♠ ♥ So, do you know how to play certain suit combinations? ♦ ♣
On Thursday, 5th January 2017, Board 23 was
played at nine tables. At one table the contract was
3NT by West. The defence could have taken the first
five tricks but perhaps the heart suit became
blocked. But how declarer subsequently scored a
ninth seems difficult to comprehend!
At another table South played in a contract of 4♥x
and scored a bottom for aggressive bidding but
perhaps could have saved a trick in the play and
saved a few match-points.
At all other tables the contract was played in
spades. What is interesting, and relates to the title
of this article, is the varying number of tricks made.
Declarers made 9, 10, 11 or 12 tricks and I suspect
that the chief determining factor was the play of the
trump suit.
The specific combination in this instance involves xx
opposite AJ9xxx but also applies to similar
combinations, e.g. xx – AJ9, xxx – AJ9, xx – AJ9x,
etc. where the total number of cards held by
declarer is eight or less.
There are two legitimate lines, i.e. finesse the J on
the first round or finesse the 9 on the first round
and if it loses to the K or Q then finesse the J on the
second round.
So, which line offers the better chance of success?
1) Lead small towards AJ9(x) and play for KQx underneath the AJ9. If the defender plays low then
finesse the J.
2) Lead small towards AJ9 and if the first defender plays low then play 9 from dummy and hope that
the second defender holds Kx(x) or Qx(x), i.e. does NOT hold the 10.
This is a question of ‘odds/percentages’. And players keen to improve their game should be familiar
with the more common suit combinations and the associated odds.
For a simple finesse to be successful, e.g. (a) xx – Kx or (b) xx – AQ, the presumption is that the key
missing card is located in a specific defender’s hand. In the two above examples we assume that in (a)
the Ace is under the Kx whereas in (b) we assume that the K is under the AQ.
The odds in these instances are quite simple…50% of the time the missing key card will be held by one
defender whereas for the remaining 50% of times it will be with the other defender. So, our chances of
success are 50%.
Now, things change when we are missing TWO key cards but again the maths is quite simple. For
example, if we need to make two tricks from the following combinations what are the odds, e.g. (c) xx
– AJ10 or (d) xx - AQ10
In each case we are missing two key cards. In (c) the defenders hold the K and Q whereas in (d) they
hold K and J.
Yet, the odds of making two tricks are the same for both combinations as long as we lead a small card
towards the honour combination and finesse 10 on the first round of the suit. If the 10 loses to the K
or Q in (c) or to the J in (d), we then take another finesse on the second round.
And so, what are the odds of making two tricks? In both cases the odds are approx. 75%. Why so?
Simple! In each case all that is required is that the defender UNDERNEATH the honour holding has
EITHER of the two missing key cards.
So, we know that there is a 50% chance that they will have one of the two missing cards and if that is
so then we will score two tricks. BUT our odds improve because in the 50% of times that the defender
does NOT hold the first key card there is a 50% chance, of the remaining 50 times in 100, that they
will hold the other missing key card.
Thus in (c) if we finesse the 10 on the first round and it loses to the K or Q there is still a 50% chance
that they will hold the other key card. Thus, in percentage terms, we have 50% that they will hold one
of the two missing card and an additional 25%, i.e. 50% of the remaining 50 times in 100 which
amounts to 25% of the original 100 times that they will have the second missing key card, This gives
us a total 75% chance that the defender under the honour combination will hold ONE of the two key
cards. And that is all that is required to make two tricks.
In (d) the odds are exactly the same, i.e. we simply require that the defender under the honour
combination holds either the K or the J…in other words, one of the two missing key cards.
This knowledge can also be used when playing various other combinations.
So, now we come to xx – AJ9xxx or the other variations mentioned above. What changes slightly is
that there are now THREE missing key cards that we can take into our calculations.
Playing the defender under the honour combination to hold BOTH missing honours amounts to a mere
25%, i.e. we already know that at least 50% of the time the defender sitting OVER the AJ9
combination will hold at least the K or the Q. So, on those 50 times finessing the J on the first round
will lose to the K or Q. And half of the remaining 50 times that same defender will have the other
missing honour. Thus, in only 25 times in a 100 will the defender under the honour combination hold
BOTH K and Q.
So, let’s add the 10 to the equation. Now, all that is required is for the defender UNDER the honour
combination to hold the 10 (50% chance) and either the K OR Q (75%). This amounts to a total
chance of 37.5% which is a huge improvement over the 25% chance of the defender holding K and Q.
So, now you know. The best line to lose only one trick with a combination of xxx – AJ9(x), xxx –
AJ9xx, xx – AJ9xxx or similar variations, is to finesse the 9 on the first round of the play of the suit and
if this loses to the K or Q then finesse the J on the next round. And obviously we also need a
favourable break in the suit! In the case of board #23 last night the suit also needs to break 3-2 in
addition to finding the favourable layout of the missing key cards in order to limit losses to one trick in
the suit.
And a final warning re odds/percentages…they change if other factors need to be taken into account,
e.g. if one defender bids thus showing high card strength or a distributional hand. And of course they
are simply odds. And odds do not come with any guarantee. But, play with them, and over time you
will finish ahead of the posse!
Be aware…but use with care!
Paul J Scannell,
January 6th, 2017
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