A General Method for Valuing Bridge Hand Distributions

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Document ID: Ⓖ220315-g25

Title

A General Method for Valuing Bridge Hand Distributions


Gavin Wilson


March 22, 2015

Abstract

Do point-count valuations for hand distribution change significantly when the bidder shifts from length points to shortage points? Having identified that, apart from three-suited hands, there is little variation, a search is begun for a simple measure that takes into account both suit length and shortage. The resultant Sigma Count, equal to the combined length of the two longest suits less five and less the length of the shortest suit, is added to High-Card Points to determine the total point count of the hand. The application of the Sigma Count to very weak responding hands is then demonstrated.


Claygate Bridge Club; Gavin.Wilson@org2b.com

The author is grateful to Bernard Magee for his valuable advice and comments on this idea, and to Horst Oehler for his suggestions.

Keywords

  • Hand evaluation, Length Points, Shortage Points

Introduction

Counting up points to evaluate their hand is one of the beginner's first difficulties. (Let's overlook the confusion caused by bridge using the word point in at least two different ways, namely points scored and hand evaluation points.) Once the beginner thinks he is on top of High Card Points, he has to cope with distributional points.

It soon becomes clear to the student that there are distributional points for length, and distributional points for shortages. It seems there are extra points to be had for any suit that isn't three or four cards long.

And then the student learns that length points are for no-trump contracts, and shortage points for trump contracts. Theory appears to be telling the student you should not use both at the same time. (Ron Klinger, for example, says the partnership should use length points only until a trump fit is found, when you should switch to shortage points, or what he now calls 'ruffing points'.)

Sometimes you get into problems by awarding your hand three or even five shortage points because you are raising partner's minor suit, only to have partner suddenly switch the contract to 3NT, leaving you in a revaluation quandary. And the student soon plays hands in trump contracts where they make tricks from both ruffs and established suit length. It seems you should be accounting for both length and shortage in one measure.

There is one comforting statement that can be made about distributional evaluation: apart from the small set of three-suited hands (i.e. 5-4-4-0, 4-4-4-1, 6-4-3-0 and 5-4-3-1), the switch from length points to shortage points usually doesn't make much difference to the total.

It would be wonderful if there were a simple formula you could use which gives a fairly accurate distribution count, irrespective of whether you end up in a trump or no-trump contract. That way you would not have to switch evaluation bases mid-auction. This is one such attempt.

Background

  • Gold[1] gives the following 4 HCP hand as an example, where partner has opened 1♣ and RHO has passed. He proposes that responder should bid 1♠ whatever the vulnerability:
♠ A 10 9 3 2
8 7 6 5
6 4 2
♣ 2
  • Gold also proposes responding 1NT with the following 2 HCP hand when non-vulnerable after partner has opened 1♠ and RHO has passed:
♠ Q 2
10 7 6 5 3
8 2
♣ 7 6 5 2
Gold says the risk of responding 1NT in this second example is that partner may raise to 2NT, in which case responder should pass and the partnership will 'go off in 50s, but who cares?'.
  • Klinger[3] suggests responding 1 to opener's 1 with the following 4 HCP holding:
♠ 9 8 4
K J 8 3
3
♣ 9 7 4 3 2
  • Klinger also insists on a positive response to a 1 opening with this 11-loser, 2 HCP hand:
♠ 8 7 5
9 8 7 2
Q 3
♣ 7 5 4 2

The arguments put by the authors for some of these bids are not straightforward, and some rely on calculations about what opener's rebid might be with various types of hands. No simple principles are proposed so, without the ability to make a judgement call based on experience, the beginner and intermediate run the risk of taking a long time while they decide whether to bid or to pass. One possible consequence of their eventual passing is that their hesitation could put their partner in possession of unauthorised information.

There is a need for a simple principle which would enable responder to decide quickly whether he will be bidding or passing.

Discussion

  • Klinger[2] says we should add one length point for each card beyond the fourth in a suit. Therefore, if we take our two longest suits, we should deduct 8 ( = 2 x 4 ) from their combined length in order to calculate our length points.
  • If we are to combine length points and shortage points in the same count, we need a less aggressive evaluation than the 5-3-1 system which Klinger uses for voids, singletons and doubletons respectively. Instead we go for a 3-2-1 count. And we should apply it only to our shortest suit — not to two short suits, if we have a 6-4-2-1 shape, for example.
  • If we take the length of our shortest suit, and subtract it from 3, we get the correct shortage point count, on a 3-2-1 basis. So a singleton scores two ( = 3 -1 ) points, for example.
  • So the total distribution score is the length of the two longest suits minus 8 plus 3 — the length of the shortest suit. Take 8 away from 3 and we are left with minus 5.

For want of a better name, we call this measure the Sigma Count, for its association in statistics with distribution. The Sigma Count is what we are going to add to our HCP to derive our hand's full point count.

SIGMA COUNT =


The Combined Length of your Longest Two Suits
less
The Length of Your Shortest Suit
and less
5

If we test this formula against the most common distributions, we arrive at the following:

Hand Shape
Length Points
Shortage Points
Sigma Count
4-4-3-2
0
1
1
5-3-3-2
1
1
1
5-4-3-1
1
3
3
5-4-2-2
1
2
2
4-3-3-3
0
0
-1
6-3-2-2
2
2
2
6-4-2-1
2
4
4
6-3-3-1
2
3
3
5-5-2-1
2
4
4
4-4-4-1
0
3
2
7-3-2-1
3
4
4
6-4-3-0
2
5
5
5-4-4-0
1
5
4
5-5-3-0
2
5
5
6-5-1-1
3
6
5
6-5-2-0
3
6
6
7-2-2-2
3
3
2

As you can see, the Sigma Count is not a bad approximation for the best distribution count of most hands, on the assumptions that:

  • Most three-suited hands are going to end up in a contract with the trump suit being one of those three suits, and
  • Most hands with a 6+ card suit will end up in a contract with that long suit being trumps.

Results: Three Sample Deals

Let us try applying this idea to a few hands.

Example 1 (of 3): A hand which illustrates the danger of passing with less than 6 HCP

Illustrative Hand
North
♠ 10 8 6 5
-
Q 10 9 8 7
♣ 9 7 5 2
West
♠ J 9 7
J 10 8 5 2
J 6
♣ A K J
None Vulnerable


Dealer: South
East
♠ 4 3
A 7 6 4
A 5 2
♣ Q 10 8 6
South
♠ A K Q 2
K Q 9 3
K 4 3
♣ 4 3


The Usual 4-card Major Bidding
S
W
N
E
1 1
No 2
No 3
No 4
  1. 17 HCP with two 4-card majors: always bid the hearts first.
  2. 11 HCP, and no 4+ card suit apart from hearts. Pass.
  3. Just 2 HCP. Don't you need 6 points to respond?
  4. 10 HCP with 4 hearts and 4 clubs. Pass.

Result: N-S can make at most 5 tricks in hearts. Two off scores -100.


Recommended Bidding
S
W
N
E
1 1
No 2
1♠ 3
No 4
3♠ 5
No
No 6
No
  1. As before.
  2. As before.
  3. This hand's Sigma Count is 5 + 4 - 0 - 5 = 4. So North has a total of 6 points — just enough to show his 4-card spade suit.
  4. 10 HCP and only 4-3 in the unbid minors, so a double would be wrong. Pass.
  5. With a Sigma Count of 4 + 4 - 2 - 5 = 1, South's total point count is 18. With four spades, he can raise North's bid to the 3-level.
  6. North was minimum for his 2♠, so he will not raise South's invitational bid.

Result: N-S make 9 tricks in ♠ for a score of +140.

Example 2 (of 3): A Responding Hand with 3 HCP which should not pass

Origin: Claygate Bridge Club, 15th December 2014.

Claygate Hand #14
North
♠ Q 8 7 5 3
J 10 6 4 2
-
♣ 8 7 2
West
♠ 9 4
K 9 5 3
Q 10 2
♣ A K 10 5
None Vulnerable


Dealer: East
East
♠ 6 2
Q 8 7
K 8 6 3
♣ Q J 6 4
South
♠ A K J 10
A
A J 9 7 5 4
♣ 9 3


The Usual 4-card Major Bidding
E
S
W
N
No 1
1 2
No 3
No 4
  1. 8 HCP balanced.
  2. 17 HCP, 6-card diamond suit.
  3. 12 HCP, but West cannot double 1 with a two-card spade holding.
  4. Just 3 HCP.

Result: N-S make 8 tricks in diamonds for +90. Three pairs at Claygate ended up in the 1 contract.


Recommended Bidding
E
S
W
N
No
1
No
1 1
No
2♠ 2
No
4♠ 3
  1. North has a Sigma Count of 5 + 5 - 0 - 5 = 5, which gives him a total of 8 points. With this tally, bid your cheaper 5-card suit first.
  2. Instead of the jump-shift, some players might bid the more demure 1♠. But with 5-card support for spades, North would still jump pre-emptively to 4♠ opposite an opening hand with at least 4 trumps.
  3. 8 points opposite at least 19 means game. Slam seems unlikely.

Result: N-S make 11 tricks in ♠ for a score of +450. Three pairs at Claygate were in 3♠ and five in 4♠.

Example 3 (of 3): Opener must reverse to find his only viable game — in hearts

This hand came from a social game of 2015.

Social Game, 5th January 2015, Hand #11
North
♠ A
9 7 6 5 2
A K 7 6 5 3
♣ Q
West
♠ 10 5 4 2
Q 10 8
10 4 2
♣ J 7 6
None Vulnerable


Dealer: North
East
♠ K J 9 3
4 3
Q 9 8
♣ A K 10 5
South
♠ Q 8 7 6
A K J
J
♣ 9 8 4 3 2


The Usual Bidding
N
E
S
W
1 1
No 2
2♣ 3
No
2 4
No
2NT 5
No
3NT? 6
No
No
No
  1. 13 HCP, open the clearly longest suit.
  2. 13 HCP, but unable to double because of the heart doubleton.
  3. 11 HCP, so responds with his clearly longest suit. Some may prefer 1♠, but the implications for North's rebid are the same.
  4. What else can you bid with this strength and six diamonds?
  5. Stoppers in the unbid suits.
  6. What else can he do? If he bids 3, surely South would pass?

Result: N-S go off two in 3NT for -100.


Bidding with added Sigma Count
N
E
S
W
1
No
2♣
No
2 1
No
2♠ 2
No
3 3
No
4 4
No
  1. North has a Sigma Count of 5 ( = 6 + 5 -1 - 5 ), giving him 18 total points, so exceeds the usual 17-point hurdle needed for a reverse.
  2. Fourth-suit forcing, showing 9+ HCP and forcing to game.
  3. Showing extra length in hearts beyond the 4 cards promised by the reverse.
  4. Signing off in game after finding the 8-card major suit fit.

Result: N-S make 11 tricks in for a score of +450.

Critique

Magee[4] says that the use of this method will result in the auction being 'much too high' when no fit is found, because distribution strength is 'of no use in a no-trump contract'.

However Gold[1] gives an example of a 2 HCP hand which he proposes should not be passed. Klinger[3] gives several examples of 3 HCP hands on which he advocates not passing. There is clearly disagreement among the experts.

In my view, the biggest risk is where responder uses the Sigma Count to justify a 1NT response, and his partner goes on to bid 2NT. You may go down a trick or two, but if so, the opponents probably had a contract themselves. And this scenario is relatively rare. Overall it is the losing option far more often to pass when the Sigma Count says you should bid.

Quiz

Your partner has opened 1, and your RHO has passed. What is your Sigma Count on each of these hands? What should you bid? (Remember to cover up the answers beneath while you work out the solution.)
1.
♠ 7 6
K 9 6 3 2
7 5
♣ 8 6 3 2
2.
♠ 8 7 5
9 3
K Q 8 2
♣ 9 6 5 3
3.
♠ 9 8 6 4
-
Q J 8 5 2
♣ 7 5 4 2
4.
♠ J 8 5
8 3
Q 7 5 3
♣ J 9 7 3
  • You have a Sigma Count of 2 (=5+4-2-5).
  • Bid 4. Always raise one-of-a-major to the four-level with 5-card support and 0-10 HCP. The Sigma Count changes nothing here.
  • Sigma Count of 1 (=4+4-2-5).
  • Bid 1NT. You effectively have six points.
  • Sigma Count of 4 (=5+4-0-5).
  • Bid 1♠. You have seven points.
  • Sigma Count of 1 (=4+4-2-5).
  • Pass. With no more than five points, you should not encourage your partner..

Conclusion

The Sigma Count is a straightforward arithmetic calculation which works most of the time and can often help to clarify those bid-or-pass decisions when a bridge hand is on the borderline. It should result in a swift decision, enabling a response in tempo, thereby avoiding the transmission of unauthorised information through hesitation.

The formula condenses length points and shortage points into a single formula and, by switching to the Sigma Count, the beginner may lose the directional aid these give him with his overall declarer strategy: if he used length points to evaluate his hand, he knows he should be trying to establish long suits, and if he used shortage points, he should be going for ruffs, particularly in the short trump hand. But this small pedagogical disadvantage will be quickly overcome as the beginner obtains more experience in declaring.

Those of you who use point-count evaluation are warmly recommended to give the Sigma Count a try within your usual system.

References

  1. Gold, David (2015). Respond Often Non-vulnerable. Aylesbury, UK: English Bridge Union. English Bridge (April 2015 edition p. 13). Print.
  2. Klinger, Ron (2007). Guide to Better Acol Bridge (Third edition). London: by Weidenfeld & Nicholson in association with Peter Crawley, pp. 2-3. ISBN 978-0-297-85352-7.
  3. Klinger, Ron (2002). When to Bid, When to Pass. London: by Cassell in association with Peter Crawley, pp. 38-42. ISBN 0-304-36219-0.
  4. Magee, Bernard (2015). Private correspondence. (February 2015).