Main
Puzzles 3Y02 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 (typically just arithmetic). While some provide practical instruction, others are just novelties based on curious observations.

The answer is shown immediately after each puzzle, however, you’ll enjoy them more if you try to solve each one on your own. In other words, no peeking!

The subject of probability is more often a case of logical thinking than an exercise in mathematics. Do you have a logical mind? Assuming that any finesse is a 50-percent chance, what is the chance that at least two out of three finesses will work?

The outcome of three finesses can be thought of as two distinct cases: (1) More of the finesses win or (2) more of the finesses lose. Logically, either case must be equally likely, so it is 50 percent.

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

Each side could make four contracts (say, 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.

At rubber bridge, what is the *fewest* total number of tricks a side can take and win the rubber? And by how many points could you win that rubber? Assume there are no irregularities. Think about it.

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!

On the last board of a Swiss team match, the contract at one table was four spades doubled, which was made exactly, vulnerable. The same contract was reached at the other table, but it was *not doubled*. Amazingly, the board was a push! How can this be?

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 compared with 790 for making four spades doubled. (The IMP scale nullifies the 10-point difference.)

In a victory-point Swiss team event (playing seven-board matches) the sets of boards commonly used are Boards 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?

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 gains or losses, which means more victory points. Observe that *eight-board* matches would be fair all around. Is anyone listening?

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?

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.

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

*Two* notrump (making seven). If you racked your brains on this one, I apologize.

At duplicate bridge there are several ways to score +550. The common ways are to bid and make three notrump or five of a minor (doubled, nonvulnerable). How else can +550 be obtained?

The only other way is to defeat your nonvulnerable opponents 11 tricks! Did somebody forget to double?

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?*

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.

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?

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?

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

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 a 6 opening, etc. See the pattern? The sum is 2 + 2 + 2^{2} + 2^{3} … + 2^{29}, which is equal to 2^{30}.

Which of these bridge 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

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

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

(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.

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)

A. The probability percentages are: 9 HCP = 9.3562, 10 HCP = 9.4051, and 11 HCP = 8.9447.

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. Here is a poser: What is the most points one side could score in a rubber *if no contract by either side is defeated?*

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!

Thanks to N. Scott Cardell for a correction in my arithmetic.

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

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 change declarer’s spades to Q-J-10-9-8-7. C. 1. Assume 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.

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

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.

There are 39 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?

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?

Which is the largest?

A. The number of legal bridge auctions

B. The number of bridge deals

C. The number of feet to the nearest star.

A. The number of possible bridge auctions is almost beyond comprehension, 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!

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

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

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

A. 635 million

B. 635 billion

C. 635 quintillion

B. To be exact, 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.

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?

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.

A “Yarborough” is defined as 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

C. If you actually 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.)

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?

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.

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

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.

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?

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.

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

[Revised 9-19-02]. When I created this puzzle I thought the answer was nine, but a recent discovery brings new hope to overbidders everywhere. Congratulations to Dmitri Shabes of Walnut Creek, California, who came up with a neat construction: 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 cannot be stopped with any defense. Curiously, with a *heart* lead you can even win 11. A cold game with no points! Scary.

During a Swiss team match your opponents bid seven notrump, vulnerable, and make it (undoubled). When you compare scores the board turns out to be a push, although your teammates *did not bid any slam*. What happened at the other table?

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.)

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

A. Cash the top honors; B. Lead the two and finesse the 10. The reason for the difference is that with (1) you can benefit from a doubleton jack, whereas with (2) the 10 falls on the same trick.

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

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 (1) you could not finesse twice if you cashed the ace first.

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

A. Finesse the queen then (if it wins) cash the ace; B. Lead the four to the jack. Note that with (2) 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.

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

A. Finesse the 10 then finesse the nine; B. Cash the king and ace. The reason for the difference is that with (2) you cannot finesse twice, and a single finesse would often lose to a doubleton honor.

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

A. Run the 10 then run the nine; B. Cash the ace. The reason for finessing with (1) is to avoid a second-round guess if the ace were played first. Note that with (2) there can be no guess because if both follow low to the ace, the only K-J will be missing.

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

A. Finesse the jack then the 10; B. Same! (I reserve the right to be tricky.) In general, missing the K-Q, it makes no difference how many cards you have — always finesse twice if possible.

© 2008 Richard Pavlicek