NORTH
♠ x x |
||

WEST
♠ Q J 10 9 |
EAST
♠ x x | |

SOUTH
♠ A K 8 x x |

To get a handle on how to approach this problem, let's start with a more familiar position:

NORTH
♠ x x x x |
||

WEST
♠ Q J |
EAST
♠ x x | |

SOUTH
♠ A K 10 x x |

*all*the time (more than 93.75% of the time to be precise), declarer can't gain an advantage.

Let's say, for example, that you play the queen from queen-jack doubleton 80% of the time and declarer knows that. How should declarer play when you drop the queen? If he finesses, he wins whenever you were dealt a stiff queen. If he plays for the drop, he wins 80% of the time that you were dealt queen-jack doubleton. Since those holdings are almost equally likely, the odds are roughly 1 to .8, or about 5 to 4, in favor of finessing. Your deviating from the correct strategy does not affect declarer's proper play.

Why, then, should you play the queen only half the time? If declarer plays correctly, it doesn't matter what you do. But what if declarer is a fool? What if declarer, perversely, decides to finesse if you play jack but to play for the drop if you play the queen? You still have an edge against such a declarer, but you don't have the edge you are entitled to. This declarer will pick up a singleton jack in your hand, and he will pick up 80% of your queen-jack doubletons - a total of 1.8 out of 3 cases (the three cases being singleton jack, singleton queen, and queen-jack doubleton). He's entitled to pick up 2 out of the 3 cases by finessing, so you do have an edge. But you would have an even greater edge if you played the queen only half the time. In that case, you would hold declarer to 1.5 wins in 3 cases. While you can hold declarer to even fewer wins than that by playing the queen even less often, such a strategy would not fully punish the opposite kind of fool: the declarer who finesses if you play the queen but plays for the drop if you play the jack. By playing the queen half the time, you hold both fools to 1.5 wins out of 3. The reason you must randomize, then, is to avoid having to guess what kind of fool declarer might be. It is to guarantee you the same edge against a declarer who misplays regardless of

*how*he misplays.

How do we determine the optimum frequency with which to play each card? You must play each card with the same frequency with which you would play it if you actually did have a singleton. Since you hold a singleton queen and a singleton jack with equal frequency, you must choose the queen or jack from queen-jack doubleton with equal frequency. It doesn't always work out that neatly.

Let's look at the following slightly more complicated position:

NORTH
♠ x x |
||

WEST
♠ Q J 10 |
EAST
♠ 9 x x | |

SOUTH
♠ A K x x x |

Q J | 1/6 |

Q 10 | 0 |

J Q | 1/6 |

J 10 | 1/6 |

10 Q | 1/3 |

10 J | 1/6 |

So you should choose your play from queen-jack-ten with the same frequencies. I think most players intuitively know not to play queen-ten, since that effectively marks you with the jack. I suspect, however, few players are aware of the corrollary: that you should play ten-queen twice as likely as any other sequence.

Obviously, jack-ten-nine works similarly. Your plays should have the following frequencies:

J 10 | 1/6 |

J 9 | 0 |

10 J | 1/6 |

10 9 | 1/6 |

9 J | 1/3 |

9 10 | 1/6 |

What about queen-ten-nine? You aren't going to drop the queen at trick one or trick two, so you will play as follows:

10 9 | 1/2 |

9 10 | 1/2 |

And with queen-jack-nine? As before, let's assume there is no gain in representing a singleton. So playing an honor on the first round is pointess. In fact, it's worse than pointless, because, when you follow with the nine on the second round, you would render it unlikely that you have a doubleton. Under this assumption you would play as follows:

9 Q | 1/2 |

9 J | 1/2 |

Now we are prepared to solve the original problem: How should you card with queen-jack-ten-nine to offer declarer no inference as to whether we began with two, three, or four trumps? As before, the answer is: you must mimic the frequency with which you would choose each sequence if you actually began with a doubleton or trebleton.

To choose one example, how often should you play queen-jack? You would play queen-jack from queen-jack doubleton half the time, and you would play queen-jack from from queen-jack-ten one sixth of the time. There are six relevant doubletons and four relevant trebletons - a total of ten cases. So the frequency with which you would play queen-jack from the relevant two- and three-card holdings is 1/2 times 1/10 plus 1/6 times 1/10, or 1/15 of the time. 1/15, then, is the frequency with which you should choose queen-jack from queen-jack-ten-nine. If you work it all out, you wind up with the following frequencies:

Q J | .07 |

Q 10 | 0 |

Q 9 | 0 |

J Q | .07 |

J 10 | .08 |

J 9 | 0 |

10 Q | .13 |

10 J | .08 |

10 9 | .12 |

9 Q | .15 |

9 J | .18 |

9 10 | .12 |

To generalize, you should usually play either nine-jack or nine-queen. Somewhat less frequently, you should play ten-queen or the ten and nine (in either order). Less frequently still, you should play touching honors in either order.

This, of course, is of theoretical interest only. No one is ever going to bother with this strategy. But I find it interesting that my intuition was so wrong.

*LIKE*

ReplyDeleteSeconded. Please feel free to post such digressions at any time (not that my opinion matters!)

ReplyDeleteThanks for putting the whipped cream and cherry on my 2nd post, where I cleaned up my first quite a bit. BTW, I am very sure you have the math close enough for all practical purposes, and much cleaner than my first ramble.

ReplyDeleteI have one teeny weeny piece to add - If you have played 9-Q from QJ109, and declarer next decides to play low from hand, you MUST win the Jack in case declarer would pitch differently from dummy depending on whether you had 3 or 4 spades.

You write "you play the queen from queen-jack doubleton 80% of the time and declarer knows that. If he plays for the drop, he wins 80% of the time that you were dealt queen-jack doubleton."

ReplyDeleteThat means you're assuming he finesses whenever he sees the jack (the opposite of the fool from your next example).

You say declarer can't gain an advantage (unless you go over 93.75%) but that's not right. You mean that he can't go right over 50% of the time. Any percentage above his entitled 33% or so is an advantage. And there's nothing magical about 93.75% (except that it crosses the 50% expectation line). The accrual to declarer when you switch from playing your queen 65% of the time to 80% of the time is exactly the same as switching from 80% to 95%, even though it crosses the 93.75% border.

In many situations declarers play you for the first time, or even if you've come across them before that particular card holding is a first time experience, so there no where declarer can get any feel or table presence about what you actually hold. Moreover if declarer does know you to be on the random side ( in an attempt to reveal nothing ) he's back to guessing yet again. Many might look for tells, but if you are not want for telling.....then he's still in the dark. Yours howard Bigot Johnson. Nice article.

ReplyDeleteLen: He's not entitled to 33%. He's entitled to 67% (2 wins out of 3) by finessing whenever you drop an honor. Playing Q from QJ 80% time doesn't change that. Declarer should still finesse if you drop an honor and he still wins 2 cases out of 3. This is because it is better to pick up 100% of the singleton queens than to pick up 80% of the queen-jacks. If you drop the queen more then 93.75% of the time, however, then declarer CAN get an edge. A specific singleton is 93.75% as likely as a specific doubleton, so that is the critical point. If declarer knows you play the queen more often than 93.75% of the time, it becomes right for him to play for the drop when he sees the queen.

ReplyDeleteYou can come fairly close to this table by two approximations.

ReplyDeleteFirst card: 40-30-20-10

Second card: equally likely among choices which don't involve playing a non-touching lower card.