[Gerben42]

We all know this one:

---

Here the optimal strategy for East is to play the Jack 50% of the time and the Queen 50% of the time.

---

Suppose you are West, declarer is South who is known to hold only small cards in this key suit.

Now here is the question: What % of the time should you play the Queen? I found the answer surprising! (That tells you 50% is not the answer )

[awm]

How about: always?

Note that declarer can always plan to play low to the nine. If I play an honor, he can win and play low to nine the second time. This will succeed any time the ten and at least one of KQ is onside, and there's really nothing the defense can do about it.

If I always play an honor if I have it, how does this help declarer? When I don't play an honor he will know that KQ are offsides, but there is absolutely nothing he can do about this. If I do play an honor, the only information declarer gains is that KQ are not both offside -- he can still only pick up the suit 50% of the time (playing low to the nine and hoping ten onside).

[han]

I think that the answer is that you can pick any percentage that you like between 50 and 100. I won't explain my answer, so that others can think about it if I'm right and I don't look as foolish when I'm wrong.

[han]

The answer to the first puzzle is wrong btw, you don't need to play the Q or the J 50% of the time. If you play the jack (say) 80% of the time then the percentage play for declarer is still to play you for a stiff jack. Only when the percentage gets very close to 100% then you lose because it gets better for declarer to play for the QJ-doubleton.

[awm]

On the other hand, how about never?

Declarer doesn't know if you have an honor or not -- you would never play one! So his best chance is to put in the nine....

I think the key here is whether you would play an honor from KQx. If you would always (or often) play an honor from this holding, then you need to occasionally play an honor from QTx/KTx also. Otherwise declarer can play the ace when you split, then play low to the jack, and pick up KQx onside as well as the normal QTx/KTx holdings.

[han]

Hmm, yes, Adam is right I think, and I was wrong. My answer assumed that you always split with KQx.

[kfgauss]

Supposing we know how many cards everyone has in the suit (namely that partner doesn't have only one or two), and that our goal is to prevent declarer from taking two tricks in the suit, any percentage of the time works, you just have to make sure that your percentage with KQx(x..) is "close enough" (it doesn't have to be that close) so that declarer can't win more often than when you have the 10 plus an honor (or KQ tight, of course).

For example, if you play the Queen 40% of the time from Q10x(x..) and you play high (let's say half the time K, half the time Q when you do do it for simplicity) from KQx(x..) more than 80% of the time, then when you play the Queen on the first round, declarer does best to play you for KQx(x..) (and when you play low, declarer plays best to play you for H10x(x..)), getting more than his fair share.

Basically to see this, one can observe that if you play low always, declarer's best play is to pick up H10x(x..), and similarly if you play high always, and you're never giving anything away (because partner has more than 2 cards, so we're not telling declarer to play for KQ tight when we play low if we e.g. always played high with an honor).

Things change if declarer could have only two small and this is a trump contract and we'd like a trick in this suit when we have KQx(x..). Then we need to weigh this concern against the concern of playing high with Q10x(x..) when partner has the stiff King (and also against the concern of telegraphing KQ tight by playing high too often) since we're apparently assuming we don't know the distribution. If we always play high with KQx(x..), half the time K, half the time Q, then we'll need to play high from Q10x(x..) 50% of the time or more. How important these two situations are relative to each other presumably depends on context.

Andy

[pclayton]

1. It depends on the frequency of a defender splitting from KQx (x..). Restricted choice certainly applies.

2. It also depends on the frequency that a defender will 2nd hand high from QTx and even KTx.

The presence of the 9 in dummy makes it easier to fly from HTx by the way than AJ8.

I suppose if you set 1 and 2 you could develop an optimum strategy by factoring in RC.

Side story: in one of my first nationals I held AJ9 in dummy and xxx in my hand. When I led toward the board and LHO (a name player) didn't play an honor, I 'naturally' played the J. LHO did have QTx, but didn't give it a moments thought.

Everyone around the table just gave me a funny look

[FrancesHinden]

Gerben42, on Apr 24 2006, 11:26 PM, said:

We all know this one:

---

Here the optimal strategy for East is to play the Jack 50% of the time and the Queen 50% of the time.

[

Is it? How do you define 'optimal strategy' ?

[Free]

It depends on declarer

I think playing the Q and then small suggests you have Qx, not the T. Playing the T first suggests Tx. If you play small, declarer will play the 9 anway, so small is not a good choice.

So what's best? I think playing the Q works best. Declarer might make the mistake of not taking the Ace when he only has 2 ♠s...

[hrothgar]

FrancesHinden, on Apr 25 2006, 03:15 PM, said:

Gerben42, on Apr 24 2006, 11:26 PM, said:

We all know this one:

AT98 76 QJ K5432

Here the optimal strategy for East is to play the Jack 50% of the time and the Queen 50% of the time.

[

Is it? How do you define 'optimal strategy' ?

Nash Equilibirum

[Gerben42]

Of course I also agree that this problem is unsolvable without the human factor. Recently with the first position I played the Jack and when I won the Queen declarer told me she was sure I didn't have the Queen because as a tricky player I would play the Queen from QJ

But what if you forget about the human factor?

[Blofeld]

I agree with kfgauss' analysis.

[FrancesHinden]

hrothgar, on Apr 25 2006, 01:40 PM, said:

FrancesHinden, on Apr 25 2006, 03:15 PM, said:

Gerben42, on Apr 24 2006, 11:26 PM, said:

We all know this one:

Here the optimal strategy for East is to play the Jack 50% of the time and the Queen 50% of the time.

[

Is it? How do you define 'optimal strategy' ?

Nash Equilibirum

So why isn't playing the Queen 90% of the time from QJ doubleton also an optimal strategy? Or indeed 91% of the time (but not 92%)?

[Blofeld]

It is, isn't it?

[MickyB]

Don't think so - there will be other information (vacant spaces) that tips the balance on some hands.

[whereagles]

Well.. some of these card positions are canonical textbook bluff/counterbluff situations. For instance, with:

---

The theoretical line of play is low to the 9 and low to the jack (AJ9 in dummy). But now if LHO is..

A beginner: he will split honors when holding KQx.
A cunning player: he'll duck with KQx, but will rise with HTx.
An expert: he'll play the beginner way or the cunning way 50% of the time.

and you can play accordingly. The expert is the hardest to play against, because you never know what he did.

[Echognome]

FrancesHinden, on Apr 25 2006, 01:00 PM, said:

So why isn't playing the Queen 90% of the time from QJ doubleton also an optimal strategy?  Or indeed 91% of the time (but not 92%)?

AT98

K5432

You are missing Q, J, 7, 6

You can pick up QJ or Qx(=Jx), in either hand, or Q or J stiff if you guess.

You can only guess one way. So if you happen to play low to the A, then you ALWAYS get it right.

The question is if you play the 8, 9, or T off dummy. And you see the Q or J (which are equivalent).

The only decision you have is when you see the other low card on your left. All other holdings are equivalent. You either already went wrong or you already made two tricks.

This is just an application of conditional probability.

So let p = prob(J|QJ). Note that prob(J|J) = 1.

where | is a conditioning argument. so that prob(J | QJ) means "the probability that East plays the J when East's original holding was QJ".

So as declarer, I have seen 3 cards, the 6, 7, and J and have just the one missing card.

So if I finesse, I "win" when East has J singleton, and "lose" when East has QJ.

finesse: prob(win) = prob(J)/(prob(J) + prob(QJ))

If I play for the drop, I "win" when East has QJ, and "lose" when East has J.

play for the drop: prob(win) = p*prob(QJ)/(prob(J) + prob(QJ))

The actual ex ante probabilities of the specific holdings are:

prob(J) = 6.2174
prob(QJ) = 6.7826

So the odds are:
finesse = 6.2174/(6.2174 + 6.7826) = 0.478
play for drop = p*6.7826/(6.2174 + 6.7826) = 0.522p

So you should finesse if

0.478 > 0.522p or
p < 0.915

In words, you should finesse, if you believe East will play the J from QJ doubleton less than 91.5% of the time. (This is the number I believe Frances is referring to.)

Now, I can repeat this analysis in its entirety replacing the J with the Q and you will find that if East plays the Q on the first round, then declarer should finesse if East plays the Q from QJ doubleton less than 91.5% of the time.

So the key to the "optimal" strategy for East is that he doesn't want his percentage to be at the extremes. He shouldn't ALWAYS play the J or ALWAYS play the Q. His optimal strategy is to play the J between 8.5 and 91.5% of the time. One of those solutions is to play the J 50% of the time.

[Echognome]

I should add that when you play low to the A, you ALWAYS get it right when East holds QJ. If West were to hold QJ, you would have the same problem.

NB: Frances posed the problem, so I blame her for my long winded reply.

NBB: One can perform a similar analysis to the second problem, but it is more involved.