Well there I was lying in the French sunshine listening to the sound of the breaking Atlantic rollers surrounded by the beautiful people when inevitably, as it does at times such as this, my mind turned to ….bridge.

What, I thought, is the probability that a 5-card suit headed by AKQ opposite a small doubleton in hand would break 3-3 to supply 5 tricks. Well actually it’s pretty simple to discover that the suit will break 3-3 about 36% of the time, about 48% of the time it will break 4-2 and the rest of the time it will break 5-1 or 6-0. If both defenders follow to the first two rounds of the suit with little cards then we can eliminate the 5-1 and 6-0 breaks so we are left with 3-3 and 4-2 in the ratio of 35:48 or about 11 to 15. The new % figures are 42% for the 3-3 break and 58% for the 4-2 break. So the chance of the 3-3 break has improved but still offers a much lower chance than a straight finesse in another suit which offers a clear 50% chance of success.

S 83

H Q5432

D 752

C AQ3

S A2

H AK

D AK83

C JT852

Contract 3NT, small spade led. There are 7 top tricks and there are chances for extra tricks in hearts and clubs. You take the Ace of Spades (no point holding up) and both defenders follow to two rounds of hearts. Now we are at the crossroads – do we cross to the Ace of clubs and play the heart Queen hoping for a 3-3 break or do we chance the club finesse? If we get it wrong the defenders will take at least four spade tricks sharpish. Well on the basis of the above theory the club finesse gives a 50% chance and the heart break is now about 42% so the finesse is a much better shot.

Now let’s strengthen the North hand a little by replacing a small heart with the ten of that suit. They still lead a spade and they both still follow to the first two rounds of hearts. Is it again a 42% chance that the suit will break 3-3? Not so you may be surprised to read. In the first example all the defenders’ cards were insignificant in that it didn’t matter in what order they were played, they could false card at will. But now the defenders hold a very significant card, namely the heart Jack and neither would willingly play this card. As neither defender did play this card then as well as eliminating the 5-1 and 6-0 distributions we can now also eliminate those 4-2 distributions which include a doubleton including the heart Jack. If you list all the possible doubletons (15) you will see that one third of them contain the Jack and so we must now eliminate a third of the original 48% probability of a 4-2 break, which is now reduced to 32%. The probability of the 3-3 break is unchanged at about 35% so the ratio is now 35:32 or just over 52% in favour of the 3-3 break. So in our example hand the heart drop is now a better bet than the club finesse.

Here is another example:

AT982

K3

Needing tricks in this suit you start with the King and then lead the three up to the Ace. Incidentally that’s much better than finessing the ten because an honour doubleton in the East hand is more likely than a small doubleton. You can easily work this out for yourself by writing down all the doubletons that East can hold and counting how many contain either the Jack or the Queen. Both defenders follow with small cards- what is the chance of the Queen and Jack now dropping together in a 3-3 break? As before we can eliminate the 5-1 and 6-0 distributions. We can now also eliminate the 4-2 breaks that contain a doubleton honour (18/30) and the 3-3 breaks where one defender holds both honour cards (8/20). Having eliminated 18/30 of the 4-2 breaks we are left with 12/30 so the % is now the original 48% x 12/30 = 19%. Likewise if we eliminate 8/20 of the 3-3 breaks we are left with 12/20 and the 3-3 % is now 35% x12/20 = 21%. So the ratio of 4-2 to 3-3 breaks is 19:21 which is again just over 52% in favour of the hearts dropping.

There is another way of looking at this. Taking this suit in isolation, after both opponents have played to the King and Ace the location of all the small cards is known, the only cards missing are the Queen and the Jack. Suppose East now plays the Queen on your lead of the ten. East has now played three spades and has ten vacant places in his hand. At this point West has played only two spades and has eleven vacant spaces. The odds are therefore 11:10 or 52.4% that West holds the remaining spade honour. The same result is reached as in our previous calculation.

The reciprocal of all this is that if a defender does show up with an honour on the second round then the 4-2 distribution remains a big favourite.

Nobody would be expected to work all this out in real time at the table but it is easy to understand the principle that when both defenders play insignificant cards on the first two rounds the suit is now more likely to break 3-3 than 4-2. It’s marginal I know but if you play with the odds all of the time you will win most of the time.