It has been a long and rough road since last year I started a series of articles in BW, to arrive at general acceptance of math principles about patterns odds, at least until now to the ones referring to percentage play when declarer knows the fall of one and two suits, which are already accepted by top experts and ex-world champions as correct evaluation at the beginning and the middle of the play. The same happened with the law of balanced distributions in percentage play as it is in complimentary relation with pattern odds.
Now I want to comment on pattern distributions but taking in consideration only the 4th suit (the one with the opposed count than the pattern), as it represents the fundament in discovering the actual count of a hidden hand under distributional concepts that are not well known in deep.
About more than twenty years ago, when learning that pattern´s odds have a 75% chance of success when assigning distribution to one hidden hand, emerge a question in my mind about the odds for having same success in both hidden hands at the same time, and my first though was to follow math principles also: “the chance to find the odds for two separated events to exist at the same time equals the product of the two individual percentages”. This number says 75%*75%=56.25%, a nice result as it is bigger than 50% (the odds for a finesse), but I wanted more to know how this expectance can be explained in bridge terms.
I analyzed at that time hundreds of hands at top level tournaments finding that this numbers were approximately right, with eight possibilities of finding the 4th suit distributed between defender’s hands. In an average of six of them (75% of hands) the 4th suit is placed in different colors, and in a 25% of the hands both defenders share the same suit as their 4th. Suddenly the light was on: 75%*50%+25%*75% equals 56.25%.
The rest was easy and the complementary principles to patterns laws were at mi sight: whenever the player is looking different positions for the 4th suits in the two seen hands, there is only a 50% chance of assuming patterns to both hidden hands, but whenever the player is looking at the same suit as the 4th in the seen hands, he can assign patterns to both hidden hands with the same 75% chance to be right. The bridge schemes of the two kinds of distributions look like the following:
a) Distributional Scheme (DS26) = Odd-Even-Even-Odd (O-E-E-O)
♠ XXXX ♠ XXX
♥ XXXX ♥ XXXX
♦ XXX ♦ XXX
♣ XX ♣ XXX
b) Distributional Scheme (DS26) = Odd-Odd-Odd-Odd (O-O-O-O)
♠ XXXX ♠ XXX
♥ XXXX ♥ XXX
♦ XXX ♦ XXXX
♣ XX ♣ XXX
In the a) scheme of the seen hands, the 4th suit is placed in two different suits, hearts and diamonds, a typical condition for assigning a certain pattern to only one of the hidden hands (any one of them) with the odds favorable in relation 3/1. In the b) scheme the 4th suit is shared by the two hands, diamonds, where all other suits maintain similar type of count, the one belonging to the pattern of each hand, also a typical condition to assign patterns to both hidden hands at the same time.
Another way to look at this two rules is the one I suggest to follow when playing at the table: whenever the sums of the cards of two seen hands, suit by suit, give a DS26 of two even results and two odd ones, the player can assign a certain pattern to only one of the hidden hands with a 75% chance of success, but when he obtains a DS26 of four results of the same type, odd or even, can assign patterns to both hidden hands simultaneously, with the same 75% chance of being right.
It is important to explain, for everybody to understand the concept, that when dummy is placed at the table each active player in the hand is looking 26 cards with his own Distributional Scheme, different from the other two but interrelated between them, allowing that the aforementioned rules were accomplished the same for declarer as for each one of the defenders.
AFFINITY OF SUITS IN TWO SEEN HANDS
Studying the positions of the 4th suit in the two seen hands lead to the most important rule to follow in trying to discover in full the distributions: when you find the 4th suit in one of the hidden hands, is mandatory that the other will have his 4th suit placed in a “related suit”. I call “related suit” when dummy is placed at the table, to the one with a sum of seen cards of the same type, odd or even. For example if declarer is looking 5♠, 8♥, 7♦ and 6♣ between his hand and dummy, spades and diamonds are related suits (odd sums of cards) in the same way as hearts and clubs are also related (even sums of cards). Consequently the 4th suits of the hidden hands will be placed in spades and diamonds or hearts and clubs, but never in a mixed way.
When the player is looking four sums of cards of the same type, odd or even, the four suits are related and consequently the 4th suit of the hidden hands will be placed in any one of them, sharing the position in the same suit.
This knowledge in the play of the hand may have extraordinary consequences in the count of a hidden hand, primarily for the defender’s side but also for declarer. As a simple example let’s suppose that declarer opened the bidding with a weak two of six cards in a major and defenders assign to him a percentage distribution of 6-3-2-2 as the likely. At the time in which one of the defenders mark the position of his 4th suit, declarer’s actual distribution is an open book to his partner, as he will know for sure where is positioned declarer´s three card suit and act in coherence in the rest of the play.
The time for talking about pattern’s play in defense still has to wait in my articles, but some examples of declarer’s play I hope will clarify the two conditions:
In one of the initial rounds in the Vanderbilt Cup 2016 transmitted in BBO, this apparently innocent hand appeared in the monitor (N/S Vulnerable, hands rotated). After receiving the ♦6 lead, fourth card in West longest suit, declarer played small from dummy (maybe better if he plays the ♦J to force the Ace and aisle defender’s hands) and the ♦10 forced early the ♦K from hand, following a heart to the Ace in dummy and the ♣Q to the finesse at third trick, East playing the ♣5 and West the ♣4. Your play…?
It looks as if East was leading diamonds with five cards, as North may have played the Ace and returned small if having four. The sum of cards suit by suit (DS26) says E-O-O-E, so declarer can assign a certain pattern to only one of the two hidden hands with a 75% chance of success, and the 4th suits will be placed in ♠/♣ or ♥/♦ as the related suits.
The position after 3 tricks is one of the frequent and tough suit combinations where declarer with eight cards is missing K and 10. The percentage play says North should follow with the ♣J but declarer played small in view of the bidding (he “must” be short in clubs if doubled 1♣) and lost the contract when the ♣10 was doubleton in West hand. Playing patterns and percentages declarer should aisle West and assign to him an odd pattern, with the distribution 5-3-3-2 as the likely, and, looking at six cards in spades, seven in hearts and eight in clubs, place the two cards addend in clubs for having the better chances, which goes perfectly according to the rule of related suits with East having also an odd pattern with the 4th suit located in spades (his four carder).
Now let´s look an example of the second type from the Spingold Cup 2016:
The contract and the lead were the same at the other table, with a transfer bid from South to 3♦ before the bid of 3NT. East won the lead with the ♥Q to continue with the ♥K where the two Wests discarded different. At this table the ♠3 was played and East shift to the ♣10 was the response, applying the heat immediately. Declarer won with the ♣K in dummy and returned the diamond at fourth trick, feeling to be squeezed if he cashed the ♠A and hoping for a defensive error or a blockage in clubs to succeed, none of which was produced in the play losing the contract against a perfect defense.
Playing by the patterns and percentages, the DS26 of the seen hands says E-E-E-E and declarer can assign an even pattern to East and an odd one to West simultaneously, as he knows both hands share the 4th suit. The likely distributions are 6-3-2-2 from East and 1-5-4-3 from West where the addends 5 and 2 or 4 and 3 may be placed equally in spades or clubs as declarer is looking at six cards in both suits. If the fall of the spade suit is 5-2, West chance to have the ♠K is two and a half times better than East (71%) and if the fall is 4-3 the odds decrease to a 57% in favor of West hand but still with better chances to be looking at the ♠K in his hand. I think that declarer should have mediate both chances (a 64% in his favor) and play the ♠A at the fourth trick, discarding a heart, before shifting to diamonds, if he wants to stay with the odds and give himself a chance to win the contract.
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