Bridge Teasers

The Answer to each of the following teasers is hidden, but can be seen by highlighting the section below the red "answer" text with your mouse

Two Finesses

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?

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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.

Most Contracts

At rubber bridge what is the greatest number of contracts that can be made in a single rubber?

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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.

Fewest Tricks

At normal 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.

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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!

Finesse For a Jack

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 no trump:

1.   A K 10 3 2   opp.    Q 4

2.   A K 4 3 2    opp.   Q 10

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1. Cash the top honours; 2. 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 will fall on the same trick.

Finesse For a Queen

Assuming you have adequate entries to either hand and with no clues from the bidding and early play, what is the best play of each of these card combinations for five tricks at no trump:

1.   A K J 10 9   opp.     8 7

2.   A K J 10 9   opp.   8 7 6
 

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1. Finesse the jack then finesse the 10; 2. 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.

Finesse For a King

Assuming you have adequate entries to either hand and with no clues from the bidding and early play, what is the best play of each of these card combinations for the maximum tricks at no trump:

1.   A Q 6 5 4   opp.   J 3 2

2.   A Q 6 5 4   opp.     J 3
 

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1. Finesse the queen then (if it wins) cash the ace; 2. 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.

Queen-Jack Finesse

Assuming you have adequate entries to either hand and with no clues from the bidding and early play, what is the best play of each of these card combinations for four tricks at no trump:

1.   A K 10 9 8   opp.    7 6

2.   A 10 9 8 7    opp.   K 6

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

King-Jack Finesse

Assuming you have adequate entries to either hand and with no clues from the bidding and early play, what is the best play of each of these card combinations for four tricks at no trump:

1.   Q 10 9 8 7   opp.     A 6 5

2.   Q 10 9 8 7   opp.   A 6 5 4

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1. Run the 10 then run the nine; 2. 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 missing cards will be the K-J.

King-Queen Finesse

Assuming you have adequate entries to either hand and with no clues from the bidding and early play, what is the best play of each of these card combinations for four tricks at no trump:

1.   A J 10 9 8   opp.     7 6 5

2.   A J 10 9 8   opp.   7 6 5 4

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1. Finesse the jack then the 10; 2. 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.

Most Points

In normal rubber bridge there is no limit to the number of points that could be scored in a rubber because of 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?
 

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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 honours on each deal (150 x 21) and they win the rubber bonus (500). Total score for N-S: 14,830!

Point Count Zoo

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?

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The deck has 40 HCP, so start with 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 7, North 11 and West 18.

How To Score +270

At duplicate bridge there are two ways to obtain the rare score of +270. One is to bid one no trump and win all 13 tricks. What is the other way?

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

How To Score +550

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, non-vulnerable). How else can +550 be obtained?

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

Most Bids and Calls

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?

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There are only 35 possible bids, so no 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

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?

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

Six Notrump Bids

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.

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(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 S opening, four with a 6 H opening, eight with a 6 D opening, etc. See the pattern? The sum is 2 + 2 + 2^2 + 2^3 … + 2^29, which is equal to 2^30.

Least Likely Hand

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.
 

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(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.

Four Aces

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.
 

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(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.

Most Likely HCP

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 is equally likely.

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

Winning and Losing

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?

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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

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?

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(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

Which is the largest? (A) the number of legal bridge auctions, (B) the number of bridge deals, or (C) the number of feet to the nearest star.
 

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(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!

Greatest Score

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

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

Singleton or Void

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.
 

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(C). Most people are surprised by this, but almost four deals out of five have a singleton or void somewhere. Practical tests on this confirm that it is quite true.

Number of Hands

What is the total number of different bridge hands that could be dealt? (A) 635 million, (B) 635 billion, (C) 635 thousand billion
 

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(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.

Three-Two Fit

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?
 

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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.

Yarborough Streak

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.

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(C). If you actually tried to calculate this, you have absolutely no sense of priorities. The answer would be in the quadrillions. A little bird told me it is 20,411,783,673,914,367 to 1.