Puzzles 3Y31 Main |
| by Richard Pavlicek |

These short bridge puzzles created by Richard Pavlicek have been used as fillers in a variety of publications related to bridge. Most are based on logic or simple math. While some provide practical instruction, others are just novelties based on curious observations.

Answers can be found at the end. *No fair peeking!*

Assuming any finesse is a 50-percent chance, what is the chance that at least two out of three finesses will work?

At rubber bridge (traditional, not Chicago) what is the greatest number of contracts that can be made in a single rubber?

At rubber bridge, with no irregularities, what is the *fewest* number of tricks a side can take and win the rubber?

And by how many points would you win that rubber?

On the last board of a Swiss team match, the contract at one table was four spades doubled, made exactly, vulnerable.

The same contract was reached at the other table but *not doubled*. The board was a push! How can this be?

In a victory-point Swiss team event, the board sets commonly used are: 1-7, 8-14, 15-21, 22-28 and 29-35.

One of these sets gives its table a slight advantage over all the others. Which set is it?

On a certain deal North has as many HCP as South and East together; West has as many HCP as North and East together. East has more HCP than South and no two players have the same number of HCP. How many HCP does each player have?

At duplicate bridge there are two ways to obtain a score of +270. One is to bid one notrump and win all 13 tricks. What is the other way?

At duplicate bridge there are several ways to score +550. Common ways are to make 3 NT or five of a minor doubled (nonvulnerable). How else can +550 be obtained?

In a legal bridge auction (no insufficient bids or other irregularities), what is the maximum number of *bids?*

Also, what is the maximum number of *calls?*

Most people are aware that 3 NT is the most common contract reached. What would you guess are the next five most common contracts? And what is the *least* common contract of all?

Assuming you are the dealer and the opponents pass throughout, how many different bidding sequences are possible to reach 6 NT?

A. One thousand

B. One million

C. One billion

What are the odds against being dealt all four aces?

A. 4 to 1

B. 64 to 1

C. 256 to 1

D. 378 to 1

Puzzles 3Y31 Main | Top Short Bridge Puzzles |

It is a proved fact that in a random bridge hand, the number of even-ranked cards is more likely to be odd, and the number of odd-ranked cards is more likely to be even. Forget that, and answer just one question: What is “Never odd or even?”

In rubber bridge (traditional, not Chicago) there is no limit to the number of points that could be scored in a rubber due to the possibility of endless sets. What is the most points one side could score in a rubber *if no contract by either side is defeated?*

It should not be surprising that a hand with exactly 10 HCP is the most likely to be dealt. What is the second most likely?

A. 9 HCP

B. 11 HCP

C. 9 or 11 HCP (equally likely)

What is the *fewest* number of HCP a partnership may hold and be able to make:

A. A grand slam

B. A small slam

C. 11 tricks

What is the *most tricks* declarer could win against best defense with no HCP in either hand?

In rubber bridge it is possible to win a rubber without making a contract or defeating an opponent’s contract. How can this be done?

There are 39 generic hand patterns, ranging from 4-3-3-3 to 13-0-0-0.

A. Which hand pattern is the most common?

B. Which is more likely: 4-3-3-3 or 5-4-3-1?

C. Which is more likely: 4-4-4-1 or 6-4-2-1?

Which is the largest?

A. The number of legal bridge auctions

B. The number of possible bridge deals

C. The number of feet to the nearest star

At rubber bridge what is the greatest score (most points) that can be won or lost on a single deal?

Most players have had the bizarre experience of being declarer in a 3-2 trump fit, typically through a bidding mishap.

What is the *most tricks* that could be won?

What is the total number of different bridge hands that could be dealt?

A. 635 million

B. 635 billion

C. 635 quintillion

A “Yarborough” is a bridge hand with no card above a *nine*. What are the odds against being dealt five consecutive Yarboroughs?

A. 625 million to 1

B. 228 billion to 1

C. Who cares

Puzzles 3Y31 Main | Top Short Bridge Puzzles |

As declarer in notrump, assume you must play a suit combination of A-J-5-4 opposite K-3-2.

What is your approximate chance (percent) to win four tricks?

What is the probability that a random bridge deal will contain a singleton or void in at least one hand?

A. 37 percent

B. 58 percent

C. 79 percent

At duplicate bridge what is the *lowest* multiple of 100 that is impossible to score on any board?

Also, what would the answer have been in 1980?

In a Swiss team match your opponents bid and make 7 NT vulnerable (undoubled). At the other table your teammates *did not bid any slam*, yet the board was a push. What happened?

Which of these hands is the *least* likely to be dealt?

A. A-8-4 K-9-2 J-10-2 Q-8-7-4

B. 13 cards in the same suit

C. 13 black cards

Assuming you have adequate entries to either hand, and no clues from the bidding and early play,

what is the best play of each of these card combinations for five tricks at notrump:

A. A-K-10-3-2 opp. Q-4

B. A-K-4-3-2 opp. Q-10

Assuming you have adequate entries to either hand, and no clues from the bidding and early play,

what is the best play of each of these card combinations for five tricks at notrump:

A. A-K-J-10-9 opp. 8-7

B. A-K-J-10-9 opp. 8-7-6

Assuming you have adequate entries to either hand, and no clues from the bidding and early play,

what is the best play of each of these card combinations for the maximum tricks at notrump:

A. A-Q-6-5-4 opp. J-3-2

B. A-Q-6-5-4 opp. J-3

Assuming you have adequate entries to either hand, and no clues from the bidding and early play,

what is the best play of each of these card combinations for four tricks at notrump:

A. A-K-10-9-8 opp. 7-6

B. A-10-9-8-7 opp. K-6

Assuming you have adequate entries to either hand, and no clues from the bidding and early play,

what is the best play of each of these card combinations for four tricks at notrump:

A. Q-10-9-8-7 opp. A-6-5

B. Q-10-9-8-7 opp. A-6-5-4

Assuming you have adequate entries to either hand, and no clues from the bidding and early play,

what is the best play of each of these card combinations for four tricks at notrump:

A. A-J-10-9-8 opp. 7-6-5

B. A-J-10-9-8 opp. 7-6-5-4

Which of these holdings is most likely to cost a trick if chosen as the opening lead against 3 NT?

A. K-J-7-6

B. A-J-10-7

C. A-Q-7-6

Which of these features is least likely to occur in a random bridge hand?

A. Four aces and four kings

B. Ten-card suit

C. One card of each rank

There are 635+ billion unique bridge hands. What is the *fewest* you would have to play to be a favorite to have held the same hand twice?

A. 1 million

B. 10 million

C. 100 million

D. 1 billion

As dealer you pick up a blockbuster hand of 26 HCP: A-K-J-2 K-Q-J A-K-J K-J-3. What is the chance that partner has exactly 7 HCP?

A. 3.38 percent

B. 4.91 percent

C. 7.05 percent

D. None of the above

Which is more likely for a partnership to hold:

A. Six-card spade fit

B. Seven-card spade fit

C. Neither (equally likely)

On average, what is the highest makable contract on a bridge deal?

A. Two spades

B. Three spades

C. Four spades

**Two Finesses** The outcome of three finesses can be thought of as two distinct cases: More of the finesses win, or more of them lose. Logically, either case must be equally likely, so it is 50 percent.

**Most Contracts** Each side could make four contracts (e.g., 1 , 1 , 1 , 1 ) without completing a game, but the next contract made by either side must end the game. Hence, nine contracts could be made in one game, so a three-game rubber could produce 27.

**Fewest Tricks** On the first deal your opponents bid five clubs and take all 13 tricks (minus 140). On the second deal they play 7 NT redoubled and you hold all four aces! In a state of shock you forget to cash two of them but still beat the contract two tricks (plus 1150). The third deal is the same as the first (minus 140) which gives them the rubber bonus (minus 700). Your side took only two tricks, and you won the rubber by 170 points!

**Push Board** At the other table four spades was reached by the *opposite side* after a bidding mishap. This was set *eight* tricks (vulnerable) for 800, which is a push versus 790 for making four spades doubled. (The IMP scale nullifies the 10-point difference.)

**Best Board Set** Boards 1-7 have two deals with both sides vulnerable, and only one with neither side vulnerable. Hence they offer the greatest potential for IMP gain or loss, which means more victory points. Note that matches of *eight* boards would be fair all around. Is anyone listening?

**Point Count Zoo** The deck has 40 HCP, so we have the equation: N+S+E+W = 40. Since N=S+E and W=N+E, substitution produces: 3S+4E = 40, which has four integer solutions: S=12 E=1; S=8 E=4; S=4 E=7; S=0 E=10. The first two do not satisfy the condition that East has more HCP than South, and the last (S=0) is rejected because it gives North the same total as East. Hence, South has 4 HCP, East has 7, North 11 and West 18.

**How To Score +270** Two notrump making seven. If you racked your brain on this, I apologize.

**How To Score +550** The only other way is to defeat your nonvulnerable opponents 11 tricks! Did somebody forget to double?

**Most Bids and Calls** There are only 35 possible bids, so no legal auction could have more than 35 bids. The maximum number of calls is 319! The ridiculous auction begins: P P P 1 P P Dbl P P Rdbl P P 1 , and ends: 7 NT P P Dbl P P Rdbl P P P.

**Common Contracts** The most common contracts are, in order: 3 NT, 4 , 4 , 2 , 2 , 1 NT. As the least common contract most people would guess one of the grand slams (say, 7 ) but that is wrong. Think about it. When was the last time you played 5 NT?

**Six Notrump Bids** C. 1,073,741,824 to be exact. There are two sequences with a 6 NT opening (6 NT P P P or P P 6 NT P P P), two with a 6 opening, four with a 6 opening, eight with 6 , etc. The sum is 2 + 2 + 2^{2} + 2^{3} … + 2^{29}, which equals 2^{30}.

**Four Aces** D. The probability of one player being dealt all four aces is calculated as: 13/52 x 12/51 x 11/50 x 10/49 = 11/4165, so the probability of not getting four aces is 4154/4165. Hence, the odds would be 4154 to 11 against, or approximately 378 to 1.

Puzzles 3Y31 Main | Top Short Bridge Puzzles |

**Never Odd or Even** It’s a *palindrome*.

**Most Points** Game 1: N-S bid 1 doubled making 7 twice (690 x 2); E-W bid and make 1 four times; N-S bid and make 7 NT redoubled (1980). Game 2: N-S bid 1 doubled making 7 twice (1290 x 2); E-W bid and make 1 five times. Game 3: N-S bid 1 doubled making 7 twice (1290 x 2); E-W bid and make 1 four times; N-S bid 1 NT redoubled making 7 (2660). In addition, N-S have 150 honors on each deal (150 x 21) and they win the rubber bonus (500). Total score for N-S: 14,830!

**Nine or Eleven?** A. Percentages: 9 HCP = 9.3562, 10 HCP = 9.4051, and 11 HCP = 8.9447.

**Fewest High-card Points** A. 5. Assume declarer has A-J-10-9-8-7 8-7-6-5-4-3-2, and dummy has 6-5-4-3-2 5-4-3-2 5-4-3-2. Spades split 1-1 and hearts 3-3, so the heart suit is easily established. B. 3. Like above but with declarer’s spades: Q-J-10-9-8-7. C. 1. Declarer has J-10-9-8-7-6 8-7-6-5-4-3-2, and dummy has 5-4-3-2 5-4-3-2 6-5-4-3-2. Opponents’ spades are a singleton ace opposite K-Q, and hearts split 3-3.

**Zero High-card Points** Ten tricks. Dummy holds 5-4-3-2 — 9-8-7-6-5-4-3-2 2, declarer has 10-9-8-7-6 -- -- 10-9-8-7-6-5-4-3, one defender has Q-J A-K-Q-J-10-9 Q-J-10 Q-J, and the other A-K 8-7-6-5-4-3-2 A-K A-K. In spades, 10 tricks can be won against any defense, and with a *heart* lead you can even win 11. A cold game with no points! Scary.

**Winning at Losing** However unlikely, it is possible for one side to be set many times (-50) while claiming honors (+100 or +150) to produce a greater total than the opponents after they win two games and the rubber bonus.

**Hand Patterns** A. 4-4-3-2 is by far the most common. It occurs 21.56 percent of the time. B. 5-4-3-1 occurs 12.93 percent of the time, and 4-3-3-3 only 10.54. C. 6-4-2-1 occurs 4.70 percent of the time, and 4-4-4-1 only 2.99. Are you surprised?

**Huge Numbers** A. The number of possible bridge auctions is incomprehensible, a 48-digit number! The number of possible bridge deals is in the octillions, a 29-digit number. To put these into perspective, consider that the number of feet to the nearest star is only an 18-digit number. Better start walking!

**Greatest Score** Many would answer 7,600 (down 13, redoubled, vulnerable), but don’t forget the possibility of 150 honors: 7,750!

**Three-Two Fit** All 13! Say, declarer has A-Q-J A-K-Q-5 4-3-2 4-3-2, and dummy has K-2 4-3-2 A-K-Q-5 A-K-Q-5. If both enemy hands are 4-3-3-3 (with four spades), declarer can win all the tricks, even after a trump lead.

**Number of Hands** B. There are 635,013,559,600 unique bridge hands. Nobody actually counted them, but it is easy to calculate as the number of combinations of 52 items taken 13 at a time.

**Yarborough Streak** C. If you tried to calculate this, you have absolutely no sense of priorities. The answer would be in the quadrillions. (A little birdie told me it is 20,414,133,359,114,717 to 1.)

Puzzles 3Y31 Main | Top Short Bridge Puzzles |

**Chances Are** To win four tricks you need two favorable things: a 3-3 break and a finesse. The chance of a 3-3 break is about 36 percent, of which you will succeed only half the time: 18 percent.

**Singleton or Void** C. Most people are surprised by this, but almost four deals out of five have a singleton or void somewhere. I have made practical tests on this, and it holds quite true.

**Impossible Score** 1500. Every multiple of 100 through 1400 can be scored in one way or another. When I first posed this puzzle (circa 1980) the answer was 2400 because the scoring of doubled, nonvulnerable sets was then 100, 300, 500, 700, … 2500. When they changed the scoring, nobody seemed to care that it screwed up my puzzle.

**Score This One Up** Your teammates must have played in one of a minor, doubled and redoubled, making six. This produces a score of +2230, which is a push when compared to your score of -2220. (The IMP scale nullifies your 10-point gain.)

**Least Likely Hand** A. Because it is one specific hand. Note that “13 cards in the same suit” comprises *four* specific hands, hence four times as likely, and “13 black cards” comprises millions of hands.

**Jack Finesse** A. Cash the top honors; B. Lead the two and finesse the 10. The reason for the difference is that with (A) you can benefit from a doubleton jack, whereas with (B) the 10 falls on the same trick.

**Queen Finesse** A. Finesse the jack then finesse the 10; B. Cash the ace then finesse the jack (and the 10 if necessary). The reason for the difference is that with (A) you could not finesse twice if you cashed the ace first.

**King Finesse** A. Finesse the queen then (if it wins) cash the ace; B. Lead the four to the jack. Note that with (B) it is impossible to win all five tricks, and leading toward the jack gains when the opponent in front of the jack has K-x.

**Queen-Jack Finesse** A. Finesse the 10 then finesse the nine; B. Cash the king and ace. The reason for the difference is that with (B) you cannot finesse twice, and a single finesse often loses to a doubleton honor.

**King-Jack Finesse** A. Run the 10 then run the nine; B. Cash the ace. The reason for finessing with (A) is to avoid a second-round guess if the ace were played first. With (B) there can be no guess if both follow low to the ace, because only K-J are missing.

**King-Queen Finesse** A. Finesse the jack then the 10; B. Same! In general, missing the K-Q, it makes no difference how many cards you have — always finesse twice if possible.

**Worst Lead** B. A-J-10-7. In fact it is the worst of all four-card holdings, costing a trick over 38 percent of the time.

**Rarest Hand** A. Of 635+ billion possible bridge hands, only 1,086,008 have four aces and four kings; 10,455,016 have a 10-card suit; and 67,108,864 have one card of each rank.

**Same Hand Twice** A. Only 938,252 to be exact. Most people are surprised by this, but the math is analogous to the classic “Birthday” problem. Despite the massive number of hands, the chance of all being unique falls under 50 percent after 938,252 hands.

**Seven High-card Points** D. In fact it is *zero*. You have all the kings and jacks, so partner cannot have any *odd* number of HCP.

**More Likely Fit** C. If one side has six spades, the other must have seven, and vice versa, so either case is equally likely.

**Highest Contract Average** B. Four spades can be made (by either side) about 7.9 percent of the time, which narrowly tops 3 at 7.6 percent. (Other suits, being outranked, are much less likely.)

Puzzles 3Y31 Main | Top Short Bridge Puzzles |

© 2008 Richard Pavlicek