Swiss teams puzzle

If it takes 30 seconds to shuffle, and 30 seconds to deal, how long does it take 2 people to make 3 boards at the start of a swiss match?

Answer below...



For most people, 2 minutes. Both people shuffle and deal a board, which takes 1 minute, then one of them makes the other board which takes another minute.

But, by scheduling better it only takes 90 seconds. Call the players A & B. Then:

0-30: A shuffles 1, B shuffles 2
30-60: A deals 1, B shuffles 3
60-90: A deals 2, B deals 3.

My partner is very slow, so I have to think about these things.

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Skip bid

In order to better convey the impression that I'm actually thinking after a skip bid, and also to introduce random variation in the length of a hesitation, I (when I don't actually have anything to think about) add up the spots in my hand.

That's all.


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After 2nt rebid

Over at Becker's Bridge Blog he posts his system after 1m-1M-2N.

I play a sort of hybrid between transfers and what he describes:

(Also applies only over 1m-1M-2N)

3C is a relay to 3D. Might be left there. Any subsequent bid, though, show opener's minor:

3N = mild slam try with 4m
3M = 5M, 4m
3oM = 4M, 5m

3D is a xfer to hearts (opener should break with preference for spades after 1m-1S-2N-3D)

3H is a xfer to spades (after 1m-1H-2N-3H, responder could be 4=4 or 4=5 in the majors, so opener bids 3N with support for neither, 3S with 3 *hearts* and past 3N with 4 spades)

3S shows the other minor

4C RKC for responder's suit

4D shows 6 of the other minor

I think showing some good supporting hands below 3N is useful, but I also like his idea of 1m-1M-2N-3C-3D-3M as showing a 6+M slam try. My style requires going beyond 3N with that hand, though that doesn't strike me as a big cost. On the other hand, I think it's more important to suggest opener's minor and still be able to stop in 3N; Becker's system doesn't seem to have that sequence at all.


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From the Reisinger

I watched this hand on Vugraph, deal 20 from the first final session of the 2009 Reisinger.

♠	10 9 5
♥	8 6 2
♦	A J 10 5
♣	Q 5 3
♠	K 8 2
♥	A Q J 7 3
♦	8
♣	A K 9 6

West	North	East	South
Pass	Pass	2♠	3♥
All Pass

Lead: ♠Q

The panel had a strong consensus that North is not supposed to raise here. I agree.

After some hesitation, East ducked the lead to declarer's K. Declarer crossed to ♦A and lost a heart finesse. A diamond came back, ducked in dummy, East played the K and declarer ruffed. After drawing a 2nd round of trumps (both following, East with the 10), declarer has 9 tricks and is left to try for a 10th.

Declarer left the last trump outstanding and successfully ruffed a club, finding West with 1=3=5=4 distribution.

This seems pretty likely to work, but I wonder if it's possible to do better. As I write this, I haven't done the analysis, so don't be prejudiced by my presentation. An analysis of an alternate line is below the fold...


This is the end position:

♠	10 9
♥	8
♦	J 10
♣	Q 5 3
♠	8 2
♥	J 7
♦
♣	A K 9 6

I think we can treat it as certain that spades started 1=6. Playing clubs as declarer did succeeds when clubs are 3-3, or when long clubs are with the remaining heart -- this is only possible with West being 1=3=5=4 or 1=3=4=5 (or, very remotely and henceforth ignored, 1=3=3=6). I'm also ignoring whatever inference is available from East's failure to insert the ♦9.

The diamond position creates some interesting loser-on-loser possibilities, even though short a dummy entry.

My first alternate line was to draw the last trump and exit a spade, aiming for a minor suit squeeze. If the defense wins 2 spades, you're set up and just need the squeeze to operate. If they win 1 and play a diamond, you can just pitch your losing spade and you'll get a diamond trick now or later. But, East can thwart this by winning the spade and playing a club. You can still try ♣A, ♣Q, and try to throw West in with a diamond. This will work if clubs are 3-3, or if East has short clubs but at least 1 honor and West has the ♦Q.

This is too much work, so I'm going to start by assuming clubs are 4=2 (everyone makes when they're 3-3). Restricted choice applies to both the ♦Q and the ♥9. If East is 6=3=2=2 there are 3 possible holdings (6 small diamonds, 1/2 for restricted choice in hearts). 6=2=3=2 without ♦Q is 15 (6c2 diamonds). 6=2=3=2 with ♦Q is 3 again.

Declarer goes down in the first case, but makes in the other 2 (18 total). Meanwhile, with 2 clubs East has at least one honor 60% of the time, so my alternate line will make the overtrick in 60% of the first 2 cases (but fail in the 3rd). That's only 10.8 total. Misdefense is plausible but not plausible enough to make up this difference.

A different alternate line is to draw the last trump, play 3 clubs ending in dummy, then pitch a spade on a diamond honor. If clubs are 3-3, this makes 11 tricks (West will be diamond tight and must have ♦Q). If clubs are 4=2, this makes 10 tricks when West has ♦Q, 9 otherwise (same odds as declarer's line, though different losing cases). When clubs are 5=1 this only makes 9 tricks (unless East has a stiff honor, then change tacks and win the 2nd club in dummy). Of course, declarer at the table also is held to 9 tricks when East has the 3rd heart with a stiff club.

This is much closer than my first idea. Comes out ahead when clubs are 3-3, 20 positions x 6 diamond positions = 120. Comes out behind when East is 6=2=4=1 with a small stiff club (4) x (20 Kxxx diamonds + 7.5 KQxx diamonds) or a stiff club honor (2) x 7.5 KQxx diamonds (never mind that AJxxxx/Tx/KQxx/J might open 1S). That's 125 positions.

So, declarer still comes out slightly ahead (125:120 in the positions that matter), but it's very close. If the offense hadn't held the ♥8 it would have swung the other way (since East 6=3=2=2 would be twice as likely without restricted choice applying to the ♥10.


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