Study 7Z75 Main |
| by Richard Pavlicek |

One of the perplexing areas in bridge theory is how probability percentages change as the play progresses. Percentages calculated at the beginning of a deal, called *a priori* percentages, are not set in stone. As you garner information from trick to trick, percentages change — sometimes drastically. Exactly *how much* they change is often difficult to determine, not only because of subjective factors but because of misconceptions in probability theory.

This study is from declarer’s perspective. To simplify matters, I will assume there are no inferences from the bidding. In real life this is rarely true because even a *lack* of enemy bidding conveys some information, such as ruling out wild distributions. Nonetheless, I want to focus on how percentages change with card play, so bidding deductions will be ignored.

For example, suppose you have a straight finesse through East for a king. Before any cards are played this is obviously 50 percent, but as play continues it may become greater or smaller according to what you discover about the distribution. If you subsequently learn that East has five cards in the suit and West only three, the finesse is now 5:3 odds in your favor, or 62.5 percent.

During the play, information you obtain about the enemy distribution is of two basic kinds: (1) *extracted* information, which is extremely valuable and helpful, and (2) *volunteered* information, which is suspect and often useless.

Extracted information typically occurs when an opponent shows out of a suit, revealing the exact layout. The information about the *suit led* (not the suit discarded) then becomes a known fact and can be used in full to determine subsequent odds and percentages. Defenders may try to deceive you, but they cannot hide the inability to follow suit.

Volunteered information is that which an opponent *chooses* to tell you, such as the cards played in following suit, and the choice of leads and discards. This information is tainted. Any meaning you might attach to it could be the deliberate plot of a defender trying to fool you. An experienced declarer learns to be skeptical of anything that comes too easily. It’s like listening to a sales pitch — you may get some facts now and then, but there’s a lot of bull in between.

Volunteered information in the form of the *opening lead*, while still suspect to some extent, is more reliable than a lead chosen after dummy is in view, or a discard. In general, declarer does well to take the opening lead at face value. At least it may help you from becoming paranoid if you believe *something*.

To illustrate how the percentages change, consider the play of the following 6 contract. (Thanks to David Burn of London, England, who posted this problem on the rec.games.bridge newsgroup.)

6 South E-W Vul | Q J 6 2 A 10 9 8 7 J A 8 2 | West Pass Pass Pass | North2 4 NT 6 | East Pass Pass All Pass | SOUTH1 4 ^{1}5 | ||

Lead: 10 | East plays K | ||||||

A K 9 8 4 3 Q 6 3 A 8 4 5 | 1. splinter bid |

You count nine top tricks, and it will be easy to ruff two diamonds for 11. Therefore, all you need is a second heart trick, and there are 16 possible distributions of that suit:

# | West | East | Percent | # | West | East | Percent |
---|---|---|---|---|---|---|---|

1 | KJxxx | — | 1.9565 | 9 | Jxxx | K | 2.8261 |

2 | KJxx | x | 8.4783 | 10 | Jxx | Kx | 10.1739 |

3 | KJx | xx | 10.1739 | 11 | Jx | Kxx | 10.1739 |

4 | KJ | xxx | 3.3913 | 12 | J | Kxxx | 2.8261 |

5 | Kxxx | J | 2.8261 | 13 | xxx | KJ | 3.3913 |

6 | Kxx | Jx | 10.1739 | 14 | xx | KJx | 10.1739 |

7 | Kx | Jxx | 10.1739 | 15 | x | KJxx | 8.4783 |

8 | K | Jxxx | 2.8261 | 16 | — | KJxxx | 1.9565 |

There are six ways (A-F) to play the heart suit, and the following table shows the percent chance of each:

Option | Winning Cases | Percent |
---|---|---|

A. Win the ace then let the 10 ride | 4-10 12-16 | 69.2174 |

B. Win the ace then lead to the queen | 4-6 8-16 | 69.2174 |

C. Run the 10; if loses to jack, run the queen | 1-9 13-16 | 76.8261 |

D. Lead to the queen; if loses, finesse the 10 | 1-5 9-16 | 76.8261 |

E. Finesse the 10; if loses to jack, run the queen | 1-12 16 | 77.9565 |

F. Run the queen; if loses, finesse the 10 | 1-12 | 76.0000 |

The total percent of each option is found by adding the percentages of each case in which the option succeeds. Option E (finesse the 10, then run the queen) is evidently best. Note that its edge over the near-equivalent Option F occurs in Case 16 when West is void — hence, you will know to change horses, whereas leading the queen allows no recovery.

It should be apparent why Options A and B are equal. Basically, it’s just a guess whether to play East for the king or jack, so the winning and losing cases balance out. Similarly, for Options C and D. In practice, however, it is usually better to choose A over B, and C over D, since there is *some chance* that if East has the king he will hop with it. Admittedly, the chance of such misdefense is close to zero against an expert, but it grows considerably against weaker defenders.

The previous *a priori* percentages are interesting and offer good general advice, but in actual play they are almost never accurate. When was the last time you needed to play a card combination *before* the opening lead? It’s like aiming a bow and arrow at a distant moving target; by the time the arrow gets there, the target will be gone. In this case you will always have more information before you play hearts, so the original percentages will not apply.

Even if you extract no information about clubs or diamonds, you will certainly know the exact trump division; and even if they break 2-1, one defender will have more space for hearts. In that event the changes will be slight, so Option E still rates to be best.*

*Actually, if *East* has the shorter spades, Options C, D and E become exactly even in theory. Given the equal choice, the edge would go to Option C because of the slight extra chance that East might misdefend and hop with the king.

Now consider a more extreme scenario, which is what actually occurred in the problem deal. While drawing trumps and ruffing diamonds, East *shows out* on the third diamond, pitching a club. It also happens that West shows out on the second trump, pitching a diamond. You now have extracted information in two suits. West is known to have begun with one spade and seven diamonds, and East with two spades and two diamonds. Hence, you know eight of West’s cards, leaving five unknown spaces; and four of East’s cards, leaving nine unknown spaces.

What about East’s *club* discard on the third diamond? Doesn’t this account for a known card in East? No! This information was *volunteered* and has no bearing on the 5:9 proportion of unknown cards between West and East. Essentially, the only information given by the club discard is that East is not void in clubs, but you knew that anyway. He would never pitch a heart in view of dummy, so the club pitch means nothing.

Given the 5:9 ratio of unknown cards, let’s recalculate the percentages:

# | West | East | Percent | # | West | East | Percent |
---|---|---|---|---|---|---|---|

1 | KJxxx | — | 0.0500 | 9 | Jxxx | K | 0.4496 |

2 | KJxx | x | 1.3487 | 10 | Jxx | Kx | 5.3946 |

3 | KJx | xx | 5.3946 | 11 | Jx | Kxx | 12.5874 |

4 | KJ | xxx | 4.1958 | 12 | J | Kxxx | 6.2937 |

5 | Kxxx | J | 0.4496 | 13 | xxx | KJ | 1.7982 |

6 | Kxx | Jx | 5.3946 | 14 | xx | KJx | 12.5874 |

7 | Kx | Jxx | 12.5874 | 15 | x | KJxx | 18.8811 |

8 | K | Jxxx | 6.2937 | 16 | — | KJxxx | 6.2937 |

Option | Winning Cases | Percent |
---|---|---|

A. Win the ace then let the 10 ride | 4-10 12-16 | 80.6194 |

B. Win the ace then lead to the queen | 4-6 8-16 | 80.6194 |

C. Run the 10; if loses to jack, run the queen | 1-9 13-16 | 75.7243 |

D. Lead to the queen; if loses, finesse the 10 | 1-5 9-16 | 75.7243 |

E. Finesse the 10; if loses to jack, run the queen | 1-12 16 | 66.7333 |

F. Run the queen; if loses, finesse the 10 | 1-12 | 60.4396 |

Wow! A complete turnaround from worst to first. Options A and B now tie for best. This is obvious if you think about it: East is likely to have longer hearts; so finessing through East is more likely to work. As suggested before, the edge would go to Option A because of the slim extra chance that East may hop with the king in Case 11.

Since you took the effort to ruff two diamonds, why not ruff two clubs as well? This might reveal even more information. Indeed, if West showed out of clubs, this would give a complete count of the distribution.

Suppose both opponents follow suit as you ruff two clubs. The obvious deduction is that *three* more cards are known for each player, so the unknown space ratio is now 2:6 instead of 5:9. Indeed, West cannot have more than two hearts, so half of the original distributions are now impossible. Based on the 2:6 ratio, the percentages are recalculated below. (Case numbers are the same as before.)

# | West | East | Percent | # | West | East | Percent |
---|---|---|---|---|---|---|---|

4 | KJ | xxx | 3.5714 | 12 | J | Kxxx | 10.7143 |

7 | Kx | Jxx | 10.7143 | 14 | xx | KJx | 10.7143 |

8 | K | Jxxx | 10.7143 | 15 | x | KJxx | 32.1429 |

11 | Jx | Kxx | 10.7143 | 16 | — | KJxxx | 10.7143 |

Option | Winning Cases | Percent |
---|---|---|

A. Win the ace then let the 10 ride | 4 7-8 12 14-16 | 89.2857 |

B. Win the ace then lead to the queen | 4 8 11-12 14-16 | 89.2857 |

C. Run the 10; if loses to jack, run the queen | 4 7-8 14-16 | 78.5714 |

D. Lead to the queen; if loses, finesse the 10 | 4 11-12 14-16 | 78.5714 |

E. Finesse the 10; if loses to jack, run the queen | 4 7-8 11-12 16 | 57.1429 |

F. Run the queen; if loses, finesse the 10 | 4 7-8 11-12 | 46.4286 |

Comparing this with the 5:9 table shows the options are in the same order, but the rich got richer and the poor got poorer. Option A is now close to 90 percent. Does this seem a little suspicious to you? If so, read on.

I do not believe the above percentages are true because of the way information was obtained. What actually happened for you to adopt a new percentage table? Not much. All you did was lead clubs a few times and both opponents followed suit, which is more or less expected. A *show-out* would have been meaningful, but it didn’t happen. Essentially, this supports the distinction between extracted and volunteered information.

Are the percentages from the 5:9 table still true then? No! They could hardly be right because half of the cases are now impossible. To find the true percentages you must eliminate the impossible holdings from the 5:9 table and *prorate* the percentage of each possible holding to obtain a 100-percent total. I seem to be chart-crazy, so let’s look at one more with these calculations made.

# | West | East | Percent | # | West | East | Percent |
---|---|---|---|---|---|---|---|

4 | KJ | xxx | 5.2632 | 12 | J | Kxxx | 7.8947 |

7 | Kx | Jxx | 15.7895 | 14 | xx | KJx | 15.7895 |

8 | K | Jxxx | 7.8947 | 15 | x | KJxx | 23.6842 |

11 | Jx | Kxx | 15.7895 | 16 | — | KJxxx | 7.8947 |

Option | Winning Cases | Percent |
---|---|---|

A. Win the ace then let the 10 ride | 4 7-8 12 14-16 | 84.2105 |

B. Win the ace then lead to the queen | 4 8 11-12 14-16 | 84.2105 |

C. Run the 10; if loses to jack, run the queen | 4 7-8 14-16 | 76.3158 |

D. Lead to the queen; if loses, finesse the 10 | 4 11-12 14-16 | 76.3158 |

E. Finesse the 10; if loses to jack, run the queen | 4 7-8 11-12 16 | 60.5263 |

F. Run the queen; if loses, finesse the 10 | 4 7-8 11-12 | 52.6316 |

As seen above, the total percent chance of each option now falls about midway between those from the 5:9 table and those from the 2:6 table. Even if you don’t believe the ideas expressed here, you might agree that this compromise feels right.

The point to stress is that, in the absence of extracted information, percentages determined at a previous trick remain true as long as all the previous possibilities exist. Once a specific case becomes impossible, it is knocked off the chart, and the percentages of the remaining cases increase proportionally.

As further evidence that an opponent following suit is volunteered information, suppose North (dummy) has A-Q and South has 3-2 with two tricks remaining and the K outstanding. Further, you know that each opponent started with four hearts and has two hearts left. Clearly your chance of the finesse winning is 50 percent.

Now suppose you lead the 2 and West follows low. Does this affect your chances? One might surmise that after West plays, he has only one unknown card, while East has two unknown cards. Therefore, the missing king is more likely to be with East by 2-to-1 odds. Of course this is nonsense, because West’s play was *volunteered*. The low heart was his choice, not yours. West would routinely play low from K-x, so the information is meaningless, and your chances are still 50 percent. In other words, following suit is *volunteered* information, so the previous percentages do not change.

[Addendum October 2012: Several readers pointed out a flaw in my analogy. If West has K-x, he will realize his king is dead, because declarer will always finesse the queen. Therefore, his only chance to win a trick is to *play the king*, hoping declarer doesn’t notice or calls for the queen too quickly. Yes, we’ve all seen that happen, and their argument is valid, which shows that almost nothing in bridge is black and white. Nonetheless, the general principle of not drawing information from volunteered plays is sound. On a humorous note, if declarer assumes the king is offside from West’s play, he should *play the ace*. Perhaps East, expecting the queen to be finessed, will act too quickly and *play the king*. But then, he might not even have the king if West reversed his strategy — as spy vs. spy continues.]

Study 7Z75 Main | Top How Percentages Change |

© 2003 Richard Pavlicek