Interesting strategic puzzle

[Gerben42]

We all know this one:

 

[hv=n=sat98&w=s76&e=sqj&s=sk5432]399|300|[/hv]

 

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

 

[hv=n=sat98&w=s76&e=sqj&s=sk5432]399|300|[/hv]

 

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 :)

[Finch]

We all know this one:

 

[hv=n=sat98&w=s76&e=sqj&s=sk5432]399|300|[/hv]

 

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 :rolleyes:

 

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]

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 :rolleyes:

 

But what if you forget about the human factor?

[Blofeld]

I agree with kfgauss' analysis.

[Finch]

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:

 

[hv=n=saj9&w=s&e=s&s=sxxx]399|300|[/hv]

 

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]

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.

[Guest Jlall]

Math people...

[Finch]

There's a reason I started down this route.

 

The QJ example, the AJ9 puzzle, and the when-to-cover-from-Kx puzzle all have the same issue: the 'optimal' strategy in terms of trick expectation is usually to play alternative cards with anything from a range of probabilities. Playing each of the Q and J with 50% probability is 'a' rather than 'the' optimal strategy (apologies to Hannie, by the way, who also pointed this out earlier but I missed it).

 

I think there's a mildly interesting bridge point here, rather than a probability theory point.

 

If you have discussed this with your partner, or played together a lot, you may have an agreement about how often you play each card, and as such it should be disclosed to the opponents. I want to take the Kx question as a good example: dummy has QJ98x, declarer is known to have Axx, when should you cover from Kx? Suppose in addition declarer cannot pick up a 4-1 break for some reason (needs to take a ruff in the short hand later, say). If I remember rightly, the answer is to cover anything between 0 and 1/3 of the time.

 

Now, if you play an infinite number of boards it doesn't matter which of the optimal strategies you play. But if you never cover from Kx, and reveal this to declarer, declarer will always get it right when you have K10 doubleton, because that's now the only relevant holding on which you cover. He'll go wrong more often when you have Kx, which compensates you in expectation calculations.

 

But if I'm playing a match against a team I expect to beat, I want to minimise random swings and rely on my skill. So I don't want to lose a game swing because I happen to have K10 doubleton and oppo get that right for certain.

 

I'm not going to use game theory terminology, because I'll probably get it wrong (I was a fluid dynamicist), but I think the optimal strategy against a worse team should satisfy two criteria:

 

i) You cannot improve your trick expectation by changing it (this I think is the 'nash equilibrium' point)

ii) You minimise the difference in trick expectation between your alternative plays

 

So in fact, I think playing the Q or the J from QJ with 50% probability _is_ the optimal strategy after all.

 

Unless I'm playing against a much better team than my team. Then I want to make things swingier. So I should tell them that I always play the Q from QJ doubleton (and pray for singleton queens), that I never cover from Kx (and pray for no K10 doubletons) and so forth.

 

I may have drivelled on for too long here, but am I making any sense at all?

[whereagles]

I may have drivelled on for too long here, but am I making any sense at all?

Not much. It's kinda pointless because it might decide 1 match out of 5000 or so.

[hrothgar]

Now, if you play an infinite number of boards it doesn't matter which of the optimal strategies you play. But if you never cover from Kx, and reveal this to declarer, declarer will always get it right when you have K10 doubleton, because that's now the only relevant holding on which you cover. He'll go wrong more often when you have Kx, which compensates you in expectation calculations.

 

...

 

I may have drivelled on for too long here, but am I making any sense at all?

The academics who have studied game theory and cards have typically focused on poker. With this said and done, there are a number of examples from poker which are directly applicable to some of the questions that we're discussing.

 

For example, there's been some very interesting analysis done surrounding 5 card draw that focused on the optimal number of cards to draw if you get dealt three of a kind. I've seen a couple good proofs that demonstrate that the equlibirum solution is to draw one card. (If you draw two cards, your chance to improve your hand. However, the math works out such that its more advantageous to conceal whether you have three of kind, two pair, or are drawing to complete a four flush)

 

The reason that I brought this up is that the same equilibirum solution also included a number of so-called mixed strategies. It turns out that optimal play requires deliberately randomizing your behaviour when you have a four-flush. You want to draw to the four-flush X pecent of the time and fold (1-X) percent of the time. The authors of the paper suggested that the "best" way to randomize is to use the entropy inherent in the your hands. (If you hold the following card combinations draw, if you hold anything else fold)

 

The same principles could easily be applied to the game of bridge. For example, suppose that you need to decide whether or not to rise with the King. Create the following rule:

 

Think back to the previous hand. If the length of your Spade suit was equal to the length of your Heart suit, cover. Otherwise duck.

 

As I recall, this rule will lead you to cover roughly 17% precent of the time.

 

For what its worth, this types of practices create some interesting disclosure issues. From my perspective, player who use these types of mixed strategies can be required to describe the existence of a mixed strategy to the opponents and register their "keys" with the Tournament Directors. However, their partner can't have any information regarding the particular key that is being used on any given day.

[han]

No problem at all Frances, you explained things much better than I did. (and I was way to lazy to compute the 91.5%)

 

Matt, it would be nice if you could explain your calculations without using sophisticated notation. I think that this is something that many bridge players are able to understand, but not if they are not familiar with the notation.

[cherdano]

But if I'm playing a match against a team I expect to beat, I want to minimise random swings and rely on my skill. So I don't want to lose a game swing because I happen to have K10 doubleton and oppo get that right for certain.

 

I'm not going to use game theory terminology, because I'll probably get it wrong (I was a fluid dynamicist), but I think the optimal strategy against a worse team should satisfy two criteria:

 

i) You cannot improve your trick expectation by changing it (this I think is the 'nash equilibrium' point)

ii) You minimise the difference in trick expectation between your alternative plays

 

So in fact, I think playing the Q or the J from QJ with 50% probability _is_ the optimal strategy after all.

 

Unless I'm playing against a much better team than my team. Then I want to make things swingier. So I should tell them that I always play the Q from QJ doubleton (and pray for singleton queens), that I never cover from Kx (and pray for no K10 doubletons) and so forth.

 

I may have drivelled on for too long here, but am I making any sense at all?

It does make sense to me. However, I would formulate it more simply: you want to maximize the winning probability. A game swing in against a weak team may increase your winning chances by 5% and your chances to lose by 10%, so you can just use these numbers instead of expected IMPs.

 

Then you can again define an optimal strategy as a Nash equilibrium by saying a strategy from your side consists of both your plan as defender and your team mates' plan as declarer, and similarly for the opposing side.

(For those who don't know the term: Nash equilibrium is a situation where either side would lose if it changed its strategy.)

 

For the QJ example, I would expect that there is still a range of probabilities that are all optimal strategy, just that it is getting smaller.

 

This sounds like enough fun to analyze completely that I might do so on a rainy afternoon...

 

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