Main Puzzle 8J01 by Richard Pavlicek
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.
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? |
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.
© 2013 Richard Pavlicek