Tim Bourke The Master Carver
Once upon a time two craftsmen sat at a dimly lit workbench earning a scant wage from the richpilgrim trade by carving lucky turtles from wood, lucky turtle after lucky turtle, day after day. Maybe they should have kept some of that luck for themselves. One turned to the other and said
‘Look at this piece of wood, all knurled and knotted. How can anyone make a turtle out of that?’
‘How can they send us wood like that? There should be a law against it,’ commented his companion. “Throw it in the trash bin.’
Just then the master carver entered the workshop.
‘Let me have a look at that. Interesting. Let’s see what I can make of it.’
The master took away the defective specimen and in a few days everyone was marveling at an impressive figurine of a fierceeyed eagle with wings outspread, sharp talons clutching a forlorn fish, and looking altogether as unturtlelike as could be.
Tim Bourke is the master carver of bridge hands. Give him a contract that appears hopelessly misconceived and he can turn something ugly into something that everyone admires. He reads the grain of the hand and turns its odd features into something aptly beautiful. Here is an example that the detractors of Precision might cherish, as the auction has given away information that will allow the opponents to defend double dummy.
♠ A852  ♠ Q3  2 ♦ *  2NT 
♥ QJ8  ♥ AK3  3 ♦**  4NT 
♦ 6  ♦ AKQ9743  5 ♥  7 ♦ 
♣ A9872  ♣ 4  Gulp!  
7 losers  4 losers  * short in diamonds,  1115 HCP 
** 4=3=1=5  
Critics of the losing trick count will note that it predicts 13 tricks, but that it is off the mark as there is an inevitable spade loser. They chortle too soon. He who chortles last chortles best. The opening lead is the ♣K. Do you see how one might make 7♦?
Tim, the master carver, looks at this configuration and wonders what can be made of it. There is a chance to utilize the clubs and spades if the player holding the ♠K is subjected to a squeeze in the black suits. How to create an ending that produces that effect? That is where the master excels: he can envision the patterns that others miss.
The ♣A is won and a club is ruffed immediately, an essential move. Both defenders follow with low cards. Six diamonds are cashed to see if the defenders give away something by their discards, but they have been well informed by the auction so there is nothing to be read from their selection. Here is Tim’s visualization of the 4card ending after the ♥A and ♥K have been cashed and all hearts have been accounted for:

North  
♠  Q3  
♥  3  
♦  3  
♣  —  
West  East  
♠  K9  ♠  J107  
♥  —  ♥  —  
♦  —  ♦  —  
♣  J3  ♣  10  
South  
♠  A  
♥  Q  
♦  —  
♣  98 
The ♥3 is led from dummy and West is caught in a trump squeeze, and he knows it! Interchange the EastWest hands and East gets squeezed. The squeeze will be effective when the defender with the ♠K also was dealt 4 or more clubs. The same ending will be effective if the lead were a passive heart or a trump.
Tim has moved a stage further in analysis, as he has written a program that calculates the probability of success of alternative strategies, although probabilities are not critical when there is just one winning option. There are 3 major influences on the probabilities once play commences: 1) the opening lead, (2) the order of play, and (3) the discards. A passive diamond lead can be thought of as the standard trump lead against a grand slam. As it is expected, there is no surprise in that lead and little to be gathered from the implications of that choice. The lead of the ♣K is relevant, especially when declarer has admitted to holding 5 cards in the suit including the ace. The gift of knowledge revealed during the auction is being repaid in part by information from the opening lead.
Let’s assume both defenders have followed to 2 rounds of clubs. Suppose that on the first 2 rounds West has played the ♣K followed by the ♣3 and East has played ♣5 followed by ♣6. We must rule out a lead from ♣K3. Here are the remaining possibilities.
Club Split  5 – 2  4 – 3  4 – 3  3 – 4 
Remnants  KQJ103 – 65  KQJ3 – 1065  KQ103 – J65  KQ3 – JT65 
Plausible Plays  2  2  2  2 
The plausible plays are the same for each remaining combination, provided we assume that East will choose randomly between his 2 low cards, and West will always lead the ♣K and follow on the second round with his low card. There is no need to signal as thanks to the Precision auction both defenders know the distribution fully. In cases where the plausible plays reach equality, vacant place analysis can be used to calculate the probabilities exactly. The situation in the club suit is such that the clubs have been reduced by 2 rounds of play from 7 to 3. Here are the resulting possible distributions:
Original Split  70  61  52  43  34  25  16  07 
One Round Played  50  41  32  23  14  05  
Two Rounds Played  30  21  12  03  
Combinations Remaining  1  3  3  1  
The number of combinations remaining would be the normal weights to be applied to the club splits when there has been no restriction on the way the clubs were dealt or played. However, in the hand being considered, the opening lead has provided information that drastically changes the probability weighting of the various club splits. There is 1 combination remaining from a possible 52 split, 2 from a 43 split and 1 from a 34 split. As the plausible plays are equal, one may apply probability weights equal to the number of combinations remaining in each category.
The club weights can be applied to Tim’s 4card end position where the heart split has been revealed and the spade split is unknown. The diamonds have split 23 and the hearts have been seen to split 43, so the red suits taken together were dealt 6 and 6.
Red Suits  6 – 6  6 – 6  6 – 6 
Clubs  5 – 2  4 – 3  3 – 4 
Spades  2 – 5  3 – 4  4 – 3 
Club Weights  1  2  1 
♠ & ♣ Weights  126  420  210  Total = 756  (100%) 
♠ K in the West  36  180  120  Total = 336  (44.4%) 
♠ K in the East  90  240  90  Total = 420  (55.6%) 
♠ K with 4+ clubs  36  180  90  Total = 306  (40.5%) 
♠ K without 4 clubs  90  240  120  Total = 450  (59.5%) 
The expected number of clubs in the West is 3.89 and the expected number of spades is 3.11, for a total of 7. As the red suits are known to split 66, the blacks must split 77.
A Vacant Place Interpretation
In the 4card ending the odds that the ♠K is in the West is exactly 4 out 9. It is as if there are 9 black cards missing, 2 clubs and 7 spades, yet there are only 8 cards still outstanding. Why? The implications on the opening lead have placed a restriction on the West hand to having been dealt 3, 4, or 5 clubs, and, as we know the reds are split 66, which places a corresponding restriction on the spades to having been dealt 4, 3, or 2 in the West. The West player has followed to 2 rounds of clubs, so the lower bound on the number of clubs in the West has to be satisfied by one of the 3 missing clubs leaving the 2 remaining clubs free to be dealt to either defender in a random fashion. We have filled a vacant place with one of the missing clubs.
Once the clubs are placed, the spades fall into line. The 2 spade discards are not relevant directly. One has seen the heart discards which have enabled an exact count in that suit. The assumption is that the spade discards are ‘free’ and have not altered the probability of the holdings in that suit, because they have not transmitted any significant information, or, rather, because we have chosen to ignore whatever information there was transmitted. There are still enough insignificant spades remaining to satisfy the indirect constraint imposed by way of the club suit.
The Plausibility of Plays
The vacant place interpretation is a consequence of Bayes’ Theorem applied to the play in the club suit wherein one took account of the number of plausible plays. Because after 2 rounds the plausible plays were equal for all remaining possible club combinations, one can use the vacant place argument. The two methods are exactly equivalent: advancing a vacant place argument is the same as claiming the plausible plays are equal.
If we were to alter the probabilities of the club combinations by changing the number of plausible plays, then this nice interpretation would no longer be strictly valid. Consider the permutations in play where West has been dealt: ♣KQJT3. He decides on the second round to play one of the touching honors, giving him a choice of 3 equally plausible plays, and the partnership 6 in all on the 2 rounds. One might also claim that West could have lead a deceptive ♣Q, ♣J or ♣10, giving him 4 possible equivalent leads and 4 possible plays on the second round. Considerations of this kind greatly reduce the probability of the 52 split. We no longer achieve the vacant place probability weights of 121.
A Maximally Likely Guess
Early in the play declarer should consider what is most likely on the basis of the play so far. This approach helps in the visualization and the planning of the play. The initial crude estimate may not be far off if the suits split as evenly as one hopes. Here are the final 3 choices of distribution shown in full with their initial weightings and their true weightings in the 4card ending:
I  II  III  
♠ 2 – 5  ♠ 3 – 4  ♠ 4 – 3  
♥ 4 – 3  ♥ 4 – 3  ♥ 4 – 3  
♦ 2 – 3  ♦ 2 – 3  ♦ 2 3  
♣ 5 – 2  ♣ 4 – 3  ♣ 3 – 4  
Crude Weights  36  100  100 
True Weights  30  100  50 
After the trumps are seen to split 23 declarer could take note that Conditions II and III have equal probability greater than Condition I in the ratio of 100:36 Therefore, it is more probable that the ♠K doesn’t sit with the longer clubs.
♠ K with longer clubs  2 x 36 + 3 x 100 + 4 x 100  = 772  (46.7%)  
♠ K without longer clubs  5 x 36 + 4 x 100 + 3 x 100  = 880  (53.3%)  
On the basis of the opening lead, declarer might judge that Condition II is more probable than Condition III. If declarer were to consider the maximally likely distribution, Condition II, by itself the probability of the ♠K being with 4 clubs would be 42.9%, not a favorable percentage, but one not without hope. The crude initial guess of 3 out of 7 is not far removed from the 4card ending estimate of 4 out of 9 (44.4%).
I am hoping you can give me an email address I can use. I just bought your book on bridge probability and I would like to ask a question/comment. TIA.
I am reachable at rfm65@telus.net
I welcome comments as I feel this is the start of an ongoing process to which many can contribute.