probabilities, a priori and a posteriori

[Wim Heemskerk]

In the last months there has been some discussion about the handling of
suit combinations in "Bridge", the monthly magazine by the Dutch Bridge
Federation (NBB).
I could use some help on the following combination:

A10xx

Kxx.

What is the optimal strategy for three tricks?

The Encyclopedia of bridge and Roudinesco in his superbe "Dictionary of
suit combinations both give as solution: "Play the King and lead to the Ten
to finesse (unless East plays Queen or Jack). This is slightly better than
playing King and Ace."
I agree with this. The point is that despite of the fact that there are
eight Qx and Jx combinatons in the East hand and only six QJxx combinations
in the West hand (which means that if a 4-2 distribution is certain, you
should play King and Ace), there are four QJxxx combinations. These 5-1
combinations make the finesse a little better. All is well up to this
point, but....
In the last issue of "Bridge" a reader stated that it is not correct to
consider the a priori probabilities. The decision wether to play the Ten or
the Ace has to be made at the moment that East has played one small card
and West has played to small cards. At this moment a lot of the original
possible distributions are ruled out (no more 6-0, no more 5-1 with Q or J
singleton, no more 4-2 with West having Qx or Jx etc.) This reader now
states that the best play is the Ace (and not the Ten).
Since I was sure the man was wrong I began writing down all conditional
probabilities. But my arithmetics say he is right....?? I must be making
some kind of mistake, but I can't find it. I am aware of the fact that the
odds are changing during the play. For example: having xxx opposite AKQ10
it is known that after playing Ace and King and both followed, the odds for
a 3-3 distribution have increased a lot and by now it is better to play for
the drop instead of finessing the Ten.
My problem is: if playing King and Ace is better than playing King and Ten,
what is the use of alle these a priori tables. I mean, the moment of
decision always comes at the second round. I am a little confused and I
will be very grateful to the person who can explain to me why the second
round finesse is better after all!!

Wim Heemskerk
The Netherlands

[bobby goldman]

"Wim Heemskerk" <w.hee...@tip.nl> wrote:

>A10xx

>Kxx.

>Wim Heemskerk
>The Netherlands

Assuming the original (a priori) analysis is correct (don't feel
like doing the math), then the INTERNAL events in that suit have not
changed the conclusion.
I suspect the "reader" who argued for a "current" analysis made the
mistake of not allowing for RESTRICTED CHOICE adjustments.

External events can and do affect the probabalities. Thus a side
suit splitting 6-2 could alter the answer of how you play a suit; but
the appearance of a specific X within the suit does not.
(mathematically at least; might based on psychological assumptions)

Best way to deal with these probabilities is to do a comparison, a
priori, of RELEVANT probabilities....then one doesn't land in the
entanglement of restricted choice adjustments vis-a-vis the play of a
specific card.

Bobby Goldman

[David desJardins]

Wim Heemskerk <w.hee...@tip.nl> writes:
> A10xx
> Kxx.
> What is the optimal strategy for three tricks?

> In the last issue of "Bridge" a reader stated that it is not correct to

> consider the a priori probabilities. The decision wether to play the Ten or
> the Ace has to be made at the moment that East has played one small card
> and West has played to small cards. At this moment a lot of the original
> possible distributions are ruled out (no more 6-0, no more 5-1 with Q or J
> singleton, no more 4-2 with West having Qx or Jx etc.) This reader now
> states that the best play is the Ace (and not the Ten).

But the reader is obviously wrong. All you have to do is write down the
possible West holdings:

(A) No longer possible: QJxxxx Qxxxx Jxxxx xxxx QJ Qx Jx Q J x void
(B) Finesse gains a trick: QJxxx QJxx
(C) Finesse costs a trick: Qxxx Jxxx
(D) Finesse is immaterial: QJx Qxx Jxx xxx xx

A priori, there are the following number of hands of these types
(probabilities computed by dividing each number by the total number of
West hands, 10400600):

(A) 2842400 (27.3%)
(B) 1511640 (14.5%)
(C) 1343680 (12.9%)
(D) 4702880 (45.2%)

A posteriori, hands of type (A) are ruled out, so there are the
following number of hands of the remaining types (probabilities computed
by dividing each number by the total number of West hands consistent
with the play so far, 7558200; the same numbers can also be derived from
Bayes' law):

(B) 1511640 (20.0%)
(C) 1343680 (17.8%)
(D) 4702880 (62.2%)

Clearly, the relative advantage of (B) over (C) doesn't change.

> My problem is: if playing King and Ace is better than playing King and Ten,

> what is the use of all these a priori tables.

The use of the tables is that they tell you the right thing to do
without being confused by readers who write in to magazines with
incorrect analyses. There are literally hundreds of wrong ways to do
the analysis, and it's too time-consuming to individually debunk each
one. I could guess what the unnamed reader did wrong (most likely, the
reader is ignoring one of the effects of restricted choice), but
frankly, I'd rather not spend the time. That's why it's nice to have
published sources to direct such mistaken individuals to.

David desJardins
--
Copyright 1996 David desJardins. Unlimited permission is granted to quote
from this posting for non-commercial use as long as attribution is given.

[Theo Groen]

"Wim Heemskerk" <w.hee...@tip.nl> wrote:

>In the last months there has been some discussion about the handling of
>suit combinations in "Bridge", the monthly magazine by the Dutch Bridge
>Federation (NBB).
>I could use some help on the following combination:
>

>A10xx
>
>Kxx.
>
>What is the optimal strategy for three tricks?
>

>The Encyclopedia of bridge and Roudinesco in his superbe "Dictionary of
>suit combinations both give as solution: "Play the King and lead to the Ten

>to finesse (unless East plays Queen or Jack). This is slightly better than
>playing King and Ace."
...

>My problem is: if playing King and Ace is better than playing King and Ten,

>what is the use of alle these a priori tables. I mean, the moment of
>decision always comes at the second round. I am a little confused and I
>will be very grateful to the person who can explain to me why the second
>round finesse is better after all!!
>
>Wim Heemskerk
>The Netherlands

IMHO I think it works as follwos: during the play information becomes
available. The a priori probabilities give the probability of a
specific distribution of cards, as well as the total number of
permutations possible. Usually that includes east/west hands
reversed.
As information becomes available, a number of combinations are ruled
out and no longer need be considered. Example: When opponents started
with QJ5432, and after 1.5 tricks West has shown the 2 and 3 and East
the 4, then only combinations have become relevant where East holds
the 5. In this particular case we are interested in East starting
with 1 or 2 cards in the suit, so we know the domain we have to
analyze has been reduced to:
QJ32 - 54 (1.61%)
Q532 - J4 (1.61%)
J532 - Q4 (1.61%)
QJ532 - 4 (1.21%)

Combination 2 and 3 ask for the A to be played in trick 2, whereas
combination 1 and 4 ask for the finesse. No restricted choice
principles applied, as Mr. Goldman suggested.

By the way, the ratio drop/finesse is equal to 1.14, which is exactly
the quotient of (12+12)/(9+12). You get to this using a vacant slot
analysis as follows: deal two more cards to West (after all we assume
he has the longer holding in the suit). Then 9 is the number of slots
West will hold to receive the third and last outstanding card
(finesse). 12 is the number of vacant slots East holds (still after
1.5 tricks) to receive the Q, J (drop) and 5 (finesse) respectively.
-- Theo Groen
"Nostalgia is not what it used to be"

[Ramsey]

In article <01bbf29e$26f278c0$LocalHost@default>, Wim Heemskerk
<w.hee...@tip.nl> writes

>I could use some help on the following combination:
>
>A10xx
>Kxx.
>
>What is the optimal strategy for three tricks?
>
>The Encyclopedia of bridge and Roudinesco in his superbe "Dictionary of
>suit combinations both give as solution: "Play the King and lead to the Ten
>to finesse

> I agree with this. but....

>In the last issue of "Bridge" a reader stated that it is not correct to

>consider the a priori probabilities. This reader now

>states that the best play is the Ace (and not the Ten).
>

I agree with the 'reader'. At the point of decision there are 8
combinations of the missing QJx. Of these 4 (x/QJ; Q/Jx; J/Qx; -/QJx)
make no difference as you either can or cannot make 3 tricks whatever
you do. This leaves 4 combinations. With 2 you should finesse (QJx/-
and QJ/x) and the other 2 you should play the Ace (Jx/Q and Qx/J).

So you should finesse if p(QJx+QJ)>p(Jx+Qx) and in the absence of any
other information this can be simplified to finesse if p(QJx)>p(Jx) as
p(QJ)=p(Qx)=p(Jx).

However you 'know' about 4 cards in the hand under the finesse and only
1 card in the other hand so the 6th card in the suit is more likely to
be in the latter hand. So p(Jx)>p(QJx) and playing the Ace is better.

However this is one of those combinations where it is almost
inconceivable that you don't have a good reason for choosing one option
rather than the other. Even if the opponents bidding, lead and play
give absolutely no guide you still have the fact that if you play the
Ace you have to get back to hand to lead the suit again and get back to
dummy to enjoy it. You also have the possible advantage with the
finesse of enjoying your winners without losing a trick. And finally
you also have the guide of which hand you prefer to lose a trick to.

Somewhere in all that there will be a reason to play one way or the
other regardless of the exact odds. My guess is that because the
finesse is so much more flexible and gives you a chance of 3 tricks
without a loser it will normally be the 'correct' play.
--
Ramsey
sjri...@sjrindex.demon.co.uk

[Ramsey]

In article <r887mm4...@aleph.CS.Princeton.EDU>, David desJardins
<de...@CS.Princeton.EDU> writes

>Wim Heemskerk <w.hee...@tip.nl> writes:
>> A10xx
>> Kxx.
>> What is the optimal strategy for three tricks?
>
>

>But the reader is obviously wrong. All you have to do is write down the
>possible West holdings:

I am always suspicious when someone says 'obviously' :)

>
>(A) No longer possible: QJxxxx Qxxxx Jxxxx xxxx QJ Qx Jx Q J x void
>(B) Finesse gains a trick: QJxxx QJxx
>(C) Finesse costs a trick: Qxxx Jxxx
>(D) Finesse is immaterial: QJx Qxx Jxx xxx xx
>
>A priori, there are the following number of hands of these types
>(probabilities computed by dividing each number by the total number of
>West hands, 10400600):
>
>(A) 2842400 (27.3%)
>(B) 1511640 (14.5%)
>(C) 1343680 (12.9%)
>(D) 4702880 (45.2%)
>
>A posteriori, hands of type (A) are ruled out, so there are the
>following number of hands of the remaining types (probabilities computed
>by dividing each number by the total number of West hands consistent
>with the play so far, 7558200;
>

>> My problem is: if playing King and Ace is better than playing King and Ten,

>> what is the use of all these a priori tables.
>
>The use of the tables is that they tell you the right thing to do
>without being confused by readers who write in to magazines with
>incorrect analyses. There are literally hundreds of wrong ways to do
>the analysis, and it's too time-consuming to individually debunk each
>one.

One of the ways to do it wrong is to treat the x's as equals and I
suspect that is what you have done :). At the decision point you have
seen 3 specific small cards and they have to be specifically taken into
account.

Theo Green, in a parallel post, gives the figures as:

QJ32 - 54 1.61%
Q532 - J4 1.61%
J532 - Q4 1.61%
QJ532 - 4 1.21%

If you treat the x's as equals then there are 6 combinations of line 1
to 4 on each of the other lines. For simplicity multiply line 1 by 1.5
to get 2.41%. The finesse is best for lines 1 and 4 (3.625%) and the
Ace best for lines 2 & 3 (3.22%). The ratio in favour of playing the
finesse is 3.62:3.22 = 1.124 which is the same as your 14.5/12.9.

Am I right?

Incidentally, both Theo Green & I in my original post failed to take
account of restricted choice. If East holds 54 you could argue that he
could have played either card at trick 1 whereas with the other 3
holdings the 4 had to be played. So line 1 should be reduced by half to
0.81% making the play of the Ace an even better bet.

Now, where did I put the aspirin?
--
Ramsey
sjri...@sjrindex.demon.co.uk

[David desJardins]

Wim Heemskerk <w.hee...@tip.nl> writes:
> A10xx
> Kxx.
> What is the optimal strategy for three tricks?

> In the last issue of "Bridge" a reader stated that it is not correct to

> consider the a priori probabilities. The decision wether to play the Ten or
> the Ace has to be made at the moment that East has played one small card
> and West has played to small cards. At this moment a lot of the original
> possible distributions are ruled out (no more 6-0, no more 5-1 with Q or J

> singleton, no more 4-2 with West having Qx or Jx etc.) This reader now

> states that the best play is the Ace (and not the Ten).

But the reader is obviously wrong. All you have to do is write down the
possible West holdings:

(A) No longer possible: QJxxxx Qxxxx Jxxxx xxxx QJ Qx Jx Q J x void

(B) Finesse gains a trick: QJxxx QJxx
(C) Finesse costs a trick: Qxxx Jxxx
(D) Finesse is immaterial: QJx Qxx Jxx xxx xx

A priori, there are the following number of hands of these types
(probabilities computed by dividing each number by the total number of
West hands, 10400600):

(A) 2842400 (27.3%)
(B) 1511640 (14.5%)
(C) 1343680 (12.9%)
(D) 4702880 (45.2%)

A posteriori, hands of type (A) are ruled out, so there are the
following number of hands of the remaining types (probabilities computed
by dividing each number by the total number of West hands consistent

with the play so far, 7558200; the same numbers can also be derived from
Bayes' law):

(B) 1511640 (20.0%)
(C) 1343680 (17.8%)
(D) 4702880 (62.2%)

Clearly, the relative advantage of (B) over (C) doesn't change.

> My problem is: if playing King and Ace is better than playing King and Ten,

> what is the use of all these a priori tables.

The use of the tables is that they tell you the right thing to do
without being confused by readers who write in to magazines with
incorrect analyses. There are literally hundreds of wrong ways to do
the analysis, and it's too time-consuming to individually debunk each

[Nick Straguzzi]

Ramsey <sjri...@sjrindex.demon.co.uk> wrote:

>In article <r887mm4...@aleph.CS.Princeton.EDU>, David desJardins
><de...@CS.Princeton.EDU> writes

>>Wim Heemskerk <w.hee...@tip.nl> writes:
>>> A10xx
>>> Kxx.
>>> What is the optimal strategy for three tricks?
>>
>>

>>But the reader is obviously wrong. All you have to do is write down the
>>possible West holdings:
>

>I am always suspicious when someone says 'obviously' :)

In this case though, that 'someone' is right. The odds have not
changed, and the finesse is still a better bet.

>One of the ways to do it wrong is to treat the x's as equals and I
>suspect that is what you have done :). At the decision point you have
>seen 3 specific small cards and they have to be specifically taken into
>account.
>

>If you treat the x's as equals then there are 6 combinations of line 1
>to 4 on each of the other lines. For simplicity multiply line 1 by 1.5
>to get 2.41%. The finesse is best for lines 1 and 4 (3.625%) and the
>Ace best for lines 2 & 3 (3.22%). The ratio in favour of playing the
>finesse is 3.62:3.22 = 1.124 which is the same as your 14.5/12.9.
>
>Am I right?

No. The x's *are* all equal, at least under the conditions of the
problem.

If you know your opponents' signaling tendencies, then perhaps you can
draw some inferences on the fall of the spot cards. For example, if
you know LHO will *always* play up the line with QJ653, and he drops
the three and then the six on the first two rounds, then you can rule
out this holding and adjust your calculations accordingly. And, since
the finesse is only a very slight favorite to begin with, it doesn't
take much inferencing to swing the odds in favor of the drop.

But that's not how the problem was posed. With QJ653, the 6, 5, and 3
are effectively equals and can be played at random. And, if LHO is
truly playing his x's at random, the finesse is a ~2% favorite.

The clearest and most straightforward way to prove this is to do what
David did: enumerate all the possibilities.

A harder way, but still reasonable, is to calculate the odds using
restricted choice. Just remember that both LHO *and* RHO are subject
to RC in how they choose to play their spot cards. That's the fallacy
here:

>Incidentally, both Theo Green & I in my original post failed to take
>account of restricted choice. If East holds 54 you could argue that he
>could have played either card at trick 1 whereas with the other 3
>holdings the 4 had to be played. So line 1 should be reduced by half to
>0.81% making the play of the Ace an even better bet.

Right, but if West had 632 he had THREE ways to choose two cards to
play to the first two tricks. The math can get a bit complicated, but
the result is the same: with nothing else to go on, finesse.

The bottom line is when you're dealing with spot cards, restricted
choice cuts both ways and tends to cancel itself out.

>Now, where did I put the aspirin?
>--
>Ramsey
>sjri...@sjrindex.demon.co.uk

Nick

--
Nick Straguzzi |
Mullica Hill, NJ | "Nice fresh chicken, too."
str...@voicenet.com | - Shemp Howard

[Theo Groen]

Theo....@inframatica.nl (Theo Groen) wrote:

>"Wim Heemskerk" <w.hee...@tip.nl> wrote:
>
>>In the last months there has been some discussion about the handling of
>>suit combinations in "Bridge", the monthly magazine by the Dutch Bridge
>>Federation (NBB).

>>I could use some help on the following combination:
>>

>>A10xx
>>
>>Kxx.
>>
>>What is the optimal strategy for three tricks?

While time went on and more postings came in, I have been looking for
the union of the approach using the a priori probabilities and actual
probabilities. (Using a lot of decimals make the numbers come to
live.) Applying restricted choice (or free choice as some appear to
call it) I came to the following reasoning, and I hope it makes sense:

1.) Watching the 5, 4 appear from West and the 2 appear East from
missing QJ5432: who has the missing 3?
Finesse wins:
a) QJ543 vs 2; Probability = 1.211180; After 1.5 trick West has shown
2 out of 3 equivalent small, therefore divide by 3 --> P'=0.403727
b) QJ54 vs 32; P = 1.614907; East has shown 1 out of 2 eqv small,
therefore divide by 2 --> P'=0.807453
Odds in favor of finesse: 1.211180 (!)
Drop wins:
Q543 vs J2; P = 1.614907; East has shown 2 out 3; div. by 3 -->
P'=0.538302
J543 vs Q2; P = ditto.
Odds in favor of drop: 1.076605.
Finesse/drop = 1.125000 (!)

2.) Watching XX'es and x'es, using a priori percentages
Finesse wins:
XXxxx vs x: P = 4 x 1.211180 = 4.844720
XXxx vs xx: P = 6 x 1.614907 = 9.689442
Odds in favor of finesse: 14.534162
Drop wins:
Xxxx vs Xx: P = 8 x 1.614907 = 12.919256
Finess/drop = 1.125000
QED?

[Ramsey]

The finesse is better than the Ace.

I concede, gracefully.
--
Ramsey
sjri...@sjrindex.demon.co.uk

[David Stevenson]

Ramsey wrote:

>I am always suspicious when someone says 'obviously' :)

Yes, I must remember that. It is obviously reasonable. :)

--
David Stevenson da...@blakjak.demon.co.uk
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