, Solution, problem 21:2004

## Problem no. 21, 2004

### The problem

In his first book on the Law of Total Tricks, To Bid or Not to Bid, Larry Cohen wrote that you should consider doubling the opponents for penalty when you hold four trumps. This is what he wrote on the subject (page 204):

"It is that same 4-1 split which takes a trick away from them that adds a trick if our side were to play the hand. The 4-1 split doesn't reduce the Total Trick count, of course, but it frequently swings the balance one trick in favor of the defending side. If they play the hand, and trumps break 3-2 for them (say we have xxx opposite xx), then our side would have two losers in that suit if we played the hand. However, with xxxx opposite a singleton our side would have only one loser, thus an extra trick. Since the total number of tricks is fixed, one more trick for us on offense means that the opponents will take one less trick if they play the hand with a 4-1 break."

Sounds convincing, doesn't it!

Well... No. As a matter of fact, there are lots of errors in the reasoning above. Your mission, if you choose to accept it, is to find those errors and report them back to me – if possible accompanied with illustrious deals. The more errors you have found, the better chance you have to win one of our books.

### Solution

The first error, which also is the biggest one, is to assume that the total number of tricks is related to the total number of trumps in some mystic way (as the Law of Total Tricks claims). A change in the distribution which gives one side more tricks if they declare the hand may result in their taking more tricks on defense as well, but it is equally likely that it won't. Take a simple example as this one:

1)

 Q J 8 9 6 8 7 4 3 J 10 3 2 A K 10 9 6 3 3 2 10 2 8 7 5

Here, it doesn't matter how East-West's cards are distributed. South always has six spade tricks and nothing more. South couldn't care less if the spades are 5-0, 4-1 or 3-2.

But for East-West the division of their spades is very important. They have all high cards in the remaining suits, so their result will be 11, 12 or 13 tricks, depending on how many spade losers they have.

The next error is to forget that all players have thirteen cards. If we gain one trick because of the opponents' cards being distributed 4-1 instead of 3-2, it might result in our losing one trick in another suit, as here:

2)

 Q J 8 9 6 5 8 7 4 3 kn 10 3 A K 10 9 6 3 2 10 6 A K Q 5

South can take nine tricks in a spade contract; he loses only two tricks in hearts and two tricks in diamonds. If we change the layout by swapping one of North's diamonds with one of South's hearts, the result will be that hearts now are 4-1. South then loses only one heart trick. But it doesn't mean he will take one more trick – since he now loses three diamond tricks instead the previous two. The swap only meant one loser was moved from hearts to diamonds. And the swap has no bearing for East-West either. How many tricks they lose is only dependant on how their black suits are divided.

The difference between 3-2 and 4-1 may also be negative, so that it is better with 3-2 than 4-1, as here:

3)

 J 10 3 9 8 7 5 4 K Q 4 3 2 A 7 4 5 2 K Q 10 4 2 A J 8 A 9 J 10 6 3 2 A 8 6 J 10 7 K Q 9 8 6 7 6 5 3 K Q 9 5

Hearts are 4-1, which looks good to North-South. Wrong! If West leads a trump, South will end up losing four heart tricks and the other three aces. But if South were allowed to swap one of his hearts for one of North's clubs, so that North-South have 3-2 in hearts instead of 4-1, and the defense is the same, they will then lose only three heart tricks plus the other three aces. It may, indeed, be better with 3-2 than 4-1 in the opponents' suit.

For East-West, the heart division also has a role to play: if it is 4-1, East-West can take ten tricks, but if it is 3-2 they can take eleven. Not as it was thought! When hearts are 4-1, both sides lose a trick compared with hearts 3-2. In the first case, the total number of tricks is sixteen, in the second it is eighteen.

When Jean-René Vernes formulated the Law of Total Tricks in his famous article in The Bridge World 1969, one of the things he wrote was this: "How the cards of one suit are divided between two partners ... has a very small, but not completely neglible, effect." But, as we have seen, that claim has no base in facts.

In these three examples, I have shown that a change in the distribution, so that one suit is 4-1 instead of 3-2, may result in one side gaining, but only in their own contract; that neither side gains anything; or that both sides do. The fourth case, that the 4-1 split means one side gains a trick both on offense and on defense (as Cohen says it should be), is also possible, of course, but hardly the only possible outcome.

Finally, there is a fifth case, which I suggest you think about yourself. I don't have a solution to it. Suppose that we have eight spades and they have eight hearts. Let us also assume that Cohen was right (which I just have shown that he wasn't), so that a 4-1 split in the opponents' suit means that we gain a trick no matter which side declares the hand. Now, the problem is what happens if both suits are 4-1. Then, both sides will think "Good, we are 4-1 in their suit. That means one trick is transferred from them to us in both offense and defense." The problem I can't solve is: Which side is right?

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