Game Theory

[Gareth Birdsall]

Has anyone considered a problem such as the following -
how do you defend the following 7NT contract?

2
765432
AK32
32

A
AKQJ1098
Q54
AQ
7NT by South. Lead: DJ.

Declarer's best line is fairly clear. Win DA, play D to Q,
then run your 8 major suit winners discarding a club from
dummy. Then play a diamond to see if they break.

With the lead in dummy:
-
-
3
3

-
-
-
AQ
If East started with the diamond length, then we must play a
club to the ace (show-up squeeze) since East is known to hold
at most one club with his remaining diamond. So declarer
only has a choice of plays if West holds the diamond length
- and for simplicity assume he started with 4.

Declarer can either play West to have started with the CK
(which will now drop, West having been squeezed), or play
East for the CK by finessing.
If we disregard the fact that we have seen some discards then
the finesse is better than the drop. We can think of this in
terms of empty spaces: West has 4 diamonds, so 9 empty
spaces. East has 2 diamonds, so 11 empty spaces. So East is
11:9 more likely to hold the CK.

But declarer has at least a partial count of the hand and
perhaps he can do better by using this information.

The defenders have a couple of options:
i) Discard all cards in the major suits keeping 1 diamond
and 3 clubs in the above end position. This gives declarer a
complete count of the hand.
ii) Keep 1 diamond, 2 clubs and 1 spade in the end position.
Declarer only gets a partial count of the hand (e.g. if West
has discarded 5 clubs then he knows West started with either
xxxxx or Kxxxxx clubs).

Should the defenders always follow strategy i) or ii)? Or
should they follow a mixed strategy (e.g. when West holds
Kxxxxx they play i) with probability p, and they play ii)
with probability 1-p)?
And what values of p should they choose for each holding West
could have started with?

It's easy to analyse if the defenders always follow strategy
i).
Declarer always gets a complete count of the hand so he should play
the defender with the longer clubs to hold the King. E.g. if
West started with 6 clubs, the odds are 6:3 in favour of the
drop.

Suppose instead the defenders always follow strategy ii).
Now declarer always gets a partial count of the hand (e.g.
knows West started with either xxxxx or Kxxxxx clubs).
It's easiest to rationalise the best play in terms of empty
spaces.
If West plays n low clubs, then he has shown up with 4
diamonds and n clubs: 9-n empty spaces. East has 2 diamonds
and 8-n low clubs: 3+n empty spaces.
So if West plays just 0, 1, or 2 clubs then play for the drop
since West has more empty spaces than East. If West plays 3
clubs it's a guess, and if West plays 4 or more clubs, finesse.

So when West holds K, Kx or Kxx, the defence might do better
to give declarer a complete count of the hand (but can't
always do this or declarer will be tipped of as to why he has
been given a complete count).
So we can see that for each holding, h, that West can hold,
the defence should keep a spade with probability p_h and give
declarer a complete count with probability 1-p_h.
So what are the best (for the defenders) choices of p_h for
each holding, h, that West can start with?

And can we choose our p_h's carefully so that declarer can do
no better than just always taking the finesse regardless of
our discards?

Cheers
Gareth

[Eric Schwartz]

Gareth Birdsall wrote:

> Has anyone considered a problem such as the following -
> how do you defend the following 7NT contract?
>
> 2
> 765432
> AK32
> 32
>
> A
> AKQJ1098
> Q54
> AQ
> 7NT by South. Lead: DJ.
>
> Declarer's best line is fairly clear. Win DA, play D to Q,
> then run your 8 major suit winners discarding a club from
> dummy. Then play a diamond to see if they break.

I'm really not capable of following the mathematical argument that
follows in detail, but it seems to me that there is one important point
that has been overlooked:
how do the defenders know to discard all their spades without exchanging
informative signals? Doesn't that affect declarer's chances? (Declarer
might need to leave the diamond suit intact to maximize the defense's
problems.)

Eric

[chudh]

I posted a similar problem a year ago, and I stated
that the defenders should keep a "13th" irrelevant
card to foul up declarer's odds.

This is a good hand.

[jurg...@verizon.net]

On Mon, 04 Mar 2002 21:31:47 GMT, chudh <chud...@worldnet.att.net>
wrote:

>I posted a similar problem a year ago, and I stated
>that the defenders should keep a "13th" irrelevant
>card to foul up declarer's odds.
>

> This is a good hand.

Yes, it is. And another brilliant idea to help those declarers who
can't count.
West holding the long Diamonds and the cK must keep an 'irrelevant'
Spade!
And if East holds the cK he must come down to the King and an
irrelevant Spade!
Why not just pitch the cK? That's much simpler.

[David desJardins]

Gareth Birdsall <gj...@statslab.cam.ac.uk> writes:
> Has anyone considered a problem such as the following -
> how do you defend the following 7NT contract?
>
> 2
> 765432
> AK32
> 32
>
> A
> AKQJ1098
> Q54
> AQ
>
> 7NT by South. Lead: DJ.

You can solve this exactly, assuming double-dummy defense (and assuming
that down 2 and down 1 are equally good for the defense). But it's
pretty messy, and it's not clear how relevant the solution is in the
real world.

As you say, the only relevant case is when West is long in diamonds.
But declarer's and defender's strategies will depend on the actual
diamond split, which declarer will discover during play. For
simplicity, I'll consider only the case where West has exactly four
diamonds, but the other cases can be analyzed in exactly the same way.

There are 20 subcases, depending on how many clubs West has (0 to 9),
and whether West has the club king (Y or N), with the following relative
frequencies:

Y(0) 0 N(0) 55
Y(1) 165 N(1) 1320
Y(2) 2640 N(2) 9240
Y(3) 12936 N(3) 25872
Y(4) 25872 N(4) 32340
Y(5) 23100 N(5) 18480
Y(6) 9240 N(6) 4620
Y(7) 1540 N(7) 440
Y(8) 88 N(8) 11
Y(9) 1 N(9) 0

Two subcases are impossible: Y(0) and N(9). In case Y(9), West shows
out on the spade ace, so the squeeze is marked. In case N(8), East's
club king pops up at trick 12. In the remaining cases, declarer has a
possible guess.

In case Y(k), West has the club king and k-1 small clubs. West's last
two cards must be the club king and a diamond. East can keep two small
clubs, or a small club and a spade. Suppose that East keeps a spade
with probability y(k)/Y(k). (Of necessity, y(8) = Y(8), since East only
has one club to keep.)

Similarly, in case N(k), West has k small clubs. East's last two cards
must be the club king and a small club. West can keep a diamond and a
club, or a diamond and a spade. Suppose that West keeps a spade with
probability n(k)/N(k). (Of necessity, n(0) = N(0), since West can't
keep a club if West isn't dealt any, and n(8) = 0, since West can't keep
a spade if West is only dealt one.)

Now, suppose that declarer reaches trick 12 and there is a spade still
outstanding. Also suppose that West has discarded k clubs. Then there
are two possibilities:

West has the club king: y(k+1) chances
East has the club king: n(k) chances

Declarer will choose whichever is larger. Similarly, if no spades are
outstanding, then there are two possibilities:

West has the club king: Y(k+1)-y(k+1) chances
East has the club king: N(k+1)-n(k+1) chances

Again, declarer will choose whichever is larger.

So, the defenders want to choose y(k) and n(k) to minimize the following
expression, which gives the number of cases in which declarer succeeds:

P = max(y(1), n(0))
+ max(y(2), n(1))
+ max(y(3), n(2))
+ max(y(4), n(3))
+ max(y(5), n(4))
+ max(y(6), n(5))
+ max(y(7), n(6))
+ max(y(8), n(7))
+ max(Y(1)-y(1), N(1)-n(1))
+ max(Y(2)-y(2), N(2)-n(2))
+ max(Y(3)-y(3), N(3)-n(3))
+ max(Y(4)-y(4), N(4)-n(4))
+ max(Y(5)-y(5), N(5)-n(5))
+ max(Y(6)-y(6), N(6)-n(6))
+ max(Y(7)-y(7), N(7)-n(7))
+ max(Y(8)-y(8), N(8)-n(8))
+ Y(9)

The default strategy of finessing for the club king (except in case Y(9)
when East shows out) succeeds with probability N(0)+N(1)+N(2)+N(3)+N(4)+
N(5)+N(6)+N(7)+N(8)+Y(9), which is the same as choosing the second term
in each max() in the previous formula. The defenders would like to
choose the y(k) and n(k) so the declarer can't do any better.

That means satisfying the following simultaneous inequalities:

y(1) <= n(0)
y(2) <= n(1)
y(3) <= n(2)
y(4) <= n(3)
y(5) <= n(4)
y(6) <= n(5)
y(7) <= n(6)
y(8) <= n(7)
Y(1)-y(1) <= N(1)-n(1)
Y(2)-y(2) <= N(2)-n(2)
Y(3)-y(3) <= N(3)-n(3)
Y(4)-y(4) <= N(4)-n(4)
Y(5)-y(5) <= N(5)-n(5)
Y(6)-y(6) <= N(6)-n(6)
Y(7)-y(7) <= N(7)-n(7)
Y(8)-y(8) <= N(8)-n(8)

Rearranging, and plugging in actual values for Y(k), N(k), n(0), n(8),
and y(8), gives:

n(1)-1155 <= y(1) <= 55
n(2)-6600 <= y(2) <= n(1)
n(3)-12936 <= y(3) <= n(2)
n(4)-6468 <= y(4) <= n(3)
n(5)+4620 <= y(5) <= n(4)
n(6)+4620 <= y(6) <= n(5)
n(7)+1100 <= y(7) <= n(6)
0+77 <= 88 <= n(7)

Since each y(k) only appears once, we can choose any y(k) between these
upper and lower bounds, as long as there is some valid choice in the
range [0,Y(k)]. Also, each n(k) must be in the interval [0,N(k)]. This
gives the following necessary and sufficient constraints on the n(k) for
a solution to exist:

n(1)-1155 <= 55
n(2)-6600 <= n(1)
n(3)-12936 <= n(2)
n(4)-6468 <= n(3)
n(5)+4620 <= n(4)
n(6)+4620 <= n(5)
n(7)+1100 <= n(6)
0+77 <= n(7)

n(1)-1155 <= 165
n(2)-6600 <= 2640
n(3)-12936 <= 12936
n(4)-6468 <= 25872
n(5)+4620 <= 23100
n(6)+4620 <= 9240
n(7)+1100 <= 1540
0+77 <= 88

n(1) <= 1320
n(2) <= 9240
n(3) <= 25872
n(4) <= 32340
n(5) <= 18480
n(6) <= 4620
n(7) <= 440
0 <= 11

Note that the second set of constraints and the third set come out to be
equivalent, which makes things a bit easier. Simplifying again:

n(1) <= 1210
n(2) <= n(1)+6600
n(3) <= n(2)+12936
n(4) <= n(3)+6468
n(5) <= n(4)-4620
n(6) <= n(5)-4620
n(7) <= n(6)-1100
77 <= n(7)

n(1) <= 1320
n(2) <= 9240
n(3) <= 25872
n(4) <= 32340
n(5) <= 18480
n(6) <= 4620
n(7) <= 440

Now we can solve these by maximizing each n(k) in turn:

n(1) = 1210
n(2) = 7810
n(3) = 20746
n(4) = 27214
n(5) = 18480
n(6) = 4620
n(7) = 440

The corresponding constraints on y(k), for these choices of n(k), are:

55 <= y(1) <= 55
1210 <= y(2) <= 1210
7810 <= y(3) <= 7810
20746 <= y(4) <= 20746
23100 <= y(5) <= 27214
9240 <= y(6) <= 18480
1540 <= y(7) <= 4620
77 <= 88 <= 440

So an optimal strategy for the defense, when West has exactly four
diamonds, is for the player without the club king to keep a spade with
the following probability:

Y(0) impossible N(0) no choice
Y(1) 55/ 165 = 33% N(1) 1210/ 1320 = 92%
Y(2) 1210/ 2640 = 46% N(2) 7810/ 9240 = 85%
Y(3) 7810/12936 = 60% N(3) 20746/25872 = 80%
Y(4) 20746/25872 = 80% N(4) 27214/32340 = 84%
Y(5) 23100/23100 = 100% N(5) 18480/18480 = 100%
Y(6) 9240/ 9240 = 100% N(6) 4620/ 4620 = 100%
Y(7) 1540/ 1540 = 100% N(7) 440/ 440 = 100%
Y(8) no choice N(8) no choice
Y(9) irrelevant N(9) impossible

This defense ensures that declarer can do no better than to finesse
(except when East shows out).

I don't know if there's a simpler description of an optimal defense. I
also don't know for sure that such defenses exist for other cases (e.g.,
West has 5 or 6 diamonds). I don't know how to find out except by
repeating the above analysis.

> Should the defenders always follow strategy i) or ii)? Or
> should they follow a mixed strategy (e.g. when West holds
> Kxxxxx they play i) with probability p, and they play ii)
> with probability 1-p)?
> And what values of p should they choose for each holding West
> could have started with?

See above. Note, the defense above is not unique; EW can vary the
defense slightly and still declarer can do no better than to finesse.

> And can we choose our p_h's carefully so that declarer can do
> no better than just always taking the finesse regardless of
> our discards?

Yes, see above.

David desJardins

[sonia.zakrzewski]

> I'm really not capable of following the mathematical argument that
> follows in detail, but it seems to me that there is one important point
> that has been overlooked:
> how do the defenders know to discard all their spades without exchanging
> informative signals? Doesn't that affect declarer's chances? (Declarer
> might need to leave the diamond suit intact to maximize the defense's
> problems.)
>
> Eric

I don't think this is really a problem (in this hypothetical scenario).
The defense could play "low card count" i.e. they signal the length of
the small cards in ca suit. Hence they would show an even number with
both xxxx and Kxxxx and an odd number with both Kxxx and xxxxx. Since
they know the count in hearts and diamonds this would give them a full
count and the player without the CK can then choose whether to keep a
spade in the end position or not.

"Low card count" is normally more useful when for example playing in
notrumps and needing at least 4 tricks from this suit, an otherwise
entryless dummy holds

AQJ10x AKJ10x
xxxx Kx or xxx Qxx
xx xx

West must give a true count signal so East can hold-up correctly, but
then declarer may be able to read a standard count signal and drop the
remaining honour.

[sonia.zakrzewski]

David desJardins wrote:
>
> Gareth Birdsall <gj...@statslab.cam.ac.uk> writes:
> > Has anyone considered a problem such as the following -
> > how do you defend the following 7NT contract?
> >
> > 2
> > 765432
> > AK32
> > 32
> >
> > A
> > AKQJ1098
> > Q54
> > AQ
> >
> > 7NT by South. Lead: DJ.
>
> You can solve this exactly, assuming double-dummy defense (and assuming
> that down 2 and down 1 are equally good for the defense). But it's
> pretty messy, and it's not clear how relevant the solution is in the
> real world.
>
> As you say, the only relevant case is when West is long in diamonds.
> But declarer's and defender's strategies will depend on the actual
> diamond split, which declarer will discover during play. For
> simplicity, I'll consider only the case where West has exactly four
> diamonds, but the other cases can be analyzed in exactly the same way.

> {snip}

>
> So an optimal strategy for the defense, when West has exactly four
> diamonds, is for the player without the club king to keep a spade with
> the following probability:
>
> Y(0) impossible N(0) no choice
> Y(1) 55/ 165 = 33% N(1) 1210/ 1320 = 92%
> Y(2) 1210/ 2640 = 46% N(2) 7810/ 9240 = 85%
> Y(3) 7810/12936 = 60% N(3) 20746/25872 = 80%
> Y(4) 20746/25872 = 80% N(4) 27214/32340 = 84%
> Y(5) 23100/23100 = 100% N(5) 18480/18480 = 100%
> Y(6) 9240/ 9240 = 100% N(6) 4620/ 4620 = 100%
> Y(7) 1540/ 1540 = 100% N(7) 440/ 440 = 100%
> Y(8) no choice N(8) no choice
> Y(9) irrelevant N(9) impossible
>
> This defense ensures that declarer can do no better than to finesse
> (except when East shows out).
>
> I don't know if there's a simpler description of an optimal defense. I
> also don't know for sure that such defenses exist for other cases (e.g.,
> West has 5 or 6 diamonds). I don't know how to find out except by
> repeating the above analysis.
>

> See above. Note, the defense above is not unique; EW can vary the
> defense slightly and still declarer can do no better than to finesse.

Yes this was the conclusion I came to. I like your method of using the
number of cases n(k) and y(k) which allows you to find feasible n(k) in
turn. I tried using probabilities instead of number of cases which made
solution to the inequalities less systematic, but possible by trial and
error (but it's probably easier to find a nice looking solution by trial
and error).
Interestingly enough, the region of probabilities for which the defence
is optimal is really quite small. In particular your whole number
percentages don't give an optimal solution, but your exact fractions do!
(my trial and error method involved specifying a defensive strategy and
then using a simple excel spreadsheet to calculate declarer's best line
in each case - so I just plugged your strategy in.)

Here's a few percentages I bashed out on the computer.
Defensive strategy Chance of success
of declarer's optimum line
Always i), complete count of hand 0.6121
Always ii), partial count 0.5805
i) or ii) each 50% of the time 0.5959
Best strategy ?

Obviously declarer can never be forced to take a line worse than:
Declarer always take the finesse 0.5500

i.e. 0.5500 means that the finesse is better than the drop in
55% of hands where it makes a difference (which is exactly
the 11:9 odds indicated by empty spaces).

In fact the defenders can follow a strategy which makes
declarer's optimal strategy to finesse regardless. There
are many of these strategies, but I exhibit one which only
requires tossing a coin 3 times.

West holds Give declarer a complete count with probability:
Kxxxxxx 0.125
Kxxxxx 0.125
Kxxxx 0.375
Kxxx 0.5
Kxx 0.875
Kx 0.875
K 0.875
xxxxxxx 0.5
xxxxxx 0.5
xxxxx 0.5
xxxx 0.5
xxx 0.5
xx 0.5
x 0.5
(If West has void or Kxxxxxxx then it's impossible to give
declarer a complete count)
This strategy is also optimal if diamonds split 5-1 or 6-0.

So when it becomes clear this is the position, the defenders
should toss a coin exactly 3 times (exactly 3 to avoid giving declarer
any information about their holdings). Then the player without
the CK defends in the appropriate way.

In fact the act of tossing the coins would be regarded as
directing your partner's defence, so better for both defenders
to toss 3 coins each before every hand in case this situation
should arise.

Cheers
Gareth

[sonia.zakrzewski]

"sonia.zakrzewski" wrote:

> Interestingly enough, the region of probabilities for which the defence
> is optimal is really quite small. In particular your whole number
> percentages don't give an optimal solution, but your exact fractions do!

Sorry, small is not the right word. Your construction of an optimal
strategy necessarily produces one on the boundary of the optimal region
- so naturally a small deviation from it is likely to produce a
non-optimal solution.

Cheers
Gareth

[David desJardins]

"sonia.zakrzewski" <sonia.za...@ntlworld.com> writes:
> I don't think this is really a problem (in this hypothetical scenario).
> The defense could play "low card count" i.e. they signal the length of
> the small cards in ca suit. Hence they would show an even number with
> both xxxx and Kxxxx and an odd number with both Kxxx and xxxxx. Since
> they know the count in hearts and diamonds this would give them a full
> count and the player without the CK can then choose whether to keep a
> spade in the end position or not.

I think this is wrong. Suppose the defenders signal "low card count",
and they don't keep any spades in the end position. Then declarer will
always guess right, because the "low card count" reveals whether East's
last card is a small club or the club king (the only choices).

So your signal would only work if the opponents always keep a spade.
But, in that case, we know that declarer can sometimes do better than
the finesse.

David desJardins

[jurg...@verizon.net]

This is rather a nice problem - thanks for the effort.

There is an interesting potential fallacy: One might think at first
glance that all declarer has to do is determine who has the greater
club length. If more clubs are on the right you finesse; if more are
on the left you play for the drop.
If this were so no defensive strategy could have a positive effect
except when the King is with precisely 4 small clubs.

But this view is right only when the count is exact. When the count is
approximate, e.g. if W has shown 2 small and may or may not have the K
or another small club in addition, the empty space argument that
Gareth gave is the answer: Now the hand with fewer known clubs (and
more empty spaces) is more likely to hold the K.

Another way of saying the same thing: As long as the defender plays
insignificant cards and no suit is exhausted the only information
gained is that the number of unknown cards and the number of remaining
spaces is reduced and odds change very little or not at all. But as
soon as a significant card falls or the count of a suit is complete,
the odds can change drastically.

Jurgen

On 04 Mar 2002 19:06:21 -0800, David desJardins

[Ed Colley]

"sonia.zakrzewski" <sonia.za...@ntlworld.com> wrote in message
news:3C84A6C2...@ntlworld.com...

I would have posted an analysis, but DDJ beat me to it, so I'll
provide some rules commentary instead.

> So when it becomes clear this is the position, the defenders
> should toss a coin exactly 3 times (exactly 3 to avoid giving declarer
> any information about their holdings). Then the player without
> the CK defends in the appropriate way.
>
> In fact the act of tossing the coins would be regarded as
> directing your partner's defence, so better for both defenders
> to toss 3 coins each before every hand in case this situation
> should arise.

Indeed. In fact, even silently tossing a coin behind a screen
is (probably) illegal:

Law 40: A player is not entitled, during the auction and play
periods, to *any* aids to his memory, calculation or technique.

I think John Probst has a scheme (for more common-or-garden
instances) whereby he calculates a "random" number from his
spot cards. I suppose he would have to keep it secret, lest
one of his partners found out, and it became an agreement.

- Ed

[John (MadDog) Probst]

In article <a65vqc$c5v$1...@paris.btinternet.com>, Ed Colley
<e...@september.org.uk> writes

I have an algorithm based on spot cards that plays the Q or J with
slightly unequal frequency (say c. 52/48 - I won't say which way) in a
restricted choice position. If I hold a QJ then I know which one I will
play first before the auction even starts, should it prove to be
important. It's just routine for me. Proddy knew I had an algorithm,
and Gordon knows this too but neither of them has worked it out.

I came to this after reading a poker book discussing saddle point
bluffing frequencies, and realised that certain bridge hands may lend
themselves to a similar treatment. cheers john

> - Ed
>
>

--
John (MadDog) Probst| . ! -^- |icq 10810798
451 Mile End Road | /|__. \:/ |OKb ChienFou
London E3 4PA | / @ __) -|- |jo...@asimere.com
+44-(0)20 8983 5818 | /\ --^ | |www.probst.demon.co.uk

[richard e. willey]

On Thu, 7 Mar 2002 02:37:32 +0000, "John (MadDog) Probst"
<jo...@asimere.com> wrote:

>I came to this after reading a poker book discussing saddle point
>bluffing frequencies, and realised that certain bridge hands may lend
>themselves to a similar treatment. cheers john

Let me guess?
"Poker Strategy, Winning with Game Theory" by Nesmith C. Ankeny?

Wonderful book. I reommend it highly for anyone interested in either
Poker or Restricted Choice

[jurg...@verizon.net]

On 04 Mar 2002 19:06:21 -0800, David desJardins
<de...@math.berkeley.edu> wrote:

>Gareth Birdsall <gj...@statslab.cam.ac.uk> writes:
>> Has anyone considered a problem such as the following -
>> how do you defend the following 7NT contract?
>>
>> 2
>> 765432
>> AK32
>> 32
>>
>> A
>> AKQJ1098
>> Q54
>> AQ
>>
>> 7NT by South. Lead: DJ.
>
>You can solve this exactly, assuming double-dummy defense (and assuming
>that down 2 and down 1 are equally good for the defense). But it's
>pretty messy, and it's not clear how relevant the solution is in the
>real world.
>

In the real world there are two black Kings (and the Q of spades) for
the defense to worry about.

Assuming that the defense cannot signal without giving the hand away,
can declarer find the King of Clubs in all cases if the defenders
attempt to protect both kings?

This is easy if both Kings are in the same hand but a little harder if
they are split.

What is the right strategy for the defense, not knowing that the sK is
irrelevant?

Jurgen

>
> David desJardins

[Gareth Birdsall]

"John (MadDog) Probst" wrote:
> I have an algorithm based on spot cards that plays the Q or J with
> slightly unequal frequency (say c. 52/48 - I won't say which way) in a
> restricted choice position. If I hold a QJ then I know which one I will
> play first before the auction even starts, should it prove to be
> important. It's just routine for me. Proddy knew I had an algorithm,
> and Gordon knows this too but neither of them has worked it out.

It shouldn't really matter how often you play Q from QJ: as long as it's
somewhhere between about 10% and 90% it's better to finesse for stiff
honour no matter which is played. So it seems to me you'd be best off
playing one particular honour say Q from QJ with probability around
80%. This way it's still an optimal strategy, but if you play against
declarers who notice you play the Q much more often than the J (you'd
play the Q about 1.5 times as often as the J) - they may decide to play
for the drop (if you play Q) which would give you an extra edge over a
50:50 player.

Cheers
Gareth

[Gareth Birdsall]

Ed Colley wrote:
> So when it becomes clear this is the position, the defenders
> > should toss a coin exactly 3 times (exactly 3 to avoid giving declarer
> > any information about their holdings). Then the player without
> > the CK defends in the appropriate way.
> >
> > In fact the act of tossing the coins would be regarded as
> > directing your partner's defence, so better for both defenders
> > to toss 3 coins each before every hand in case this situation
> > should arise.
>
> Indeed. In fact, even silently tossing a coin behind a screen
> is (probably) illegal:

How about looking at the second hand on your watch? I can't see how you
could enforce a law against this. It also has the benefit you can
simulate an event with quite complex proabilities e.g. 7/60, 1/3, 1/5
etc - That's the real reason why there are 60 seconds in a minute.

Cheers
Gareth

[David desJardins]

"Jurgen" <jurg...@verizon.net> writes:
>>> 2
>>> 765432
>>> AK32
>>> 32
>>>
>>> A
>>> AKQJ1098
>>> Q54
>>> AQ
>

> What is the right strategy for the defense, not knowing that the sK is
> irrelevant?

My opinion is that you can't answer such questions without defining
exactly what the defenders *do* know. And you also need to know if
East knows what West knows. And if West knows that East knows what West
knows. Etc.

David desJardins

[David desJardins]

Gareth Birdsall <gj...@statslab.cam.ac.uk> writes:
>> Law 40: A player is not entitled, during the auction and play
>> periods, to *any* aids to his memory, calculation or technique.
>

> How about looking at the second hand on your watch? I can't see how you
> could enforce a law against this.

I think many people would say that this is illegal under L40. The fact
that the prohibition can't easily be enforced doesn't mean it doesn't
exist. Many things that the laws prohibit are very difficult to detect
and punish.

David desJardins

[jurg...@verizon.net]

Yes - you get an infinite recursion if declarer attempts to deduce
anything from the sequence of discards.

On 07 Mar 2002 14:19:38 -0800, David desJardins

[John (MadDog) Probst]

In article <3c8782d1....@News.CIS.DFN.DE>, richard e. willey
<rwi...@isi.com> writes

"Play poker, give up work and sleep till noon", Fox (I think)

[Danil]

David desJardins says...

See Bridge Movie #20 for a brief discussion of this (The Bridge World,
August 1984).

[Jeremy Rickard]

Gareth Birdsall <gj...@statslab.cam.ac.uk> writes:

> "John (MadDog) Probst" wrote:
> > I have an algorithm based on spot cards that plays the Q or J with
> > slightly unequal frequency (say c. 52/48 - I won't say which way) in a
> > restricted choice position. If I hold a QJ then I know which one I will
> > play first before the auction even starts, should it prove to be
> > important. It's just routine for me. Proddy knew I had an algorithm,
> > and Gordon knows this too but neither of them has worked it out.
>
> It shouldn't really matter how often you play Q from QJ: as long as it's
> somewhhere between about 10% and 90% it's better to finesse for stiff
> honour no matter which is played. So it seems to me you'd be best off
> playing one particular honour say Q from QJ with probability around
> 80%.

This might (theoretically) cost if the problem for declarer is a bit
more complicated than just the question of how to pick up this suit
for no losers.

For example, if declarer has slight extra chances (of an endplay,
perhaps) after wrongly playing for the drop, but these extra chances
are not enough to tip the balance away from the finesse if you play
each card with equal frequency from QJ, then you don't want to create
a situation (by playing the Q more often) where declarer can gain by
switching to this strategy when you play the Q.

Jeremy.

> This way it's still an optimal strategy, but if you play against
> declarers who notice you play the Q much more often than the J (you'd
> play the Q about 1.5 times as often as the J) - they may decide to play
> for the drop (if you play Q) which would give you an extra edge over a
> 50:50 player.

--
Jeremy Rickard
Email: j.ri...@bristol.ac.uk
WWW: http://www.maths.bris.ac.uk/~pure/staff/majcr/

[Gordon Rainsford]

John (MadDog) Probst <jo...@asimere.com> wrote:

> In article <3c8782d1....@News.CIS.DFN.DE>, richard e. willey
> <rwi...@isi.com> writes
> >On Thu, 7 Mar 2002 02:37:32 +0000, "John (MadDog) Probst"
> ><jo...@asimere.com> wrote:
> >
> >
> >>I came to this after reading a poker book discussing saddle point
> >>bluffing frequencies, and realised that certain bridge hands may lend
> >>themselves to a similar treatment. cheers john
> >
> >Let me guess?
> >"Poker Strategy, Winning with Game Theory" by Nesmith C. Ankeny?
> >
> >Wonderful book. I reommend it highly for anyone interested in either
> >Poker or Restricted Choice
>
> "Play poker, give up work and sleep till noon", Fox (I think)

Two out of three for you, John.

--
Gordon Rainsford

London UK

[John (MadDog) Probst]

In article <1f8qa09.dais8flj3qm4N%gordonr...@btinternet.com>,
Gordon Rainsford <gordonr...@btinternet.com> writes

After yesterdays effort I would have said all three! (I scored 45%
playing with Gordon - he was about 55%)

[Stephen Pickett]

Gordon Rainsford wrote:

Sounds like John's choice was somewhat restricted, Gordon.
--
Stephen Pickett, PO Box 44538, Vancouver BC Canada V5M 4R8
Telephone: (604) 874-7327, Fax: (604) 874-7326, ICQ UIN#212132
Go see BRidgeBRowser at http://www.microtopia.net/bridge/

[Thomas Dehn]

"Gareth Birdsall" <gj...@statslab.cam.ac.uk> wrote:
> "John (MadDog) Probst" wrote:
> > I have an algorithm based on spot cards that plays the Q or J with
> > slightly unequal frequency (say c. 52/48 - I won't say which way) in a
> > restricted choice position. If I hold a QJ then I know which one I will
> > play first before the auction even starts, should it prove to be
> > important. It's just routine for me. Proddy knew I had an algorithm,
> > and Gordon knows this too but neither of them has worked it out.
>
> It shouldn't really matter how often you play Q from QJ: as long as it's
> somewhhere between about 10% and 90% it's better to finesse for stiff
> honour no matter which is played.

Let me guess: you flunked "Introduction to Statistics" ;-)

Thomas

[David desJardins]

Gareth Birdsall <gj...@statslab.cam.ac.uk> wrote:
>> It shouldn't really matter how often you play Q from QJ: as long as
>> it's somewhhere between about 10% and 90% it's better to finesse for
>> stiff honour no matter which is played.

Thomas Dehn <thomas...@arcor.de> writes:
> Let me guess: you flunked "Introduction to Statistics" ;-)

Gareth's comment seems right to me. I think I have as much training in
statistics as anyone posting here. What is it that you object to?

David desJardins

[Thomas Dehn]

If an opponent is known to play the Q only 10% of the time when
he has QJ doubleton, then it normally is correct to play hin for
QJ when the J appears and there are five missing
cards in that suit. Here, QJ doubleton in his hand
a priori has 3.4%, whereas J singleton in his hand
a priori has 2.83%, and the break-even point approximately
is at 17%.

If there are six missing cards in
that suit, the QJ doubleton a priori has 1.67%, whereas
J singleton a priori has 1.21%, and the break-even
point approximately is when the opponent plays the
Q from QJ doubleton at least 27% of the time.

It gets even worse if there are seven
missing cards in the key suit.

Furthermore, an evil defender might have played
an honour from QJx(x) to
mislead declarer.

Thomas

[David desJardins]

Thomas Dehn <thomas...@arcor.de> writes:
> It gets even worse if there are seven missing cards in the key suit.

I see your point. But how often do you find yourself deciding between
stiff J and QJ when you are missing 7 cards and there has been no
bidding to help?

> Furthermore, an evil defender might have played an honour from QJx(x)
> to mislead declarer.

I don't see how this affects the odds. You can't pick up QJx(x) offside
anyway, so what difference does it make whether you play for the finesse
or the drop?

David desJardins

[Thomas Dehn]

"David desJardins" <de...@math.berkeley.edu> wrote:
> Thomas Dehn <thomas...@arcor.de> writes:
> > It gets even worse if there are seven missing cards in the key suit.
>
> I see your point. But how often do you find yourself deciding between
> stiff J and QJ when you are missing 7 cards and there has been no
> bidding to help?

7 cards is not that likely, but with five or six
cards missing such decisions do occur quite frequently

> > Furthermore, an evil defender might have played an honour from QJx(x)
> > to mislead declarer.
>
> I don't see how this affects the odds. You can't pick up QJx(x) offside
> anyway, so what difference does it make whether you play for the finesse
> or the drop?

When you have another suit you could play for
winners. For example, you might have
AKT9xx opposite xx in one suit, and xx
opposite AKT9xx in another suit, and you need
anything from five to eight tricks without losing one.

Thomas

[Mark Goddard]

"Thomas Dehn" <thomas...@arcor.de> wrote in message
news:a6aqol$crbc7$1...@ID-57266.news.dfncis.de...

ROTFLMAO

>
>
> Thomas
>

[Ashok]

In article <a6b60g$d3r3n$2...@ID-57266.news.dfncis.de>, thomas...@arcor.de says...

>
>
>"David desJardins" <de...@math.berkeley.edu> wrote:
>> Gareth Birdsall <gj...@statslab.cam.ac.uk> wrote:
>> >> It shouldn't really matter how often you play Q from QJ: as long as
>> >> it's somewhhere between about 10% and 90% it's better to finesse for
>> >> stiff honour no matter which is played.
>>
>> Thomas Dehn <thomas...@arcor.de> writes:
>> > Let me guess: you flunked "Introduction to Statistics" ;-)
>>
>> Gareth's comment seems right to me. I think I have as much training in
>> statistics as anyone posting here. What is it that you object to?
>
>If an opponent is known to play the Q only 10% of the time when
>he has QJ doubleton, then it normally is correct to play hin for
>QJ when the J appears and there are five missing
>cards in that suit. Here, QJ doubleton in his hand
>a priori has 3.4%, whereas J singleton in his hand
>a priori has 2.83%, and the break-even point approximately
>is at 17%.
>
>If there are six missing cards in
>that suit, the QJ doubleton a priori has 1.67%, whereas
>J singleton a priori has 1.21%, and the break-even
>point approximately is when the opponent plays the
>Q from QJ doubleton at least 27% of the time.

I don't know why my fellow-examiner David let you get
off the hook that readily!

Gareth's statement obviously assumed the context to be
one where the defenders have QJxx and declarer has
A and K under the defender with Q, J, or QJ. And Gareth's
statement is substantially correct.

Note that the key aspect in the above context is that the
optimal play for the declarer involves first playing A (or K).
Your explanation here appears to assume that the optimal
first play in the suit for the declarer remains the same
when he has 8 or 7 cards in the suit. That is, in fact,
not the case.

>Thomas

My grade for you in Introductory Decision Theory is an F,
but because of David's lenience, right now it is
"Incomplete." For a passing grade, submit a full analysis
of the equilibrium strategy pair for the following
situation:

Dummy: AKT98
Declarer: 765

Your analysis must cover all the important holdings
for the defenders.

For extra credit, you can try

Dummy: AKT98
Declarer: 76

Ashok

[David desJardins]

"Ashok" <adhar...@hotmail.com> writes:
> I don't know why my fellow-examiner David let you get
> off the hook that readily!

He's right that one can construct situations (esp. involving a whole
hand, not just one suit) where it's important to play either honor from
QJ with probability very near to 50-50. I'm sure these weren't the
situations Gareth was thinking of, but then he didn't do a good job of
qualifying the positions he had in mind.

> Note that the key aspect in the above context is that the
> optimal play for the declarer involves first playing A (or K).
> Your explanation here appears to assume that the optimal
> first play in the suit for the declarer remains the same
> when he has 8 or 7 cards in the suit. That is, in fact,
> not the case.

It depends on the arrangement of the cards. With

Dummy: AT2
Declarer: K3

needing 3 tricks in the suit, there certainly doesn't seem to be much
choice except to cash the king, and if RHO plays an honor then guess
whether to play for the drop or finesse.

In practice, I don't think it's particularly important to play exactly
50-50 in these situations, because even if you don't, declarer isn't
likely to know which you favor.

David desJardins

[DavJFlower]

>Subject: Game Theory
>From: Gareth Birdsall gj...@statslab.cam.ac.uk
>Date: 04/03/02 18:57 GMT Standard Time
>Message-id: <3C83C389...@statslab.cam.ac.uk>

>
>
>Has anyone considered a problem such as the following -
>how do you defend the following 7NT contract?
>

> 2
> 765432
> AK32
> 32
>
> A
> AKQJ1098
> Q54
> AQ

>7NT by South. Lead: DJ.
>
>Declarer's best line is fairly clear. Win DA, play D to Q,
>then run your 8 major suit winners discarding a club from
>dummy. Then play a diamond to see if they break.
>
>With the lead in dummy:
> -
> -
> 3
> 3
>
>
> -
> -
> -
> AQ
>If East started with the diamond length, then we must play a
>club to the ace (show-up squeeze) since East is known to hold
>at most one club with his remaining diamond. So declarer
>only has a choice of plays if West holds the diamond length
>- and for simplicity assume he started with 4.
>
>Declarer can either play West to have started with the CK
>(which will now drop, West having been squeezed), or play
>East for the CK by finessing.
>If we disregard the fact that we have seen some discards then
>the finesse is better than the drop. We can think of this in
>terms of empty spaces: West has 4 diamonds, so 9 empty
>spaces. East has 2 diamonds, so 11 empty spaces. So East is
>11:9 more likely to hold the CK.
>
>But declarer has at least a partial count of the hand and
>perhaps he can do better by using this information.
>
>The defenders have a couple of options:
>i) Discard all cards in the major suits keeping 1 diamond
>and 3 clubs in the above end position. This gives declarer a
>complete count of the hand.
>ii) Keep 1 diamond, 2 clubs and 1 spade in the end position.
>Declarer only gets a partial count of the hand (e.g. if West
>has discarded 5 clubs then he knows West started with either
>xxxxx or Kxxxxx clubs).

>
>Should the defenders always follow strategy i) or ii)? Or
>should they follow a mixed strategy (e.g. when West holds
>Kxxxxx they play i) with probability p, and they play ii)
>with probability 1-p)?
>And what values of p should they choose for each holding West
>could have started with?
>

>It's easy to analyse if the defenders always follow strategy
>i).
>Declarer always gets a complete count of the hand so he should play
>the defender with the longer clubs to hold the King. E.g. if
>West started with 6 clubs, the odds are 6:3 in favour of the
>drop.
>
>Suppose instead the defenders always follow strategy ii).
>Now declarer always gets a partial count of the hand (e.g.
>knows West started with either xxxxx or Kxxxxx clubs).
>It's easiest to rationalise the best play in terms of empty
>spaces.
>If West plays n low clubs, then he has shown up with 4
>diamonds and n clubs: 9-n empty spaces. East has 2 diamonds
>and 8-n low clubs: 3+n empty spaces.
>So if West plays just 0, 1, or 2 clubs then play for the drop
>since West has more empty spaces than East. If West plays 3
>clubs it's a guess, and if West plays 4 or more clubs, finesse.
>
>So when West holds K, Kx or Kxx, the defence might do better
>to give declarer a complete count of the hand (but can't
>always do this or declarer will be tipped of as to why he has
>been given a complete count).
>So we can see that for each holding, h, that West can hold,
>the defence should keep a spade with probability p_h and give
>declarer a complete count with probability 1-p_h.
>So what are the best (for the defenders) choices of p_h for
>each holding, h, that West can start with?

>
>And can we choose our p_h's carefully so that declarer can do
>no better than just always taking the finesse regardless of
>our discards?
>

>Cheers
>Gareth
>

Assuming that the bidding has revealed declarer's exact hand, so the defenders
do not need to signal, then there is an interesting point that does not seem to
have been picked up:

If the club K is with the long clubs, then it is in the defenders' interest to
reveal the distribution, and vice versa.

But a suspicious declarer might reject the Greek gift . . .

Dave Flower