Main Study 4B73 by Richard Pavlicek

1.

A J 10 7 6 | |

5 4 3 2 |

Proper play is to finesse the jack, then finesse the 10.

2.

A Q 10 7 6 | |

5 4 3 2 |

Proper play (for maximum) is to finesse the queen, then play the ace.

See the paradox? Each holding has the same number of cards, yet with Example 1 you should finesse twice, and with Example 2 you should finesse only once.

The reason is that in Example 1 you are missing *equal* cards (K-Q). If East held K-Q doubleton, he had a choice of plays — he could win either the king or queen. In Example 2, if East held K-J doubleton there is no such choice — he would always win the king.

Consider Example 1. Assume the first finesse loses to an honor, and West follows low twice (else there is no problem). If you finesse again you will succeed when East began with a singleton honor (two possible holdings). If you play the ace you will succeed when East began with K-Q doubleton (one possible holding). Finessing twice gives you two chances instead of one.

Below is a table of the relevant holdings for Example 1:

West | East | Probability |
---|---|---|

9-8 | K-Q | 6.8% |

Q-9-8 | K | 6.2% |

K-9-8 | Q | 6.2% |

Notice that K-Q doubleton occurs slightly more often than either the singleton king or queen, but *combined* the singletons are far more likely. Hence, if you finessed twice every time without even looking at East’s card, you would have a success ratio of 12.4 to 6.8. If you finessed just once, you would be a big loser in the long run.

The classic argument against this is that once East wins, say, the king, he cannot have a singleton queen. Certainly true, but the crux is that with K-Q doubleton, East can tell you *whatever he wants*. When you observe the king you know it was either a *forced* card with a singleton or a *chosen* card from K-Q. The odds favor the forced card because he might have played the other one if he had a choice.

Perhaps a better way to look at this is by *events*. When the first finesse loses (and West follows low twice) there are *four* possible events:

East Won | Probability |
---|---|

King from K-Q | 3.4% |

Queen from K-Q | 3.4% |

King singleton | 6.2% |

Queen singleton | 6.2% |

The percentage apportionment of the first two events is arbitrary and depends on your opponent. If East varies his play equally, we have the same ratio in favor of finessing. For example, if we see East play the queen, the odds are 6.2 to 3.4 that it’s the singleton.

For practical purposes, it makes little difference how your opponent tends to play with K-Q doubleton. Any percentage lost in one event is gained in another. The only exception might be against a player who *always* plays a particular card, say the king. Then, if the queen appeared, you would have a lock in finessing; or if the king appeared, you would have a slight edge in playing for K-Q doubleton. Nonetheless, it is foolish to consider this exception since there is no way to label a player with such a habit.

Whether you accept the reasoning behind restricted choice or not, it has proved to be valid in practice. Experts know this and consistently go with the odds. The bottom line: *Always finesse twice* unless there are overriding circumstances relating to the complete deal. Or to quote from a late, not-so-great president, “Trust me; I am not a crook.”

Now look at a table of the relevant holdings for Example 2:

West | East | Probability |
---|---|---|

9-8 | K-J | 6.8% |

J-9-8 | K | 6.2% |

The above are the only possible layouts when the queen loses to the king and West follows low twice. *No choices* are involved (unless you want to consider the moronic play of the jack). Clearly, playing the ace on the second round is better, with odds of 6.8 to 6.2 in your favor.

Example 2 is also interesting from another perspective. Playing for the maximum dictates the loss of *two tricks* if East has a singleton king. If you can afford to lose one trick but not two, there is an effective safety play: *Cash the ace first*, then return to the South hand and lead toward the Q-10-7-6.

3.

A J 7 6 5 | |

K 4 3 2 |

Proper play is to cash the king and ace.

Next consider that four missing cards are expected to split 3-1 more often than 2-2. This seems to contradict the rule. If a 3-1 break is actually more likely, why not take the finesse? Good question.

The explanation lies in the fact that there are two ways for a suit to split 3-1. This is easily seen by listing the possible distributions with their approximate percentages.

West | East | Probability |
---|---|---|

0 | 4 | 5% |

1 | 3 | 25% |

2 | 2 | 40% |

3 | 1 | 25% |

4 | 0 | 5% |

Note that the 2-2 break is more likely than any specific 3-1 break; however, the *combined* chance of a 3-1 break totals 50 percent, which is the more likely occurrence.

Consider Example 3 again. After cashing the king and leading toward dummy, suppose West follows low twice (else no problem) and East plays the 10. Only two holdings are possible: East either has a singleton 10 (original 6.2 percent chance) or Q-10 (original 6.8 percent chance) so the latter is favored.

Another way to explain this is by the “available-space” theory, which has many applications in bridge. At the moment of decision, West will have followed twice and East only once, so East will have one extra space in his hand; hence he is more likely to hold the missing queen.

A similar logic applies to:

4.

A Q 10 4 | |

K 3 2 |

Proper play is to cash the top honors.

Missing six cards, it is well known that a 4-2 break is more likely than 3-3, so one might conclude that the finesse is correct. No, because at the moment of decision West will have played three low cards and East two (else there is no problem) which eliminates all but two possible layouts. Since East has played one less card, the available-space principle makes him more likely to hold the missing jack.

5.

A 10 7 6 5 | |

K 4 3 2 |

Proper play is to cash the king. Then, if the queen or jack drops from East, finesse against West for the missing honor.

Quite a difference! Because of the equality of the missing Q-J, this is almost like Example 1. An identical table could be constructed (except with Q-J instead of K-Q) to show the superiority of the finesse.

Now consider this holding, which is similar to Example 4:

6.

A Q 9 4 | |

K 3 2 |

Proper play is to cash the queen and king. Then, if the jack or 10 drops from East, finesse against West for the missing honor.

Again, the normal technique of playing for the drop is overridden by restricted choice. In this case we are comparing the chances of East having J-10-x, J-x or 10-x. An event table would look like this:

East Played | Probability |
---|---|

Jack from J-10-x | 3.6% |

10 from J-10-x | 3.6% |

Jack from J-x | 6.5% |

10 from 10-x | 6.5% |

The 3.6 percents assume that East varies his play equally with J-10-x, in which case the odds are 6.5 to 3.6 in favor of the finesse. The exact odds when East plays the jack as opposed to the 10 will vary with the defender’s habits, but for practical purposes the third-round finesse is always the better play.

7.

A Q 9 5 4 3 | |

K 2 |

Proper play is to cash the top honors, *even if East drops the jack or 10*.

If the king catches an honor from East, it may seem correct to finesse. Wrong! This is because East is not limited to a singleton honor or J-10 doubleton. It could also be a routine falsecard from J-10-x.

© 2013 Richard Pavlicek