# Probability and Defenders’ Card Play

**Probability and Defenders’ Card Play**

Probabilities play a central role in decision making during the play of a bridge hand. Declarers benefit in the long run from choosing the path that is most likely to lead to a good result. There are two kinds of probability to consider: the probability of the random deal, and the probability of the play. The first kind involves the a priori odds, and the second involve everything that happens once the first call is made. Players are familiar with the a priori odds, but are less clear on how those odds are altered by subsequent actions. The use of vacant places to decide the most likely location of a missing queen is well understood, but the transition of the criteria from the initial conditions to the current conditions is not. This process is governed by Bayes’ Theorem. As a demonstration of the application we shall work through a simple problem in declarer play. From this the reader will hopefully add to his theoretical understanding of a process that has largely remained hidden to the average player.

**Use of Weights** Probabilities of the deal are related to the number of card combinations available under the various conditions. Suppose the conditions are designated A, B, C, and the associated combinations are NA, NB, and NC.

_____NA + NB + NC = M, the total number of card combinations currently available.

The critical characteristics are the proportions between NA, NB, and NC .

_____(NA/M) + (NB/M) + (NC/M) = 1

_____PA + PB + PC = 1

We associate the probabilities PA., PB, and PC with the corresponding ratios.

In this way we needn’t calculate the numbers of combinations absolutely, but only in relative terms among the all-encompassing conditions. In general when dealing with probabilities we are dealing with relative strengths and proportions, which simplifies calculations. When it comes to expressing the results in probability terms, one must keep in mind the quantities must add to 1 and must cover all the possibilities without overlap between conditions.

**A Simple Problem**

A common situation is the finesse for the trump queen, missing Qxxxx, or more exactly, Quwxy, where u,w,x, and y represent cards of equal rank that may be chosen in any order. It is important to make a distinction between the insignificant cards, as when the cards are played, the players observe particular cards, not just ‘low cards’.

Let’s consider the case where against a contract of 4 hearts spades are led and are known to split 3-5. Declarer ruffs the third round before a minor suit is played and proceeds to draw trumps, planning to finesse against the LHO who holds only 3 spades. Declarer holds ♥ KT86 and dummy has ♥ AJ95, so he plays the ♥ K and leads towards the ♥ AJ9. Here is the vacant place situation the defenders following with low cards throughout, in particular, card u followed by card y followed by card w.

Initial vacant places |
10 |
8 |

One card removed from each side |
9 |
7 |

Two cards removed |
8 |
?? |

We still expect the ♥ Q to be finessible, but how have the odds changed? Here are the situations remaining under the circumstances of the trump play so far.

Hearts Split | 4 – 1 | 3 – 2 | 3 – 2 | 2 – 3 |

Initial Cards | Quwx –y | Quw – xy | uwx – Qy | uw – Qxy |

Remainder | Qx – 0 | Q – x | x – Q | 0 – Qx |

Probability of Choice | 1 out of 6 | 1 out of 4 | 1 out of 6 | 1 out of 4 |

Weighting Factor |
2 |
3 |
2 |
3 |

The probability of choice is where Bayes’ Theorem comes into play. With a 4-1 split, there are 6 ways for defenders to choose their low cards without giving up the ♥ Q. If the cards are chosen at random, each sequence is equally likely, so the probability of their having chosen specifically u-y-w is 1/6. With a 3-2 split with the queen onside, there are 4 choices, so the probability of u-y-w having been chosen is 1 in 4. This is more likely than in the case of 4-1, so that weighs in favour of the 4-1 split in a ratio of 3 against 2. This can be expressed as a weighting factor to be applied to each initial split.

This is the essence of Bayes’ Theorem: the probability of a given combination having been dealt is affected in proportion to the probability that the observed sequence would arise from that combination. The greater the probability of the emergence of the observed sequence, the greater the probability of its source.

The weighting that accounts for differences outside the suit being played depends on the number of combinations available in the untouched suits, diamonds and clubs, of which nothing has been disclosed. The probabilities with regard to these suits are the probabilities of the deal, so directly related to the number of combinations available.

Spades Split |
3 – 5 |
3 – 5 |
3 – 5 |
3 – 5 |

Hearts Split |
4 – 1 |
3 – 2 |
3 – 2 |
2 – 3 |

Minors |
6 – 7 |
7 – 6 |
7 – 6 |
8 – 5 |

Combinational Weights |
4 |
4 |
4 |
3 |

Now incorporating the weighting due to the play in the heart suit itself, we find

Remaining Combo |
Qx – 0 |
Q – x |
x – Q |
0 – Qx |

Total Weights |
8 |
12 |
8 |
9 |

Percentage of Total |
22% |
32% |
22% |
24% |

The weights total 37. The Q is on the left for 20 out of 37 of total weights. So, the probability of the ♥ Q on the left is now 54%. Initially, before a heart was played, it was according to the vacant place ratio, 56% (10 out of 18).

One may use the ratio of the current vacant places to calculate probabilities exactly if the card play weighting factors are equal, that is, the observed sequence of play is equally probable to have arisen from the various remaining candidates. In the above situation equality would be achieved if declarer didn’t finesse on the second round, but rather went up with the ♥ A in dummy. When the RHO follows on the second round with card x, the remaining conditions are Q – 0 and 0 – Q, with corresponding weights 4 and 3, respectively. These weights represent the ratio of the current vacant places, 8 and 6.

When the weighting factors are not equal, the probability of Q on the left will not generally be equal to a ratio of real (integer) vacant places, however, one may define a fractional vacant place to achieve exact correspondence. We call this the virtual vacant place (VVP). In the above example, we would define VVP according to the requirement that 8 / (VVP + 8) equals 20/37. In this case VVP is 6.8, which lies between 7 and 6.

**A Second Demonstration**

The use of current vacant places in this manner when the observed sequence is equally likely to have arisen from each remaining split remains valid when 4 hearts are missing.

Initial vacant places |
10 |
8 |

After one round (u followed by y) |
9 |
7 |

Here are the situations remaining under the circumstances of the play so far.

Hearts Split | 3 – 1 | 2 – 2 | 2 – 2 | 1 – 3 |

Initial Cards | Qux –y | Qu – xy | ux – Qy | u – Qxy |

Remainder | Qx – 0 | Q – x | x – Q | 0 – Qx |

Probability of Choice | 1 out of 2 | 1 out of 2 | 1 out of 2 | 1 out of 2 |

Weighting Factor |
96 |
84 |
84 |
56 |

The probability of the ♥ Q being on the left is proportional to 96 + 84 = 180.

The probability of the ♥ Q being on the right is proportional to 84 + 56 = 140.

The ratio of these numbers is 9:7, exactly the ratio of the current vacant places after 1 round of hearts has been played. Magic. Note that there is a difference in the number of plays required before the equality rule takes effect depending on whether the number of missing cards in hearts is odd (5) or even (4), as reflected also in the familiar rule of thumb regarding ‘Eight-Ever, Nine-Never.’