Play/math problem

[Fluffy]

A good opponent will play the ♥J about half the time and the ♥10 also about half the time. That way he cannot be manipulated.

But they will also play it sometimes on the first round instead of the second, my intuition is that it doesn't change anything, but does it?

[keylime]

At the table, I'd probably play along the concept of symmetry, and finesse the heart, regardless of what the actual math tells me otherwise.

[kenrexford]

I must be missing something.

 

As Declarer, my last two cards appear to be one spade and one heart. RHO must save a spade, lest my small spade be good. So, of his last two cards, one of them is a spade.

 

The remaining three cards to hand out are two hearts and the club Queen. LHO followed to my small heart toward dummy, meaning that he has one of the small hearts.

 

Thus, the remaining two cards to hand out are the club Queen and the heart honor (Jack or 10, whichever).

 

I then back up one round. Every time that RHO holds both the club Queen and the missing heart honor, he is squeezed. If he has neither, then RHO is also "squeezed," in that he MUST ditch the club Queen under the Jack.

 

So, if we assume that RHO must know the club count whenever he has the Queen and a little club, plus a spade, as his last three cards, then it seems like RHO should always play the Queen if he has it. Or, at a mimimum, he has a clear option to play whichever club suits him. Psychologically, if he has no heart in his hand, he will tend very heavily toward dropping the Queen.

 

Thus, this seems like a restricted choice problem, and even moreso. If RHO does not play the club Queen, he doesn't have it.

 

The pure restricted choice may be 66% or so (he has one club, the other club, or both) for only one club, but the tendency too play the Queen in this situation is so strong that it would be probably more like 80-90% that RHO doesn't have the Queen.

 

So, I'd never hook.

[nigel_k]

Han, I still don't understand why you are considering xxxx opposite Qxxxx in clubs but not also considering Qxx opposite xxxxxx.

 

There is also a falsecard possibility for LHO. He can play Q from Qxxx to try to induce a finesse. Or he can be forced to play Q when he started with Qxx. So once you adopt the approach of considering the cases where an overall strategy succeeds for fails, I think you need to take into account Qxx opposite xxxxx as well since you can't automatically assume LHO's queen is a true card.

 

Adding in the Qxx opposite xxxxxx cases changes it from 9:8 in favour of the drop to 4:3 in favour of the finesse.

[bluecalm]

I must be missing something.

 

First thing that you are missing is that the point of the thread is to give exact or approximate odds of both play working not to point which play is correct.

 

You are also missing some things in your reasoning:

 

I then back up one round. Every time that LHO holds both the club Queen and the missing heart honor, he is squeezed.

 

This can't occur because then RHO wouldn't play a heart honor on 2nd round (having Jxx or Txx).

 

The possibilities are:

-LHO has Q♣ and x♥ or:

-LHO has ♥Hx

 

then it seems like RHO should always play the Queen if he has it

 

This as already explained by previous posts is nonsense.

 

Or, at a mimimum, he has a clear option to play whichever club suits him

 

This is why in MFA's original analysis he says that if clubs are 4-5 location of the queen doesn't matter

 

The pure restricted choice may be 66% or so

 

And now this is completely wrong.

If you want to argue that please explain how did you come up with that 66%.

[bluecalm]

Han, I still don't understand why you are considering xxxx opposite Qxxxx in clubs but not also considering Qxx opposite xxxxxx.

 

Because of this:

 

11. Trump. LHO discards a small club, you discard the Jack of clubs from dummy, and RHO also discards a small club.

 

If clubs were originally Qxx to xxxxxx then 3rd club discarded by LHO would be the queen and we would cash 13th trick with our J♣.

 

This is why 1) question is different than 3) question.

In 3) question location of club queen doesn't matter.

[hanp]

A good opponent will play the ♥J about half the time and the ♥10 also about half the time. That way he cannot be manipulated.

But they will also play it sometimes on the first round instead of the second, my intuition is that it doesn't change anything, but does it?

I'll leave this question to the people that know people.

 

At the table, I'd probably play along the concept of symmetry, and finesse the heart, regardless of what the actual math tells me otherwise.

Keep it up!

[ceeb]

Ignoring what RHO actually plays is a cop-out -- it's saying that the actual problem is too hard so I'll instead consider how to play against computers.

 

Sorry but this is nonsense.

 

We can say the same thing about the heart 10, maybe the RHO always plays the jack from ♥J10x? Then we would be sure that the 10 is from ♥10x and we can certainly finesse. Or maybe RHO always plays the 10 from ♥J10x, so that it becomes much better to play RHO for ♥J10x instead of ♥10x?

 

A good opponent will play the ♥J about half the time and the ♥10 also about half the time. That way he cannot be manipulated.

 

For the same reason a good RHO will play (from ♣Q-6th) the club queen about half the time and the fifth small club about half the time. In that case we should play for the drop whether the queen appears or a small card appears, and it is 9:8 versus the finesse.

It's a nice comparison -- the ♥J10 choice vs. the ♣Qx choice -- but are they really the same? I think it's facile to equate them and you're wrong on two counts, one psychological and one mathematical. I'll take them in that order.

 

For the typical "quack" (or ♥J10) randomization, there is at most little scope in a reasonable A/E game for out-thinking what the opponent will do or expect, because we all know and know the opponent knows our way around this particular block, ad infinitum. Therefore it's just game theory, not mind games. But even if the ♣Q-x play were in principle the same thing, it does not follow that our merely expert (not cyborg) opponent will recognize it as such. For proof, simply note the varying reactions of several indubitably expert players here including even Fred. Yes, his uncertainty about the position was from the declarer side and perhaps as defender he would have seen it more easily. But anyway that asymmetry alone legitimizes hoping to out-guess the opposition here (e.g. Ken Rexford's discussion). Experts too can be had, under pressure and when there is something a little new.

 

... it is clear that Fred assumes the opponents defend properly when he gives this problem in the A/E forum.

In fact, even in the tired old QJ randomization situation we can guess a bit better than 50% which card our opponent will play when holding both. But -- and here's the mathematical point -- in the classical case that doesn't help because the 2:1 restricted choice odds are so overwhelming that we quite properly and contemptuously dismiss mind reading the opponent unless we are Barry Crane against the rabbit. But the present case is different: The mere 8:9 nominal odds are quite likely to be a weaker reason for choosing a play than the psychological reason. The analogy between classical restricted choice randomization and this hand is a misleading one, fatally flawed.

[kenrexford]

I must be missing something.

 

First thing that you are missing is that the point of the thread is to give exact or approximate odds of both play working not to point which play is correct.

 

You are also missing some things in your reasoning:

 

I then back up one round. Every time that LHO holds both the club Queen and the missing heart honor, he is squeezed.

 

This can't occur because then RHO wouldn't play a heart honor on 2nd round (having Jxx or Txx).

 

The possibilities are:

-LHO has Q♣ and x♥ or:

-LHO has ♥Hx

 

then it seems like RHO should always play the Queen if he has it

 

This as already explained by previous posts is nonsense.

 

Or, at a mimimum, he has a clear option to play whichever club suits him

 

This is why in MFA's original analysis he says that if clubs are 4-5 location of the queen doesn't matter

 

The pure restricted choice may be 66% or so

 

And now this is completely wrong.

If you want to argue that please explain how did you come up with that 66%.

Oh! I just realized where my reasoning reads wrong. I used "LHO" for "RHO" by accident. Obviously, I MEANT that RHO is squeezed when he has the club Queen and therefore will pitch it to look squeezed.

[cherdanno]

Ken, you must be miscounting cards or something.

[kenrexford]

Maybe another explanation. RHO is "squeezed" in that he must eventually pitch a club, whether that club pitch means anything or not. He is forced to pitch it by Declarer's timing.

 

Look at it this way. For RHO's last THREE cards, we know by LHO's penultimate card that RHO's last three cards were:

 

♠big ♥honor ♣Q

♠big ♥honor ♣x, or

♠big ♥-- ♣Qx

 

On the first, he would be "squeezed" into playing the club Queen.

 

On the second, he cannot play the club Queen and would be "squeezed" into playing the small club.

 

On the third, he is "squeezed" into playing on or the other club. However, he has a choice.

 

Initially, it looks like 66% that playing for the drop is right. But, I think it is MUCH higher. For example, if RHO would always play the Queen from situation #3, then the drop is a 100% play. If he would play small every time, then the drop is roughly 50-50, adjusted by the obvious a priori's and whatever, A rough math, then, suggests to me that the likelihood of the queen being played is roughly divided in half and added to 50%. I think somehow it works on some sort of log scale or something, perhaps. But, I end up with Queen being so "obvious" to the normal player that the drop seems to be wildly right when a small club pops to the right at trick 11.

 

If the Queen had dropped, then the analysis is a lot different.

[bluecalm]

This is all wrong Ken.

 

♠big ♥honor ♣Q

♠big ♥honor ♣x, or

♠big ♥-- ♣Qx

 

On the second, he cannot play the club Queen and would be "squeezed" into playing the small club.

 

On the third, he is "squeezed" into playing on or the other club. However, he has a choice.

 

You seem to think from that somehow follows the conclusion:

 

Initially, it looks like 66% that playing for the drop is right.

 

I guess your reasoning is that small ♣ comes twice as often from x than from Qx (because from Qx defender had a choice and supposedly played randomly).

 

The problem with that reasoning and the reason all your reasoning collapses is that your 2nd and 3rd situations are not equally likely.

We could arrive at:

 

♠big ♥honor ♣x

 

point from any initial distribution which had clubs: Qxxx to xxxxx

 

and at the

 

♠big ♥-- ♣Qx

 

point from any initial distribution which had clubs: xxx Qxxxxx

 

Which is more likely ? As previously explained the 2nd (your 3rd) is about twice as likely because it leaves space for more ♥ combos (Jx - 4combos, Tx - 4combos while JTx and xxxxx is 4 combos overall).

So now for 100% of situations 33% you are in 1st case and in 66% you are 2nd case so if you see small ♣ it's equally likely it comes from 1st and 2nd case (because opponent will play it from Qx half the time).

 

I am not going to analyse it's further here. I think it's enough to show that your reasoning is off.

[kenrexford]

It doesn't matter that much what the odds are, though.

 

We would only want to finesse if the odds of RHO playing "x" from Qx multiplied by the odds of RHO having option #3 is greater than 50%. As I believe that the odds of RHO playing x from Qx are less than 50-50, then clearly the finesse is anti-percentage.

 

However, just as an example, suppose that RHO plays x from Qx 50% of the time. Because Qx is not a 100% venture, the odds still favor the drop.

 

Suppose that x from Qx occurs 80% of the time and that option #3 happens 50% of the time. Then, the odds are only 40% that the hook works.

 

If the odds of option #3 are 75% of the time, and he plays x from Qx 75% of the time, then the finesse is right (56% or so).

 

Thus, it seems that a finesse only makes sense if the chance of Qx existing is well-above 50% AND the chances of playing low from Q-x are also well above 50%. THAT is a set of propositions that I cannot imagine to be true.

[dburn]

The quickest way to answer the first of Fred's questions appears to me to be this:

 

East began with either ♥J10x and any five clubs, or with ♥10x and six clubs to the queen (if West had four hearts and three clubs to the queen, he would have been squeezed and the hand would be over).

 

There are four ways for East to hold ♥J10x and four ways for him to hold ♥10x, so the relevant ratio is the number of ways he can hold any five clubs (63) to the number of ways he can hold the queen and five low clubs (56). This gives odds of 9 to 8 in favour of playing for the drop.

[fred]

The quickest way to answer the first of Fred's questions appears to me to be this:

 

East began with either ♥J10x and any five clubs, or with ♥10x and six clubs to the queen (if West had four hearts and three clubs to the queen, he would have been squeezed and the hand would be over).

 

There are four ways for East to hold ♥J10x and four ways for him to hold ♥10x, so the relevant ratio is the number of ways he can hold any five clubs (63) to the number of ways he can hold the queen and five low clubs (56). This gives odds of 9 to 8 in favour of playing for the drop.

But there are 8 (not 4) ways that East could hold Hx of hearts!

 

Your get the right 9:8 because you also omitted a factor of 2 on the other side - there are 126 (not 63) ways that East could hold any 5 clubs.

 

Fred Gitelman

Bridge Base Inc.

www.bridgebase.com

[dburn]

So I did. Still, better to be lucky than good.