Part VIII - Evaluating Shortness
(Page of 3)

The opponents have bid up to four spades, in what you suspect is a ten card trump fit, where you hold a void.  How valuable is that void?  What is it really worth?

This, like all bidding questions, is a matter of probability, of expected values.  Let me approach this from several directions:

1.  How many high card points will partner have wasted in spades?

This is pretty easy.  There are ten spade high card points, and partner seems to hold three spades, so partner will hold, on average, 3/13 of these highs, or a little over two points in spades.  We can refine this estimate if we know more about the hand, for example:

2.  Suppose partner has shown weak-notrump values in the auction, and we hold ten high card points along with our spade void.  How much wastage will partner have?

This is more complex, but the calculations are similar.  Again, if we suspect that partner has only three spades, then a typical high card point in his hand will be in spades roughly 3/13 of the time.  However, since we hold ten points outside of spades, there are only 20 good points available, and ten points of spade wastage.  On a typical hand, an honor will be in spades, as opposed to another suit, ¼ of the time, but here, since more of the non-spade points have been accounted for, a point in partner’s hand is 4/3 as likely to be in spades as in any other suit.  Combining these estimates, partner rates to have around 4/13 of his 12-14 HCP located in spades, so around four wasted points.

If we are considering bidding on to the five level, we can anticipate roughly 9 working points in dummy, and should bid accordingly.

3.  How does our spade void affect the total trick estimate?

That, of course, is the theme of these articles, so I will answer that question in full detail.  First off, let’s determine how many defensive tricks, in spades, partner will contribute.  I will assume that our hand has been active enough in the bidding to pinpoint which hand, if any, would hold a void.  Thus declarer will never misguess the queen holding something like

KJxxx              A10xxx

nor will declarer go wrong when partner holds something like AJx.  However, we won’t allow declarer to play double dummy.  When partner holds, say, QJx, that will score a trick, even when the ace and king are behind him.  In Table 5, below, I have listed every possible three card holding for partner that might win a trump trick:

Table 5 – Expected (Wasted) Tricks from 3 Cards

 Holding Expected Tricks Number of Cases Total AKQ 3 1 3 AKJ 2 ½ 1 2 ½ AK10 2 ¼ 1 2 ¼ AKx 2 8 16 AQJ 2 ½ 1 2 ½ AQ10 2 1 2 AQ9 1 5/8 1 1 5/8 AQx 1 ½ 7 10 ½ AJ10 1 ¾ 1 1 ¾ AJ9 1 ½ 1 1 ½ AJx 1 ¼ 7 8 ¾ Axx 1 36 36 KQJ 2 1 2 KQ10 1 ¾ 1 1 ¾ KQ9 1 5/8 1 1 5/8 KQx 1 ½ 7 10 ½ KJ10 1 ½ 1 1 ½ KJ9 1 3/8 1 1 3/8 KJx 1 ¼ 7 8 ¾ K109 7/8 1 7/8 K10x 5/8 7 4 3/8 Kxx ½ 28 14 QJx 1 9 9 Q10x 5/8 8 5 Qxx ¼ 28 7

These total to 156.125 winners.  Since there are 286 ways that partner could be dealt three spades, these average out to .55 winners.  Look back at Table 1 – when the opponents hold a solid ten card fit, they rate to cash, on average, .78 winners in that suit on defense.  Since we hold a spade void, we know that they have no spade tricks to cash, so we start off .78 tricks ahead of the game.  Deduct partner’s wasted spade winners, and we are left with a plus of .23.

This is a pretty good general guideline – figure a plus adjustment of a ¼ of a trick to the total trick estimate for a void.

Next up, let’s extend this analysis to the case when we hold a small singleton opposite three unknown cards.  Making the appropriate modifications in Table 5 (the ace is no longer a wasted winner, and, since we hold a singleton, there is one less x in our case counting), partner will hold around .36 wasted winners opposite our singleton[1].  So, suppose the opponents have bid up to three hearts, in what rates to be a nine card fit, and we are contemplating bidding on to three spades.  If we hold a singleton heart, what is that worth?  How does it affect our total trick estimate?

Looking back to Table 1, a nine card solid fit will cash, on defense, 1.31 tricks.  Our singleton lowers that to 1, so we gain .31 tricks, but give back .36 winners to wastage.  This is essentially a wash.  What we gain in losers we give back in wastage.  In this case, there is no adjustment to the trick total, either up or down.

The same cannot be said when we hold a doubleton heart.  Our own loser count has climbed from 1.31 to 2, with no compensation, so that doubleton heart puts us .69 total tricks below expectation.  This explains the simulation results from last time.  We have what I like to call the Old School Rule:

• Add a quarter of a trick to the total trick estimate for a void.  Subtract off three-quarters of a trick if our side holds no singleton.

All of this loser analysis applies to our other short-suit holding.  The loser count drops with a void, and rises for flat shape, but complications arise when we convert losers to winners.  Suppose the opponents have saved in four spades over our four heart game, and we hold a singleton club.  That shortness lowers our loser count, and increases the chance that we can make five hearts, but, if we lead our singleton, it also greatly increases the potential penalty against four spades.  Bidding on might net us 450 or 650, doubling might net 800 or 1100.  These are high-risk, high-reward scenarios.  Doubling and leading our singleton might collect a juicy number, or might pickle partner’s J10xx, and lead to -590 or -690.  I have no good advice here, but, keep in mind, my loser-approach to total tricks can vastly overestimate the trick number when there are multiple ruffs available.  The more power we have shown, the more likely it is that partner will have a quick entry for ruffs.  Defending, doubled, will often be the winning decision when we have the balance of power, and a singleton to lead.  Bidding on is more attractive when partner is on lead.  If you can’t get ruffs, however, the total trick number will often be large, and defending a disaster.

I have come to the end of my analysis of total tricks.  In the next and final installment, I put these findings to work in a short bidding quiz.

[1] This is assuming, again, that we have been active enough in the bidding to help declarer guess partner’s holdings, like Qxx or AJx.  If we have been eerily quiet in the auction, then declarer will misguess these holdings, and partner’s wasted winners will increase to .47.  Apparently, information is worth a tenth-of-a-trick or so.  Of course, we can’t judge the total trick potential of a hand without exchanging information, but that exchange will help declarer guess some critical cards.  I like to think of this as the bridge players’ Heisenberg Uncertainty Principle:  Determining the total tricks on a hand increases those tricks.