Experts Prefer Endplays
In the December 2009 issue of the ACBL Bulletin Mark Horton, editor of Bridge Magazine, gives this advice to the readers: ‘As declarer, eliminate a suit if you can even if you don’t know why you are doing it. Good things can happen.’ There is no warning label; Horton leaves it to the reader not to get carried away and to use some common sense in its application. I like that. Another good piece of advice would be, ‘always make the play that has the greatest probability of success’, but here the difficulty is that a declarer may not know which choice fits the description. That is where Jeff Rubens comes in. In his recent book Expert Bridge Simplified, Rubens presents the reader with methods that allow a player to sift out the plays that stand the best theoretical chance of being successful, always assuming that the player is capable of the necessary rapid mental arithmetic. This book is on my Christmas list.
The following deal as presented by the ACBL reviewer, Paul Linxwiler, combines these two pieces of advice as it involves the possibility of a triple elimination with the estimation of the probability of success. Will Horton’s advice prove its worth on careful examination?
Opponent’s Cards  Even Split  
♠ AQJ1098  ♠ 7543  3  2 – 1 
♥ AK  ♥ J7  9  5 – 4 
♦ 73  ♦ AK  9  5 – 4 
♣ AK4  ♣ 76532  5  3 – 2 
West reaches slam in spades and receives the opening lead of the ♥10. Going to dummy in diamonds to take a winning spade finesse would assure the contract, but that play has a probability of success slightly less than 50%. The alternative is to play the ♠A, hoping but not expecting to drop the ♠K, then cashing all the top side winners before end playing a defender who started with doubletons in both black suits. The central question is how likely is it that a player holding the ♠K doubleton also was dealt a doubleton club. Rubens points out that the latter play should be chosen as the success rate is about 54%.
Linxwiler agrees, but is at pains to point out that for most readers it may be too close to call. The best approach for many may be to save energy, take the finesse, and move quickly to the next hand. I object to that. A margin of 4% is not too low to matter, and good players should attempt to make the distinction at least part of the time.
The simplest advice one can imagine with regard to going with the odds is Bob’s Blind Rule: always play for the most likely splits possible given what you know at the time of decision. How much easier can it get?
Declarer controls a division of sides 10=4=4=8 (all even), so the defenders hold a sides of 3=9=9=5 (all odd). There has been no interference and the opening lead is a nondescript ♥10, so no alarm bells are ringing, and there is no reason to assume that the suit aren’t splitting normally. Let’s look at pairs of hands composed of the most even splits in each suit, which are also the most probable combinations on the evidence so far.
I  II  III  IV 
♠ 2 – 1  ♠ 2 – 1  ♠ 2 – 1  ♠ 3 – 0 
♥ 5 – 4  ♥ 5 – 4  ♥ 4 – 5  ♥ 4 – 5 
♦ 5 – 4  ♦ 4 – 5  ♦ 5 – 4  ♦ 4 – 5 
♣ 3 – 2  ♣ 2 – 3  ♣ 2 – 3  ♣ 3 – 2 
Condition IV is less likely that Condition I by a factor of 1/3, so we ignore the possibility. For Conditions I through III the odds are 2 to 1 that the doubleton spade is in the same hand as the doubleton club. The obvious reason is that there are more possibilities when it is not required that both red suits split 5 in one hand and 4 in the other. Therefore, under Bob’s Blind Rule, declarer should adopt Rubens’ suggestion: play to the ♠A and if the ♠K doesn’t drop, play off the side suit aces and kings (as Horton suggests) then exit with a second spade hoping the winner has no more clubs. Note that it doesn’t matter which defender has the double black suit doubletons. Of course, one cannot say the odds are anything as good as 2 to1 in favor, but it appears likely that this sequence gives better odds than an initial trump finesse.
Let’s look at the suit combinations that lie at the bottom of this simple method. The first critical move is the play of a low spade towards declarer’s hand to which South follows with a low spade. What are the probabilities at this critical moment of first decision? By ignoring the implications, if any, of the red cards played to the first two tricks, one may use the initial vacant places as a rough guide to the number of combinations. These are as follows:
Spades  3 – 0  2 – 1  1 – 2  0 – 3 
Others  10 – 13  11 – 12  12 – 11  10 – 13 
Weights  11  13  13  11 
The weights are the relative number of card combinations that accompany the given splits in spades. Next we shall consider the play of low trump to which South follows with a low card. If the missing spades are denoted as K,u, and w, and the spade played is card u, this is the situation at the time of decision:
Spades  2 – 1  1 – 2  1 – 2  0 – 3  
Kw – 0  w – K  K – w  0 – Kw  
Weight  26  26  26  22  
Plausible Plays  1  1  2  2  
Adjusted Weights  26  26  13  11  Total = 76 
To obtain the current odds the combinatorial weights must be adjusted through division by the number of plausible plays available in the play of card u. If South held both cards u and w, he might have played card w instead, so the probability that he would play card u is halved. Now one can calculate the probability of the finesse winning by comparing the number of combinations for which the finesse wins (FW) to the number for the finesse losing (FL) or the drop winning (DW).
FW = 26 + 11 = 37; FL = 26 + 13 = 39; and DW = 13.
The finesse has a winning percentage of 49%, the drop, 17%. Thus, if it were just a matter of finesse versus drop, declarer would finesse, however, as the saying goes, 2 chances are better than 1. The probability of success for the subsequent endplay is the product of the probability of king now being singleton (52/76) multiplied by the probability that the doubleton club sits with the singleton ♠K. The following configuration shows the weights to be attached to the relevant conditions.
I  Spades  12  II  Spades  12 
Clubs  23  Clubs  32  
♥ & ♦  108  ♥ & ♦  99  
Weight  9  Weight  10 
The chance of the doubleton club having been dealt to the hand containing the ♠K doubleton is 10/19, slightly more than 1/2. So we find the probability, PE, that the endplay will be successful after card u appears:
PE = 13/76 + (52/76) times (10/19) = 0.53 > 0.49
The conclusion is that the endplay is the preferred strategy by about 4%. This is in close agreement with Rubens. The calculation confirms the more general advice of Horton and Bob’s Blind Rule. One might conclude that happiness in bridge as well as life depends largely on one’s ability to recognize possibilities as they arise.
Here is an obvious variation on the Rubens hand that involves an elimination play when the defenders hold 7 cards in a suit. It is more readily recognized as there will be a loser in clubs should the spade finesse fail. How does Bob’s Blind Rule fare?
Opponent’s Cards  Even Splits  
♠ AQJ108  ♠ 97543  3  2 – 1 
♥ KQ2  ♥ A74  7  4 – 3 
♦ A3  ♦ 8  10  5 – 5 
♣ A108  ♣ K953  6  3 – 3 
One of the interesting features of the above hand is that the numbers of missing cards in a suit are a combination of evens and odds. This profoundly affects the card combination probabilities as shown in the following list of the 5 most likely distributions.
I  II  III  IV  V  
♠ 2 – 1  ♠ 2 – 1  ♠ 2 – 1  ♠ 2 – 1  ♠ 2 – 1  
♥ 3 – 4  ♥ 4 – 3  ♥ 3 – 4  ♥ 3 – 4  ♥ 2 – 5  
♦ 5 – 5  ♦ 4 – 6  ♦ 5 – 5  ♦ 4 – 6  ♦ 6 – 4  
♣ 3 – 3  ♣ 3 – 3  ♣ 2 – 4  ♣ 4 – 2  ♣ 3 – 3  
Weights  1  5/6  3/4  5/8  1/2 
Given that the spades split 21, creating an imbalance in vacant places of 1, the other odd numbered suit, hearts, can split 34 and allowing the even numbered suits to attain an evensplit status (Condition I). If the longer card in hearts and spades lie to the same side, it creates a vacant place imbalance of 2, so one of the even numbered suits must split unevenly to fill the vacant places, thus reducing the numbers of possible card combinations (Conditions II thru IV). It is much more probable that the longer minor (diamonds) is split unevenly (conditions II and V). Taken all together, one sees that on these 5 common conditions, the clubs are most likely to split 33.
Usually playing in a matchpoint contest, simple is best. If the contract is 4♠, taking the spade finesse is the recommended line, as the field may make 12 tricks on this line of play. If the contract is 6♠, other factors require consideration. What proportion of the pairs will reach this slam? If fewer than half, one will score above average just by making the contract, whereas going down 1 will be a disaster. The situation is such that assuring the contract is the paramount consideration, just as it is at IMPs. Rather than maximizing the number of tricks won, one strives to minimize the number of tricks lost.
If we return to the situation where a spade is led towards the trump tenace and South follows with card u, the probability of the drop succeeding is 17%, so under those fortunate circumstances 12 tricks are assured. When the drop fails to produce the desired result, declarer must revert to the endplay hoping not to lose a trick in clubs. It would be extremely unethical at this point to mutter ‘Oh Damn!’ or your preferred equivalent. Hearts are eliminated and a spade ducked to the ♠K. A club is returned and declarer plays generally on the hope that the ♣QJ are split between North and South. Are the odds good enough to give an overall probability of success greater than 49%?
Suppose first declarer always plays for split club honors. The probability, PE, that the endplay will be successful on those grounds alone:
PE = (13/76) + (52/76) times the probability that the ♣QJ are split.
One may estimate the probability of the ♣QJ being split on an a priori basis as follows:
Split  Probability  Probability with
Split Honors 

33  35.5%  21%  
42  48.5%  26%  
51  14.5%  5%  
60  1.5%  0%  Total 52% 
In the case of a 60 split, the endplay will be successful at least half of the time. So we arrive at an estimated overall rate of success, PE, of 53% before a club is played. This is about the same chance as with the first problem considered. The next question is whether anything can be gained or lost from the enforced return in the club suit.
Suppose South wins the ♠K and returns the ♣Q. Should declarer continue to play for split honors, or does that lead make it more likely South also holds the ♣J? If so, declarer can win in the dummy and finesse against the jack presumed to be in the South hand. Here is the situation with a 33 split. The ratios are a measure of the relative conditional probabilities once the queen has been placed on the table.
South  North  Combinations  Plausible Plays  Ratio 
Qxx  Jxx  6  3?  2 
QJx  xxx  4  2  2 
With QJx South must lead an honor, but he could choose either, so there are 2 plausible plays with 4 possible combinations. With Qxx South must find North with the jack, so he can lead with equal effect any of 3 cards with 6 possible combinations. On the assumption of perfect knowledge the lead of the queen is equally probable from Qxx or QJx. This illustrates the general principle that maximum uncertainty is achieved by defenders who choose equally between equivalent cards.
Well, that’s the way one might program a computer to defend, but reality is different. The computer strategy may represent the distillation of results for defenders of wide ranging abilities, but in practice one plays against one pair at a time. By those who see only their own cards, an honor would be led only when obviously necessary, that is, when holding both queen and jack. It is (almost) certain that the lead would have been low from Qxx, in which case declarer should let the lead ride, win the jack with the king, and finesse for the queen on the way back.
An imaginative player can see that declarer must hold the ♣10 in order to spurn the spade finesse, so that leading low from Qxx is a losing proposition. Such a player will most often lead the queen from Qxx hoping declarer will play him for both honors. Against such a player declarer should assume the lead is from Qxx, win the ace and finesses North for the jack. The reason is that the queen will be played from more combinations of Qxx than of QJx. There may be a nasty surprise in store, but one must expect occasionally to pay off to an unusual play. Advice to declarers: play for split honors unless a poor player leads an honor.
The above argument is based on a 33 split which is the most likely situation even when South has shown out on the third round of hearts (Condition V above), the reason being that the longer diamonds will split unevenly more readily than the shorter clubs. Here is the rare situation where South leads the queen from Qx:
South  North  Combinations  Plausible Plays  Ratio 
Qx  Jxxx  4  1  4 
QJ  xxxx  1  2  1/2 
The odds for leading the queen from Qx greatly outweigh the odds for QJ, the reason being that a QJ doubleton is a rare occurrence. On any lead, declarer should play for split honors.
Hack again?!