Puzzle 8F05   Main

Toughest Beer in Bridge

  by Richard Pavlicek

Most bridge players, or at least those of the younger breed, are familiar with the “beer card,” a pseudonym for the seven of diamonds. Per modern folklore, if declarer wins the last trick with the diamond seven without negatively affecting his bridge result, he wins a beer. (Presumably the opponents buy the beer since they might have been able to prevent it.) In any case, or six-pack for that matter, don’t expect this bonus to appear in the Laws of Bridge any time soon.

Hold off on the brew for a moment and consider this challenge:

Construct a deal where South can make 7 NT against any defense without winning an ace, king or queen.

Just to clarify, if North or South held a suit A-K-Q-J-10-9-8, only four tricks (J-10-9-8) could be won by that suit.
The A-K-Q would have to be discarded on winners in the opposite hand.

As if the above construction weren’t tough enough, the following requirements must also be met:

No hand has a suit over seven cards.
South’s longest suit is hearts.
East’s longest suit is spades.
West has the best poker hand.
Declarer wins a beer!

There you have it! There is only one deal in the brewery, and if you find it you’re a master brewer.

Jim Munday Wins

In September 2010 this puzzle was presented as a contest. Understandably, its weirdness and difficulty were unappealing, as the dismal turnout of only 15 participants proved. Even so, it was a welcome change of pace from serious bridge. Only four persons, listed below by date-time of entry, produced the unique deal to satisfy all the conditions. Well done!

The winners were not only few but far between, spread over three continents (almost four considering the stretch to call England part of Europe). Three names are familiar from my poll and contest days; and the new kid on the block, Pavel Striz, might be my long lost cousin who was spelling challenged. At least it was no surprise that a guy named Pavel is Czech.

Winner List
1Jim MundayCalifornia
2Julian WightwickEngland
3Pavel StrizCzech Republic
4Leif-Erik StabellZimbabwe

Puzzle 8F05   MainTop   Toughest Beer in Bridge


Solving this puzzle required various deductions, beginning with scrutiny of the bizarre requirement that no trick can be won with an ace, king or queen. Indeed, this might seem impossible, but breaking it down to essence shows the way. On a similar vein, I can remember some students who couldn’t win an ace, king or queen; but I digress.

First, West (opening leader) cannot have any HCP; else he could lead that card to win the trick or force declarer to win with a queen or higher. Similarly, if East has any HCP, West must be void in that suit; else he could lead it for equal effect. Further, if West is void in East’s suit with HCP, North-South must have at least six cards there (East is limited to seven) which are wasted because they can’t win a trick. Also wasted are the A-K-Q in three other suits, which makes at least 15 wasted cards; so the remaining 11 cards cannot win 13 tricks.

Therefore, North-South must have all the HCP. Since the A-K-Q of each suit are wasted, they must be discarded; so 12 tricks must consist of North or South winning and pitching an ace, king or queen from the opposite hand. Obviously these tricks cannot be won by the same hand, so another trick is needed to cross the table, which by law requires following suit. This accounts for all 13 tricks, so the discarding must begin on the opening lead.

Since suits longer than seven are disallowed, the logical plan is to give North and South isolated two-suiters (opposite voids) except for the one card needed for communication. South’s longest suit must be hearts, so suppose we start by giving South S H A-K-Q-J-10-9-8 D 8 C A-K-Q-J-10; North will be 7-6 in the other suits; and West will have only hearts and clubs so any lead can be won by South. Let’s see: South wins the first six tricks with H J-10-9-8 C J-10 as North pitches his A-K-Qs; then a diamond is led to dummy, which is high. Yes, it works!

Well, not quite. There’s that little catch about West having the best poker hand, which doesn’t bode well with South holding a royal flush (two in fact). Thus, South’s sequences must be broken by removing the 10s, which cannot go to North; else declarer is deprived a discard opportunity and the goal can’t be met. Further, the 10s cannot go to West (with length) as South then couldn’t win a nine or eight; so they must go to East and be singleton, so that each falls under the jack.

The poker hurdle in diamonds is fixed similarly by giving South the D 10 for transition. But what about spades? If North has S A-K-Q-J-9-8, who gets the 10? It can’t be West (he must be void or a spade lead is fatal); it can’t be East (with length) as then the S 9 couldn’t win; so it must be South, who needs another card, since removing his 10s left a maximum of six hearts and five clubs. Alas, now South can’t win the six tricks necessary for North to pitch his A-K-Qs. Argh. Is the construction impossible after all? The plot thickens.

Aha! Instead of breaking the royal flush in spades by removing the 10, we can remove the queen and give it to South. Having only five winners in South is now fine, because North has one less top card to jettison. And North will now have an extra winner for South to pitch his A-K-Qs and the blank S Q. Finally!

Here is the complete deal, on which West wins the poker pot with an eight-high straight flush:

7 NT SouthS A K J 10 9 8TrickLead2nd3rd4th
H1 WC 8S A10J
D A K Q J 9 8 72 SC 92S KS 2
C3 SH J2D A10
STableS 7 6 5 4 3 24 SH 93D KS 3
H 7 6 5 4 3 2H 105 SH 84D QS 4
DD 6 5 4 3 26 SD 10C 3J2
C 8 7 6 5 4 3 2C 107 ND 93S QC 4
S Qcontinued below…
H A K Q J 9 8
D 10
Lead: C 8C A K Q J 9

No matter what West leads, North jettisons five top cards as South wins C J-9 H J-9-8. Next the D 10 is overtaken with the jack, and the D 9 allows South to pitch the S Q. This leaves the North hand high:

NT win allS J 10 9 8TrickLead2nd3rd4th
H8 NS J5H QH 5
D 8 79 NS 106H KH 6
C10 NS 97H AH 7
STableS 7 6 511 NS 8D 4C QC 5
H 7 6 5H12 ND 85C KC 6
DD 6 5 413 ND 7!
C 7 6 5CBeer!
North leadsC A K Q

North of course saves the D 7 for last, then as my friend “Dear Billy” would say: It’s Miller time.

The uniqueness of the deal can be sensed (but not proved) by trying to make a change. South’s spade cannot be the king, as this would give North a queen-high straight flush. West cannot have one more heart and one less club, as South’s longest suit would then be clubs. North cannot have one more spade and one less diamond, as East’s longest suit would then be diamonds and the D 7 couldn’t win a trick. And so on. Also note that all East-West cards (except the blank 10s) must be the lowest in their suits; e.g., swapping the H 7-8 prevents South from winning three heart tricks.


Julian Wightwick: The “best poker hand” condition is pretty, both to force the singleton 10s and to create uniqueness in the spade suit.

Thanks. Many people waste their time on useless pursuits, but I reserve mine for something meaningful.

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© 2010 Richard Pavlicek