The early play goes:
CQ, K, A, 3
Club, 9, J, 8
Heart, 9, Q, A
S3, 2, A, 9
CT, small, H2, small
S5, T, J, Q (sigh)
S8, 7, club, K (new life?)
Yes, there were some layouts where a diamond hook at trick 6 is better, but if trumps come in you’re probably making it without the D hook, so playing trumps looks right. Your bidding convinced lefty you were 5-2-4-2, so he didn’t cash the setting trick with HK (wrong of him, since on that layout the pitch he set up would be useless.) Now, here you are needing the rest with:
Dummy: --- J3 K843 ---
Declarer: 64 6 AJ2 ---
Most of you know that GIB’s algorithm for single-dummy play is to generate lots of random hands, then see how to make it double-dummy. It thus tends to miss plays that “steal” tricks in single-dummy play. Here, like a good GIB, I calculated that the only legitimate chance was for rho to have T9 tight in diamonds. Then I have 9 tricks with the backward finesse, and LHO with his presumed HK is squeezed out of his Q76x in diamonds for 10. So, I cashed my trumps pitching hearts; unfortunately no diamonds were pitched, so it was clear there had been no squeeze. Knowing it was unmakeable, I just played DK, diamond hook to try for down 1.
Did you see the play that, GIBlike, I missed? Just lead the DJ from hand. If lho has Qxx (no T), you’ll make it for sure – no way will he cover when he thinks you have AJTx and are trying to tempt him. This also retains the tiny legitimate squeeze chance I mentioned.
In practice lho had Q and T of diamonds, so nothing works, except stopping in 2 or 3 spades!