# Algebra, Bayes’ Theorem, and Vacant Places

*What would life be without arithmetic but a scene of horrors?*

–Reverend Sidney Smith (1771-1845)

This blog is for bridge players who love algebra. There are a few of us. Recently I have gained some sympathy for the Reverend’s unchristian view while recovering from broken leg. Incapacitated at home I have been kept away from the topsy-turvy turmoil of the bridge table and have found quiet comfort in returning to some theoretical work on the effect of card play on probability of the location of a missing queen. As demonstrated in a previous blog the transition of the criteria from the initial probability to the current probability is governed by Bayes’ Theorem. This entails the number of plausible plays that were available for each candidate split. When the numbers of options are equal across the board, vacant place ratios give exact probabilities.

First we shall give a general treatment to the problem of finding the queen when there are 4 cards missing in the suit, namely Quxy where low cards are denoted by u,w, and y. For sake of clarity suppose the suit to be played is hearts. Further suppose that it is known that another suit is split with N card on the left and M cards on the right. If N and M are zero, then nothing is known (or assumed) about the location of vacant places outside the heart suit (the a priori condition).

Initial vacant places | 13 – N | 13 – M |

After one round (u followed by y) | 12 – N | 12 – M |

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 |

Because the numbers of plausible plays are the same across the board, we suspect that the ratio of probability of the queen on the left (QL) to the queen on the right (QR) equal the ratio of the current vacant places, (12-N) divided by (12-M). To prove this we need consider the relative numbers of combinations available in the outside suits (clubs and diamonds).

Hearts Split | 3 – 1 | 2 – 2 | 1-3 |

Minors Split | 10 – N 12 – M | 11 – N 12 – M | 12 – N 10 – M |

Weights | (11-N)x(12-N) | (12-N)x(12-M) | (11-M)x(12-M) |

Summing the weights for Q appearing on the left and Q appearing on the right, one finds:

_____QL = (12-N)x(23-N-M), QR = (12-M)x(23-N-M), so that the ratio of QL to QR is (12-N) / (12 – M), the ratio of the current vacant places.

**Vacant Places and Combinational Weights**

One sees that the relative strengths of combinational weights are expressible in terms of products of current vacant places.

Initial vacant places | 13 – N | 13 – M |

One card removed from each side | 12 – N | 12 – M |

Two cards removed | 11 – N | 11 – M |

We may form a matrix of inclusion (designated by a √ mark) of vacant places in the weighting factors defined by their products.

Split |
3 – 1 |
2 – 2 |
1- 3 |

√13 – N √13 – M | √13 – N √13 – M | √13 – N √13 – M | |

√12 – N 12 – M | √12 – N √12 – M | 12 – N √12 – M | |

√11 – N 11 – M | 11 – N 11 – M | 11 – N √11 – M |

The pattern is apparent and reveals the connection between the splits and the vacant place inclusion. The elements that are common to all splits (the top line) do not appear in the relative weighting. This makes the calculation of weights easy.

**A Case of 5 Cards Missing**

We repeat the exercise for the case where the missing hearts are QJuxy, where J denotes the jack which will not be played voluntarily. On the first round of hearts card u appears on the left, card y appears on the right, and on the second round card x on the left.

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

Initial Cards | QJux –y | Qux – Jy | Jux – Qy | ux – QJy |

Remainder | QJ – 0 | Q – J | J – Q | 0 – QJ |

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

Split |
4 – 1 |
3 – 2 |
2- 3 |

√13 – N √13 – M | √13 – N √13 – M | √13 – N √13 – M | |

√12 – N 12 – M | √12 – N √12 – M | √12 – N √12 – M | |

√11 – N 11 – M | √11 – N 11 – M | 11 – N √11 – M | |

√10 – N 10 – M | 10 – N 10 – M | 10 – N 10 – M | |

Weights |
(11-N) x (10–N) |
(11-N) x (12-M) |
(11-M) x (12 – M) |

Summing the eights for Q appearing on the left and Q appearing on the right, one finds:

_____ QL = (11-N) x (22 – N – M) and QR = (12 – M) x (22 – N – M).

The ratio of QL to QR equals (11 – N) / (12 – M), the ratio of current vacant places after 2 cards have been played on the left and 1 card on the right.

Hopefully this little exercise gives the reader some insight into how probability and vacant places are tied together by Bayes’ Theorem. Central to the treatment is the assumption that the plausible plays are equally likely to have occurred for all splits.

Thanks very much for writing this. I am downloading it and will study it in detail when I have a little more time.