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Based on my (imperfect) understanding of the "law," it means that my partnership should bid up to the same level as the number of trumps. For instance, with four trumps over my partner's five card major, I could raise his bid to three.

But I was taught to use a combination of trumps and points. Thus, we'd want to bid four spades with 26 points and eight trumps, or with 23 points and nine trumps, or with 20 points and ten trumps. So why the "law?'

Apparently, the opponents' (theoretical) trump suit also figures into this equation. How do I count for their "trumps?" And once I've done so, how do I combine their trumps and mine into a "law" calculation? Why would I do this anyway?

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1 Answer 1

up vote 6 down vote accepted

Law of total tricks is an empirical law which states the following:

If opponents play the contract with their best suit (with A cards) as trumps, and they can make x tricks, and you play the contract in your best suit (with B cards) as trumps and you can make y tricks, then A + B = x + y.

This was first proposed by Jean-Verne Rene in a bridge world article and made popular by Larry Cohen in his book: "To Bid or not to Bid: The Law of Total Tricks."

The purpose of the law is to decide whether to play or defend and the level to play, when both sides are bidding, i.e. it is more useful in competitive auctions. Trying to apply it in constructive auctions (where opponents haven't bid) does not make much sense. (Of course, when you are raising partner with a weak hand, as a preempt, you are kind of using the law in advance).

As a simple example: Say you hold xx, Kxxx, QJx, Txxx, playing IMPS, non-vul.

Suppose partner opens 1 Heart (showing 5+ cards), RHO overcalls 1 Spade. You bid 2 Hearts and LHO bids 2S, which is passed around to you.

Now assuming opponents have 8 spades between them, if you assume the law, then there are 8+9 = 17 total tricks.

  • Now suppose 2S is making exactly. Then by the law, it means you can make 3H.
  • Now suppose 2S is making 9 tricks. Then by the law, 3H is at most down 1.
  • Now suppose 2S is down 1. Then by the law, you make 10 tricks in hearts!

In all cases, you are better off bidding 3H (even if doubled, non-vul) as opposed to letting them play in 2S.

Of course the law is based on empirical evidence, and there are many adjustments that need to be made, which you can find in Larry Cohen's book.

On the flip-side, there is a book by Mike Lawrence: I Fought the Law of Total Tricks which says that the law might actually not be as useful as people think.

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My sensse is that the "law" is basically pretty good. Now there are some "extreme" versions of it like the Bergen raises (of his partner, Marty Bergen), raise to 3 with NO HCP. That's where I, (and probably Lawrence), might have a few problems. Everything in moderation, right? –  Tom Au Jun 6 '11 at 13:09
    
@Tom: Well, I haven't read the book, but I am guessing Lawrence has a good statistical analysis to back up his claims and objections are not just some cases, but probably at a more fundamental level. –  Aryabhata Jun 6 '11 at 16:36
    
Here is a case where using the law in advance is helpful! Since you bid 2H to begin with, LHO was able to bid 2S safely; now both opponents know they share an 8-card fit, and if RHO has 6 spades they can profitably bid 3S over your 3H. If you had bid 3H at your first opportunity, you and your partner both know about a 9-card fit, and your RHO doesn't know about the spade fit. If LHO doesn't have 4 spades, they can't afford to bid 3S, and if RHO has 6 spades, they won't know about their partner's 3-card support and bid 3S. –  ruds Jun 13 '13 at 7:17
    
@TomAu: I haven't read Lawrence's book (and I highly recommend Cohen's book). Some judgment is required, but raising to 3 with no HCP can be a reasonable call. Imagine this situation: Not vulnerable vs vulnerable, your partner has dealt and bid 1H. Your hand is void Txxx T9xxx xxx, and RHO has passed. Looking at your hand, it is extremely likely your opponents have a spade fit, and if they don't have game values then your partner has a 16-count or better. 3H is likely to play well in that situation, and 3H may be enough to keep opps out of their vulnerable spade game while not going down much. –  ruds Jun 13 '13 at 7:23
    
@ruds: It depends on whether you judge your hand is preemptive (so you bid 3H over the 1S) or constructive (so bid 2H), and might reach a game/double opponents in 4S etc. The example hand I gave is borderline (because it has some defense), and would not argue with a 3H bid. In any case, it was just an example to demonstrate the usage of the law :-). In your example to Tom, you cannot bid 3H with that hand (which has 12 cards, btw), as that would show a limit raise in "standard", and you would need special agreements to have 1H - 3H be a preempt. I am guessing you play Bergen raises or similar. –  Aryabhata Jun 13 '13 at 15:08

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