Main Puzzle 8J01 by Richard Pavlicek
A simple finesse, whether it wins or loses, will often decide the fate of a contract — and sometimes the winner of a tournament. The element of luck involved tends to balance out over the long run, though bridge players have a tendency to dwell on their unlucky deals. “My finesses never work!” is a common lament when yet another contract falls by the wayside.
Suppose a contract depends on a straight finesse for a king and nothing else. Clearly this a 50-percent chance a priori. Half the time the king will be on your left; half the time on your right; so you will make the contract half the time — exactly.
Now suppose a contract depends on two finesses (again assume straight finesses for a king). The logical odds now become 25 percent (50% × 50%) that both will work, 25 percent that both will fail, and 50 percent that one works and the other fails. Well, approximately. The placement of two kings is not independent like the outcome of two coin flips. Instead, the location of one king affects the odds on the location of the other.
To illustrate, consider your prospects as South to make this slam:
6 NT | A 4 3 2 A 4 3 2 K 2 Q J 10 | ||
Q J 10 9 Q J 10 9 A Q A K 2 |
Obviously the K and K are your two nemeses. If West has both you will make an overtrick. If East has both you will fail — perhaps miserably with a diamond lead. If the kings are split you will make 6 NT exactly. Below is a table of the a priori chances:
Case | West | East | Percent |
---|---|---|---|
1 | K K | — | 24 |
2 | K | K | 26 |
3 | K | K | 26 |
4 | — | K K | 24 |
The percentages are easily calculated by combinatorics. The chance of West being dealt both kings is 13/26 × 12/25 = 24 percent. Likewise for East. The chance of West being dealt one specific king and East the other is 13/26 × 13/25 = 26 percent. Available space accounts for the 2-percent difference; once a king is placed in one hand, the other hand has more room to receive the second king. In other words, the chances are dependent.
From the table it is easily seen that your chance of success in 6 NT is the sum of Cases 1-3 = 76 percent. Or more simply, you will fail only in Case 4 (24 percent).
Now suppose we swap the North-South heart holdings:
6 NT | A 4 3 2 Q J 10 9 K 2 Q J 10 | ||
Q J 10 9 A 4 3 2 A Q A K 2 |
The previous percentages remain the same, but the failing case becomes different. The chance of success now drops to 74 percent, because you fail in Case 3 (26 percent). As any expert would be aware, the underlying principle can be summarized: Two finesses through the same hand offer better odds than through different hands. Or in the common case of needing at least one of two finesses, the chance is never 75 percent; but either 76 or 74 percent, respectively.
That was too easy. Suppose we revise the original deal to include a club finesse as well, thus involving three finesses:
6 NT | A 4 3 2 A 4 3 2 K 2 Q J 10 | ||
Q J 10 9 Q J 10 9 A Q A 3 2 |
To succeed you need at least two of the three finesses to work, which is logically 50 percent. The best way to understand this is to consider the outcome of three finesses as two distinct groups: More of them win, or more of them lose. Either group should be equally likely, at least without considering dependency. So how does dependency affect the issue now?
For the above deal, see if you can solve this three-part puzzle:
A. What is the exact chance of success? |
B. What if the N-S hearts are swapped? |
C. What if the N-S clubs (only) are swapped? |
Curiously, the chance of at least two of three finesses working is unaffected by the direction each is taken. Dependency effectively washes out, even when all three finesses are through the same hand. The chance of success in Part A, B or C is exactly 50 percent. Sorry if you wasted energy looking for some hidden particle, but it’s just not there — like the Higgs boson nobody can find.
The easiest way to verify this is with a table (see below). Note that all three kings in one hand calculates to 13/26 × 12/25 × 11/24 = 11 percent; and each balanced split calculates to 13/26 × 12/25 × 13/24 (or 13/26 × 13/25 × 12/24) = 13 percent. Part A succeeds in Cases 1,2,5,6; Part B in Cases 2,3,5,8; and Part C in Cases 1,2,3,4. The four successful cases in each part comprise one 11-percent and three 13-percent chances, totaling 50 percent.
Case | West | East | Percent |
---|---|---|---|
1 | K K K | — | 11 |
2 | K K | K | 13 |
3 | K K | K | 13 |
4 | K K | K | 13 |
5 | K | K K | 13 |
6 | K | K K | 13 |
7 | K | K K | 13 |
8 | — | K K K | 11 |
Charles Blair (Urbana, Illinois) offered an even simpler solution: No matter how you place the three kings, swapping the East-West hands will change success to failure, or vice versa. Hence this one-to-one relationship for all distributions dictates a 50-percent chance — exactly. Top Main
© 2013 Richard Pavlicek