The Theory of Total Tricks: Part II - Counting Losers
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Introduction

“(T)wo factors … will not change the Total Trick count:

1) Location of high cards.  Finesses that are onside for one pair will be offside for the other.  The Total Trick count is constant.

2) Distribution of the suits.  Bad breaks for one side translate into good breaks for the other.  The Total Trick count is constant.” (Cohen, To Bid or Not to Bid, 1992)

Bidding is usually a matter of odds, of expected values.  You bid a game because it will often make, not because it will always make.  When trumps go 4-1 or 5-0, as they will fairly often, you go set, but shrug it off.  The game was still worth bidding.

Larry Cohen argued, above, that estimating total tricks rises above the luck of the deal.  It is a scientific concept, a Law, not an expected value estimate.

Pure Hands

It is time to start developing a full theory of total tricks.  My breakthrough, if you can call it that, came by looking at hands that Mel Colchamiro and I called completely pure (Some of these ideas appeared in our Bridge World article, (Bloom & Colchamiro, December, 2011)).  A bridge hand is completely pure if we hold all the honors in our longest two suits, and no honors in our other two suits.

Here is a completely pure hand:

North
KJ32
KJ5
1087
954
South
AQ654
AQ87
932
6

Notice, if our hands are completely pure, then our opponents also hold completely pure hands.  On this hand, we will obviously take nine winners in a spade contract, or ten, if the defense messes up.  How will we fare defending a club contract?  That depends, of course, on the adverse distribution.  We don’t know that, but it is pretty easy to use probability tables to calculate the expected number of winners we could cash.

This is a crucial concept in my approach.

• Total tricks are estimated based on the expected number of winners that can be cashed.

I don’t buy Mr. Cohen’s science of total tricks – I consider all such estimates as averages based on probabilities.  In my example hand, a 3-1 or 4-0 spade split won’t stop us from taking nine tricks, but will certainly impact our defensive chances against a club contract.  I am not aiming at the total number of tricks available on a given hand, rather, I want to calculate the expected number of total tricks available over a large group of similar hands.

To estimate the winners we can cash, let me introduce a table of expected winners, based on the length of our solid suit-fit.

Table 1: Expected Number of Winners Cashing on Defense

 Length of our Solid Suit-Fit Expected Number of Winner to Cash 11 .52 10 .78 9 1.31 8 1.64 7 2.18 6 2.54

Applying this to our example hand, we held nine spades and seven hearts, and so can expect to cash 1.31+2.18 = 3.49 winners[1].  Couple that with our four losers, and we conclude:

• There are 18.51 total tricks available on this hand.

Good!  The next step is to extend this analysis to other completely pure hands.  A warning, though:  Converting losers to winners is not so completely trivial.  Go back to our example hand:

[1] I am cheating a bit when I add these probabilities.  A specific spade split affects the probability of various heart splits, and that should really be considered, but adding these gives a result that is close enough for government work, and keeps the results simple enough for practical uses.  A large simulation found that we could cash, on average, 3.468 winners, so 3.47 might be a bit more accurate.

North
KJ32
KJ5
1087
954
South
AQ654
AQ87
932
6

One and a half percent of the time, hearts will split 6-0.  That split was considered in the 2.18 trick calculation, but such a dire split would mean the defenders can score several ruffs against our spade contract.  Our four losers won’t yield nine winners – we will end up with six or seven winners.  This is rare enough that we can safely ignore the impact, but, try this example:

North
KJ32
KJ53
108
954
South
AQ65
AQ87
9632
6

There are only three losers.  Does that mean ten winners?  Suppose the defense leads a heart.  If hearts are 4-1 and we try to trump clubs in hand, we will run into many ruffs.  Ouch!  Even if hearts and spades are three-two, we give up a club, win the heart in dummy, trump a club, trump to dummy, club ruff, and are stuck in hand, and will often lose a heart ruff.  On a heart lead, we will most likely just draw trumps and take eight or nine winners, depending on the trump split.

Okay, try this variant:

North
KJ3
KJ5
108
95432
South
AQ65
AQ87
9632
6

Again, there are only three losers, but even eight winners will be a real struggle in a spade contract.

Looking at losers, rather than winners, seems correct when trying to understand total tricks, but the number of winners may be less than 13-minus-losers.  This will often be the case if our trump suit is inadequate.  In practice, loser-conversion will tend to be pretty accurate if we have nine or more trumps, but over-estimate winners when we have only an eight or seven card trump fit.  That ninth trump is incredibly important.  When we have fewer than nine trumps, we will have to adjust our total trick estimate downward.  My experience, and some simulations, suggests subtracting half a winner from the count with only eight trumps, and a full winner for a seven card fit.

Mel Colchamiro espoused a notion he called effective trumps.  Going from eight to nine trumps is very important and useful.  Going from a ten card fit to an eleven card fit?  Not so useful.  If we hold a six card spade suit, it is hard to imagine how the play will vary much if dummy has five card support instead of four card support, yet that extra trump adds a full trick to the Law estimate of total tricks.  How much impact will that extra trump really have?  First off, it makes it slightly more likely that partner will be shorter in our side suit, and cover a loser.  That would gain a trick.  Also, from Table 1, we are .26 of a trick better off on defense if we have only a ten card fit.  That extra trump will impact the total trick expectation, but by substantially less than one full trick.

It is time now to count the expected losers for a wide range of completely pure hands.  We will have to do some adjustments to convert these to winners, but the loser count is easy using Table 1.  The results, organized by suit lengths, appear in the table below:

Table 2: Expected Defensive Winners for Completely Pure Hands

 Length of Longest (Solid) Suit Length of Next Longest (Solid) Suit Defensive Winners 7 7 4.36 8 6 4.18 8 7 3.82 9 6 3.85 8 8 3.28 9 7 3.49 10 6 3.32 9 8 2.95 10 7 2.96 11 6 3.06 9 9 2.62 10 8 2.62 11 7 2.70 10 9 2.09 11 8 2.16

The first things that struck me, looking at this table, are the tight clusters of nearly identical numbers, particularly near the bottom of the table.  For example, three rows produce winners of essentially three – the 9,8 row, the 10,7 row, and the 11,6 row.  What in the world do these have in common?  The answer is pretty obvious.  In all three cases, the total number of cards in our long suits is 17.  17 cards in our two suits mean three winners to cash on defense.

If we hold 17 cards in our two long suits, then we have 9 cards in our short suits, and so the opponents also hold 17 cards in their two long suits.  Now they can look at this table as well, and predict cashing three winners on defense.  They get three defensive tricks, as do we.  So there rate to be 20 total tricks.

This is a profoundly important observation:

• For completely pure hands with 17 cards in the two long suits, there will be, on average, 20 total tricks.

Yet, on these hands, the total number of trumps will vary from 18 to 22!  Seeing this, I realized, absolutely, that the Total Trick theory of Mr. Vernes was bogus.  Trumps have nothing to do with total tricks.  Instead, we should focus on the lengths of our side’s two longest suits.

This seems like a good stopping place.  In the next installment, I will look a little more closely at the connection between total tricks and the longest suit lengths, and try to explain why the famous Law is so attractive, and yet so wrong.

Part III of The Theory of Total Tricks