Brute Force Hand Evaluation from Hard-Headed Old Timer

By BBOer roman99 (Roman L. Weil)

I’ve circulated the hand below around my club and can get no one to agree with me. I must be wrong. My intuition says otherwise so I have resorted to brute force hand analysis, as I didn’t know how to do this analytically, to see what I can find out. I’m telling you what the results show so that you can add these to your intuition to see if they persuade you to agree with me later, if you don’t now. Here’s the problem from a Silver Points in-person matchpoints duplicate. [I quit playing around 1970 long before the ACBL invented Silver Points and created new hurdles for us old timers.  Even with the grandfathering awards of some token points, the change in requirements irks.]

The auction assumes you play a relay over a 15-17 1NT opening bid commanding a 3♣ response, which the responder will deal with according to the shape and quality of his hand, presumably loaded with some minor suit cards.

The diagram shows the North hand who commanded the relay with his seven-card club suit and now must decide whether to pass 3♣ or to invite game.

What do you do?


*2♠ relayed to 3♣ which invites corrections as needed to match North's wishes in the minor suits.

Me? I’m from the old school who learned about revaluing for length and adding, approximately, one point for the 5th card and two points each for the 6th and 7th cards, so I revalue this hand upwards from its nominal high card points of 5 to about 10: King + Queen + five for length. And working points, too. I figure there is about a 2/3 chance the spade King will carry its weight and produce a trick (or otherwise help partner with his suit). Moreover, the club Queen is no quack in this context. I think this hand warrants an invitation to game. Partner will not accept if his club holding is a worthless doubleton and if his outside-club holdings are quack-like, more-or-less worthless in a minor suit game contract. For me, this is an easy raise to four Clubs.

Many I’ve shown this hand to disagree with me. Some think the risk of encountering the doubleton club and going down in four clubs is too great to take the chance of finding the fit needed to grope for five.

The most helpful comment came from modern guy BBOer Miamijd, who said,

The first thing you need to do is get a better system.   You need a system where 2♠ reveals interest in clubs and 2NT in diamonds, so that there are two possible responses.  The first one becomes negative; the second one is positive. So if opener hears 2♠ and rebids 2NT, you just sign off in 3♣. If opener rebids 3♣ (accepting the invitation) I would try 5♣. That way you don’t end up in 5♣ down when partner has a hand that doesn’t accept an invitation.

Time for me to adopt the modern convention that lowers the level of the bidding for making the key decisions. Note that with the modern convention, one gets to decide at a lower level whether the hand is worth pushing towards game. Still the holder of that 7-card club suit with the outside king needs a good way to think about it.

Miamijd often uses the losing trick count — he’s about persuaded me to start using it — and continues his evaluation of this hand as follows: This seven-loser hand in clubs has a good chance to make 11 tricks if partner has a good 1NT opener with six losers. [Those who use the losing trick count for hand evaluation add the counts for the two hands, once a trump suit has been established, and subtract that sum from 24 to estimate the trick-taking potential of the two hands.  Here 6 + 7 = 13; 24 – 13 = 11.  Here’s a good, thorough introduction, which tells you how to do it.  Wikipedia gives a good history, but is weak on how to do it.]

Before I started this column, I had no firm idea of the likelihood that partner has only a doubleton club.  Now, I do.  I decided to find out by brute force and give you a link to the results, which I’ll tell you about in a moment, after I discuss the results.

If you start with the notion that your partner has no worse than a doubleton in any suit when he opens 1NT and you hold seven cards in a suit, you can be confident that partner holds a doubleton only 20% of the time (tripleton 40% of the time; four or more 40% of the time).    Further brute force analysis, of the sort in the link, shows that the expected trick loss is just one trick if all we know are the deal probabilities.  If we know that partner accepts our club invite, meaning he has three cards to a top honor, then the expected loss drops to under one trick.  You didn’t need any detailed computations to tell you that:  If you hold ten cards in a suit Q-seventh opposite A-third or K-third, but you’re not sure which it is yet, you expect to lose no more than one trick.

Here’s the link to the computations where I show you how I derive the probabilities described in the preceding paragraph.



To make the calculations, I assume that North has seven clubs. Then, I assume that South never opens 1NT with a singleton, so has at least two clubs. Then I take the remaining four clubs and distribute them randomly among the three players West, South, and East.

To be concrete, look at the fifth row of the table below (Exhibit 1), which shows West with 2, South with 1 and East with 1. Adding South’s one to his assumed starting doubleton means that he will end this simulation with a tripleton, as shown in the header of the column. The enumeration column shows this can happen twelve ways.

This is where the brute force counting comes. If you start with four cards and distribute two to West, one to South, and one to East, then you can do that in twelve different ways. [The twelve ways to distribute four cards (2, 3, 4, 5) between three hands, two to West, one each to South and East are: 5 4 |3|2; 5 4|2 |3; 5 3|4|2; 5 3|2|4; 5 2|4|3; 5 2|3|4; 4 3|2|5; 4 3|5|2; 4 2|5|3; 4 2|3|5; 3 2|4|5; 3 2|5|4.]

The table enumerates all the ways to distribute the four clubs between West, South, and East given that North started with seven and South with two. Then I count how many of those ways leaves South with none of the four, so that he ends with his original doubleton. At the foot of the doubleton column we see 16. For tripleton, 32; For four cards, 24; for five, 8 and for all six, 1. That’s a total of 81 possible ways to distribute the four cards among the three hands. The likelihoods (probabilities) of the suit holdings for South result from using these numbers:

Doubleton: 16 of 81 = 20%
Tripleton: 32 of 81 = 40%
Four-carder: 24 of 81 = 30%
Five- or six-carder: 9 of 81 = 10%.


Exhibit 1

We’re not done yet. To know that South will hold a doubleton club 20% of the time doesn’t tell us how many club tricks we expect to lose if clubs are trumps and South holds a doubleton. To compute that takes more, even more tedious, brute force counting of the sort in the table below (Exhibit 2).

Exhibit 2

Look at the third row of the second table, which shows the situation when south holds Ax of clubs and East-West hold KJxx. There are 36 combinations of actual spot cards for distributions of those six cards where South has two and East-West have the other four in various combinations. The first entry in the third row shows K=Jxx which is the conventional notation to mean West holds the King and East holds the Jack and two spot cards. There are three combinations like that when there are a total of three spot cards, with the third being held by South in his Ax doubleton. The next entry in that row shows KJx-x which is the conventional notation to mean that West holds KJx and East holds x OR vice versa. There are six combinations like that when there are a total of three spot cards, with the third being held by South in his Ax doubleton. So brute force counts the number of combinations of each kind, with the South doubleton shown on the left of each row.

Then the analysis has to put each combination into a column for how many tricks the declarer can expect to lose for that combination. The analysis assumes the same order of play for a given holding by declarer opposite the queen-7th of clubs. [For example, in the table for the situations where declarer holds a tripleton Axx and the outstanding cards are KJx, then when they lie K=Jx, declarer can lose no clubs by laying down the Ace. But when they lie J=Kx, the winning play is to lead the Queen towards the Ace. The computations assume the declarer plays Axx opposite Queen-7th the same always.]

Doing the arithmetic at the foot of Exhibit 2 provides the result we’ve been looking for. The expected number of tricks to lose given that South holds a doubleton club is 1.46.

The same sort of tedious computations, not shown here, provides the result that when South holds a tripleton club, the expected number of tricks to lose is 1.37, when South holds four cards, the expected number of lost tricks is 0.46., and with five or six the expected number of lost tricks is 0.22. The data arrays like this:

South/Declarer has Club Length With Probability Then Expects to Lose This Many Club Tricks
Doubleton 20%
1.46
Tripleton
30%
1.37
Four Cards
40%
0.62
Five/Six Cards
10%
0.22


Now, we can with the hindsight of these computations evaluate North’s prospects. The mechanical computation of multiplying the probabilities of the layouts above by the expected club trick losses for each layout [(.20 x 1.46) + (.30 x 1.37) + (.40 x .62) + (.10 x .22)] = 0.97 tells us that our club suit expects to lose .97 trick, close enough to 1.00 to say we expect to lose 1 trick from the perspective of the North bidder deciding to play in clubs in the auction given in the bidding diagram. To summarize: a North looking at a seven-card club suit opposite a partner assumed to hold at least a doubleton can expect to lose no more than one club trick. But if we issue the invitation and South accepts, he will have a tripleton or better and then his expected club losses come from the last three lines of the table above. The expected loss is then only .70 club trick.


How about that spade King?  Could be wasted.  Could be worth a trick in the play.   As I said earlier, a partner revaluing his hand to accept a minor suit game invitation will likely disregard quack [non-]values and focus on top honors.  If he follows such a process, then a King is more likely valuable than usual.  That’s my case for thinking the invitation is a good idea.

What about the risk averse thoughts — those who worry about the possibility of the rotten club fit?   The expected loss with the worst of all possible fits is a trick and one-half.  We write off the heart and devalue the spade King.   Three and one-half down-graded tricks.  How big a risk is a raise to four? 

Show me a hand that makes four a bad bet and has the property that if I swap the non-club holdings in the hand, then a four-level contract remains a bad bet.  I just want to be sure you are not cherry picking. 

I end with the actual hand I held as South, which proves nothing except that when I held it, I thought about super-accepting.  I decided that partnership discipline demanded that I stay in lane.   I had scolded this very partner for bidding my values for me in an earlier session, so super-accepting seemed unwise for morale, even if sound bridge. 

If you are inclined to comment, tell me what you think of the old advice I still carry with me about adding length points:  one for the 5th card and two for each card after that.  I think the main lesson here is to use a system in responding to opening NT bids that enables exploration of minor suit fits without going beyond the 3 level.   A secondary lesson is that a seven-card suit  headed by the Queen facing a known doubleton, or better, expects to lose no more than one trick. 

11 comments on “Brute Force Hand Evaluation from Hard-Headed Old Timer”

  1. a note about LTC -- when used by a distributional hand, there's a grave danger of duplication of values. Stiff heart vs KQx in effect counts 4!!! winners in the heart suit. So, must use LTC with caution here.
    Even opposite a near perfect hand, there's only a 75% chance of game. (You go down with a 2-0 club split with the diamond K off). If that's the best there is, inviting is probably too much. Or, Rubens's rule -- invite only if can make opp a perfect min. Such a min exists, but it is VERY unlikely: something like:
    AQJx xx Axxx A9x. Even then, you may go down with a diamond lead and a 3-0 club split (or a 2-1 club split and an early ruff in spades). Also, does your system let partner know what's working? Perhaps if you can do this: 3C (P) 3H (you)(singleton). Even then, he may go to 3N with xxx KQJx KQxx Ax, and that doesn't fare well at all if the S K is off.
    As between your "brute strength" use of LTC and Rubens' rule, I'd go with Rubens. LTC at best works on average. Rubens' rule is tailored to the individual hand. (I once wrote a 6-page essay criticizing LTC. Mathematically, it's equivalent to an honors and shortness count (like the standard one), except that A, K, and Q are all valued at 3, and doubleton - singleton - void are valued at 3-6-9. (That last shows where duplication comes in. Awfully easy to have duplication in a suit if P counts 9 for his void and you have an honor).

  2. Anonymous is on the right track to wonder about a stiff high honor.

    When you make this leap, the landscape of all four hands changes. Also, I don't know of any follow-on conventions that ask “Any stiffs, partner?”

    This new trap door stiff carries the specter of creating Duplication of Value
    that damages the Captain's ability to cogently judge the final landing spot.

    So until your partnership comes up with a reliable way to fill this gap, may I gently suggest that one say “for overcalls, yes; for opening, no.”

  3. Since one CAN open a NT with a singleton honor (AK or Q) this analysis is incomplete. Additionally, IF partner has a singleton AC then he has "possibly" 4441 shape and 5C will fail BUT the opener will STILL have enough (unwasted) values to accept the invite over a 4C rebid---------since this will show a responders hand UNSUITABLE for a 3NT rebid.

  4. What about picturing hands for opener? If opener has four aces (s)he will think that's worth accepting a game try in clubs. But game is by no means certain and you could even lose a spade, a diamond and the KJ of clubs. And opener's 1NT could be much worse for you than four aces.

  5. I think that the calculation of clubs distribution is a variation of the Monty Hall problem...knowing that the NT opener has at least two clubs makes the distribution of the rest of the clubs non-random.

    By my calculations (based on distributing all remaining clubs randomly to open slots and then excluding all possibilities where NT opener has less than 2), when you have a 7-card club suit, NT opener will have the following:

    2 Clubs: 53.6%
    3 Clubs: 34.2%
    4 Clubs: 10.7%
    5 Clubs: 1.5%
    6 Clubs: 0.1%

  6. What a good point Henry Zee makes. No, not the praise of my suggested auction*. I refer to his phrase “these agreements do not happen often.” Happily, there is a modern fix for that, too.

    BBO sports a feature called Constraint Files. These are available at Bidding Tables and Teaching Tables.
    See http://www.charlesandgerry.com/bridge/constraints.html

    Using the Super Accept constraint, you can practice such hands as often as you and your partner can stand. With enough practice, rare hands become old home week when they actually come up.

    * A quintessentially forgivable mistake. 8-))

  7. Charles' auction is excellent. It gives you to leeway to stop at 3C, also with co-operation from partner get to game.
    Well done!!

    I also like your items 2 and items 3 ideas.

    The only issue is for old folks like me, I probably will mumble to myself when partner responded 3H, what the heck is that?? 'cause these agreements do not happen often.

  8. Super-accepts, even over minor transfers, have been around since at least the 1970's (which is when I started to learn them). And yes, both partners need to reevaluate after each call (“support points”). It's just that you, the partnership, missed two evaluation opportunities: the need to super-accept clubs and the need to evaluate for Duplication of Value.

    Let's take your suggested starting point and build on it.
    1N – 2S(1)
    3D(2) – 3H(3)
    4C(4) – 5C(5)
    all pass

    (1) club transfer, presumably 6+ cards.
    (2) Partner, in clubs my hand revalues to 18 or more and if you have shortness in diamonds, that's bad (duplication).
    (3) I have a stiff or void in hearts, that's where Duplication would be worst.
    (4) accepting at long last, preferring clubs to NT.
    (5) Sorry, partner, not a slam complement.

    You're probably getting the hint here that it's not brute force that is called for, but as is so common in this game, cooperative bidding. But of course that depends on both partners knowing what their hands are really worth and having the tools available to cogently ask or tell.

    Alert: Items 2 and 3 above are much newer ideas (than the '70s) so you'll have to discuss and practice them, but then you'll be way ahead of a field where brute force and guesswork are still in effect. 😎

  9. I think you are assuming that partner rates to hold a club doubleton 20% of the time based on the rough a-priori odds of his having a doubleton in a particular suit.

    But this is the wrong way of figuring it. The correct way is to figure out how many hands can be made with exactly two out of the *six missing clubs* (you have 7) vs all hands that can be made with three and four. I believe you'll find out that the doubleton holding is much more likely given you have seven. I'll try to figure the exact odds and post again when I have some more time.

  10. Guess we had the same teacher, Strassburgers and Slagers early on taught the 4 loosing tricks.... 2clubs...............enjoyed your column and glad to see you are active in the bridge world.

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