Strategy and Information
Let's consider a common singledummy
problem:
In this layout: 

A J 6 5 

we want to take all four tricks 
? 
N W E S 
? 


K 4 3 2 

Should we play top honors or should we finesse ? – assuming that the
suit is breaking 32:
W E 

W E 
All other breaks make our goal impossible to achieve.
Thus, for the purpose of our analysis, those breaks can be eliminated. 
Qxx–xx 
xxx–Qx 
We are facing a dilemma, since:
–
if the missing cards are divided: Qxx–xx – only finessing
works
–
if the missing cards are divided: xxx–Qx – only playing for the
drop works
and we don't know what the
actual layout is.
Let's
forget the present deal and let's concentrate on working out a rule for the
future, say 100 deals like this one. What gives us a higher ratio of success:
–
always finessing ? or
– always playing from the top ?
On that future 100 deals the missing
cards will be dealt, on average, like this:
xxx–Qx 
1098–Q7 
10 times 
40 times 
Why were all the elementary breaks assigned the same (10%) frequency
? But why not
? Is the 9, at shuffle time, anything different than the 7 or the Queen
? 
1097–Q8 
10 times 

1087–Q9 
10 times 

987–Q10 
10 times 

Qxx–xx 
Q109–87 
10 times 
60 times 

Q108–97 
10 times 

Q107–98 
10 times 

Q98–107 
10 times 

Q97–108 
10 times 

Q87–109 
10 times 
The table above
clearly indicates that:
§
always finessing we we'll succeed 60 times
§
always playing for the drop we we'll succeed 40 times
which means that we should always
finesse.
Let’s check what
happens if we deviate from the advised line and play for the drop from time to
time, say one time in ten:
1098–Q7 
10 times 
1 success 
4 
As you can see we'll succeed only 58 times in 100 deals (58%) while invariably
finessing assures 60%. 
1097–Q8 
10 times 
1 success 

1087–Q9 
10 times 
1 success 

987–Q10 
10 times 
1 success 

Q109–87 
10 times 
9 successes 
54 

Q108–97 
10 times 
9 successes 

Q107–98 
10 times 
9 successes 

Q98–107 
10 times 
9 successes 

Q97–108 
10 times 
9 successes 

Q87–109 
10 times 
9 successes 
There are, however, bridge players who
complain against the calculations presented above arguing that:
Agreed,
it's really true but only before we start playing the suit !
However,
if we play the King (and both opponents follow) and then lead a small card from South, awaiting
a card from West – the result of the calculations will be different.
At
that point there will be only two remaining cards which can be distributed in
two, equally probable, ways: Q–x or x–Q.
Thus,
finessing is as good as playing for the drop (50% for each line).
Our calculation is
significantly better as it takes into account the extra information given by
the fall of three spot cards.
That reasoning,
although superficially correct, is, however, a fallacy and can be countered in
several ways:
1) 
If those two
breaks (Qx, xQ) were equally probable that would mean that at this point of
the play the defenders could detach those two cards (the Queen and the small
card), shuffle them and redistribute again between the two hands. But it was the
entire deck that was shuffled ! 
2) 
The information
we receive watching the small cards played by the defenders doesn't increase
our knowledge about the actual distribution. Provided that (and both sides of the argument agree with that) the
defenders follow suit completely randomly, never displaying any habits (eg
always in ascending order) or regular patterns. It was clear
beforehand (ie before the play started) that at the decisive point of the
play some three small cards will have been revealed. We could equally well
ask West to show us his two small cards and East – to show us his small card,
and our chances of success wouldn't improve. 
3) 
If the point we
have just made seems vague to the reader, we recommend its more emphatic
version: There are thirteen hearts lying on the table (face down), divided into
two groups: in
the smaller one – 3 cards, in the bigger one – 10
cards. Where is the Ace ? We bet on the bigger part, of course, and are willing to accept the odds of 10:3 on. Then, our opponent shows us 9 cards
selected from the bigger stack (of course he doesn’t select the Ace to show,
if it is there). Are
we going to bet on the smaller part (three cards face down) now and accept
the odds of 3:1 on ? 
This kind of
reasoning doesn't nullify our adversaries' argument about making use of surplus
information, though. It's indeed true that :
The
calculation of the chances should take into account
not
only shuffling but also the defenders' plays !
Let's calculate the probabilities again,
this time taking into account the way
defenders play their spot cards and assuming that they do it randomly.
Let's make it: 89–7
( West followed with the 8, and then the 9 – East followed
with the 7 )
If so, the
result of the shuffling was either D98–107 or
1098–D7, and both of them are equally
probable.
Let's analyze 480 deals (that number taken to avoid fractions) featuring those two
distributions (240 for each), along with all possible ways of playing the spot
cards by defenders:
distribution 
discards 


Q98–107 240 times 
98–10 
60 times 
and we'll see that
on those 100 deals where spot cards were played 98–7 (see · in
the table): Q98–107 distribution will occur 60
times 1098–Q7 distribution will occur 40
times Thus, the probabilities remain the same as the original ones:
finessing – 60%, from the top – 40%, which means that randomly revealing
three spot cards by defenders doesn't change anything. 
89–10 
60 times 

98–7 
60 times 

89–7 · 
60 times 

1098–Q7 240 times 
109–7 
40 times 

910–7 
40 times 

108–7 
40 times 

810–7 
40 times 

98–7 
40 times 

89–7 · 
40 times 
Let's check now
what happens if defenders do not adhere to randomness of playing their spot
cards ?
Let’s assume that they always play their
spot cards in ascending order:
distribution 
discards 
success 

Q109–87 
910–7 
1 
We can notice
here an interesting phenomenon: §
In four cases (marked 1) the distribution is fully disclosed. §
In all other cases (marked 1/ 2) we can as well finesse as play for the drop (both chances are equal). eg 89–7 played points towards one of the two equally
probable distributions marked with the letter x. 
Q108–97 
810–7 
1 

Q107–98 
710–8 
1 

Q98–107 
89–7 
1/ 2 x 

Q97–108 
79–8 
1/ 2 y 

Q87–109 
78–9 
1/ 2 z 

1098–Q7 
89–7 
1/ 2 x 

1097–Q8 
79–8 
1/ 2 y 

1087–Q9 
78–9 
1/ 2 z 

987–Q10 
78–10 
1 
As we can see, there
is no constant and uniform strategy in that case. The way we play varies with
the way defenders play they spot cards.
Summing up the number of successes (1) and partial successes (1/2) we obtain
7 successes in 10 possible distributions which means that owing to
nonrandomness of defenders' plays the chance of catching the Queen rose to
70%.
Readers are invited to examine how other “cunning” ways of following
suit affect declarer's chances of success. For example:
1) West, holding xxx, conceals the
smallest card – East plays the
higher from xx.
2) West, holding xxx, conceals the middle card – East
always plays the lower from xx.
3) East plays the higher
from xx (to “suggest” Qx) – West follows suit in ascending order.
¯
Strategy and Information (2)
Let’s start with
the solutions of the exercises we suggested to the readers ending the part one:
variation 1 = 70 %
variation 2 = 80 % !
variation 3 = 90 % !!
Let’s analyze, for
example, the most impressive variation 3:


West follows suit in ascending order; East plays the higher from xx
(to suggest Qx). 
Distribution W – E 
Played W–E 
Probability of success 
Therefore,
declarer’s chance of success is indeed not less than 90% ! As we can see, „cunning” plays by
defenders don’t achieve the desired objective, quite on the contrary – declarer’s chance of success
significantly rises (!), provided he can determine (which is often
possible) the kind of trickiness employed by defenders. 

Q109 
– 
87 
910 
– 
8 
1 


Q108 
– 
97 
810 
– 
9 
1 


Q107 
– 
98 
710 
– 
9 
1 


Q98 
– 
107 
89 
– 
10 
1 


Q97 
– 
108 
79 
– 
10 
1 


Q87 
– 
109 
78 
– 
10 
1/ 2 


1098 
– 
Q7 
89 
– 
7 
1 


1097 
– 
Q8 
79 
– 
8 
1 


1087 
– 
Q9 
78 
– 
9 
1 


987 
– 
Q10 
78 
– 
10 
1/ 2 

The best defenders’ strategy is, therefore, to play randomly insignificant cards, and any
„deceptive” plays should be consider naivety, especially after the publication of this article.
It may happen, however, that even completely random
plays by defenders alter declarer’s probabilities to the extent that he has to
change his line of play. Although it doesn’t apply to the problem of “catching
the Queen missing five cards”, it applies to the following:

S played the Ace: W followed
with the Jack, E – with the three. S crossed to the table and led a
spot – E followed with the two. What now – to finesse or to play
the King ? 
Declarer looked
up the relevant tables and found there the following probabilities:
distribution QJ–32 = 6,78 %
distribution J–Q32 = 6,22 % ,
which would mean that playing from the top is better.
However, the tabular probabilities apply only to
shuffling the entire deck !
To obtain correct results we have to take into account
defenders’ plays. Let’s do it, then, assuming their plays are random:
distribution 
plays 
probability 


QJ–32 
Q–32 Q–23 J–32 · J–23 
1,695 % 1,695 % 1,685 % 1,695 % 
6,78 % 
One can see that: If J–32
appeared the probabilities are as follows: QJ–32 = 1,695 % J–Q32 = 3,110 % 
J–Q32 
J–32 · J–23 
3,110 % 3,110 % 
6,22 % 
Transforming this, so as to the sum of the
probabilities adds up to 100%, we obtain that at the crucial point of the play:
distribution QJ
= 35,28 %
distribution J–Q32=
64,72 % !!!
Since defenders played their cards randomly (thus, in
optimal fashion) we could expect that nothing would change. It turned out to be
the opposite: the probabilities were altered so significantly that the optimal
strategy changed. Finessing on the second round of the suit is best (64.72%),
winning by a huge margin over playing from the top which originally (based on
shuffling alone) might be considered best (52%).
We could arrive at the conclusion that „something will
change” in a different way, reasoning like follows:
If West had QJ, then – assuming he plays his cards randomly – he could
equally well drop the Queen as the Jack. When he drops the Jack (as he did) the
probability of his holding QJ decreases (and at that – exactly by half).
The reasoning outlined above is known in bridge
literature as the
Principle of Restricted Choice (PRC):
Any play chosen by a player makes it more
likely that his choice was restricted,
ie he had not an equally good play as an
alternative.
The wording of the PRC isn’t precise enough, though,
and it can lead us astray if we face a more subtle situation. When in doubt,
it’s always better to conduct a precise tabular analysis.
The problem of correcting the probabilities is too
complex to be enclosed in one universal rule. Let’s content ourselves with the
following general advice:
The calculation of probabilities should
take into account not only the fact that the deck was shuffled but also all
calls and plays made by the opponents and, generally – EVERYTHING THAT HAS
HAPPENED AT THE TABLE !
Yes – everything. For example, the facts established
on occasion of the following deal:
AJxx AQxx 
S plays 7NT on a diamond lead, preceded by a detailed
inquiry about the bidding and a… long huddle by W. The NS bidding revealed strong
fits in both ¨ and ©, so W „had” to lead one of these
suit for safety reasons. 


K10xx KJxx 
Who has the Heart Queen ?
E,
for sure, if we assume that W had reasons to think before making the
opening lead !
The reasoning is simple:
If W had the Heart Queen he wouldn’t be thinking that long before leading.
If we don’t take the huddle into account,
the chance of E having the Queen is 2 / 3.
Check it yourself !
If it turns out that it’s W who has the Queen
after all, don’t make a fool of yourself by complaining about the huddle (“What
were you thinking about?”, “Why were you inquiring about the bidding”)!!! But
that is another issue.







