# The Bridge Goddess Thumbs Her Nose

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*by Matthew Kidd*

This is board 10 from the Friday morning open pairs at the Palm Springs regional. Bill Grant and I swiftly arrived in 6♥, the choice of 71% of the pairs (46 of 65) across five sections. Three pairs overbid to 7♥ and eight tried for the matchpoint top in 6N.

A club lead gave me two pitches but that only allowed a spade and a near certain fifth round diamond winner to be pitched without solving the problem of losing the third round of diamonds.

To make this slam either trump or diamonds must be brought in without losing a trick. For trump to work the suit must split 3-2 with the queen onside or 1=4 with the singleton queen or ten offside, where in the latter case, restricted choice will lead you to play for the singleton ten. Using the *a priori* odds of a 3-2 (67.8%) and 4=1 (14.1%) split respectively, trump will come in ½ × 67.8% + (2/5) * 14.1% = 39.6%. Bringing in diamonds without loss requires a lucky Qx in either hand or for LHO to have a singleton queen for a total chance of (2/5) * 67.8% + (1/5) * 14.1% = 29.9%. Don’t be thrown off by the ♦9 and ♦8 in dummy. The percentage play is still to bang down the ace and continue with the king unless the queen drops in which case you come to your hand with the jack and finesse.

The chance of both suits failing is (100% - 39.5%) × (100% - 29.9%) = 42.3% which means the slam has a 57.67% chance, making it the percentage bid. A slightly more accurate calculation (details below) yields almost exactly the same answer. All these calculations ignore the very rare chance of a ruff on the opening lead.

When the hand was over and I found LHO had started with ♥Qxx, I shrugged and said it was a good slam but not today’s slam. No one else made it either.

And yet double dummy there is a perverse line that happens to work on this hand. Declarer can strip dummy and his hand of the black suits, taking any finesses not handed to him by the opening lead, strip West of diamonds by playing off the top two, and cash three rounds of trump. In with the ♥Q, West must give declarer a ruff-and-slough, whereupon the third round diamond loser vanishes.

Funny, nobody found this line.

Three of the eight daring souls in 6N, not having to worry about trump, setup diamonds to make their contract. Locals John and Bette Strauch were in that elite set.

The probability calculations above are not entirely accurate because the suit split probabilities for the red suits are slightly correlated. If this worries you, we can do better by calculating the joint probabilities of relevant red suit distribution pairs.

Let **(n,k)** denote the number of different ways to choose **k** objects (cards) from **n** objects where the order of the objects is immaterial. Recall that:

When hearts are say 3=2, symmetry is broken and this means that the probability of 3=2 and 2=3 diamond splits are no longer equal. Here and above we follow the *Bridge World* convention of using an equals sign to indicate a specific distribution, i.e. 2=3 means LHO holds two and RHO holds three, while a dash, i.e. 3-2, indicates 3-2 either way.

The opponents hold five hearts, five diamonds, and 16 black cards between them. As example, the total number of hands where LHO has two cards in each red suit is given by (5,2) × (5,2) × (16,9). The total number of hands LHO can hold is (26,13). So the probability of West holding two cards in each red suit is:

A series of similar calculations, easily done with a small program, yields this table:

♥ 0=5 | ♥ 1=4 | ♥ 2=3 | ♥ 3=2 | ♥ 4=1 | ♥ 5=0 | |

♦ 0=5 | 0.01 | 0.09 | 0.42 | 0.77 | 0.55 | 0.12 |

♦ 1=4 | 0.09 | 1.05 | 3.85 | 5.50 | 3.09 | 0.55 |

♦ 2=3 | 0.42 | 3.85 | 11.00 | 12.37 | 5.50 | 0.77 |

♦ 3=2 | 0.77 | 5.50 | 12.37 | 11.00 | 3.85 | 0.42 |

♦ 4=1 | 0.55 | 3.09 | 5.50 | 3.85 | 1.05 | 0.09 |

♦ 5=0 | 0.12 | 0.55 | 0.77 | 0.42 | 0.09 | 0.01 |

Adding up all the cells gives 100% since all cases are covered. And it is symmetric along the diagonal because we have two five cards suits against us and no other knowledge to break the symmetry.

For each cell in the table we need to calculate the probability that we lose a trick in both diamonds and hearts. For example, consider the case ♦ 2=3 and ♥ 1=4. Diamonds work 2/5 of the time when the queen drops and hearts work only when the queen or ten drops singleton, also 2/5 of the time. So both suits fail 3/5 × 3/5 = 9/25 = 36% of the time. Here is the full failure table where the values are in percent.

♥ 0=5 | ♥ 1=4 | ♥ 2=3 | ♥ 3=2 | ♥ 4=1 | ♥ 5=0 | |

♦ 0=5 | 40 | 60 | 40 | 60 | 100 | 100 |

♦ 1=4 | 40 | 48 | 32 | 48 | 80 | 100 |

♦ 2=3 | 40 | 36 | 24 | 36 | 60 | 100 |

♦ 3=2 | 40 | 36 | 24 | 36 | 60 | 100 |

♦ 4=1 | 40 | 60 | 40 | 60 | 100 | 100 |

♦ 5=0 | 40 | 60 | 40 | 60 | 100 | 100 |

Note that if hearts break 5=0, the final column, we are lost regardless of what happens in diamonds because we will lose two trump tricks. To get the total probability of failure, multiply the value in each cell of the first table by the corresponding cell of the second table and sum up the result to get 42.22%. So the probability of success is 57.78%.

This result is very close to the 57.67% answer we got using the *a priori* probabilities initially. This shows that the correlation between the red suit distributions is insignificant. Typically the correlations are not very significant both because the changes in the split probabilities are small and because the impact of these changes partly cancels in computing the total probability.

Note the difference between 57.67% and 57.78% is real. It is not due to round off error. Though many numbers above are only shown to three decimal places, the actual calculations were done using fully accurate *a priori* suit split probabilities.