[ceeb]

Edit -- posted the below in too much haste. Withdraw.

hanp, on Aug 25 2010, 01:10 PM, said:

There are 3:2 times as many 5-4 splits as 6-3 splits. (9 choose 4 compared with 9 choose 3).

The last question is easiest. If we don't have the club jack then RHO having 10x or Jx + 6 clubs is 4:3 times as likely as J10x plus 5 clubs. So in that case we should hook.

Agree. The possible layouts & probabilities (scaled from the prior probabilities) are just these two
xx hxxx xxxx xxx (4/7) xxxx hx x xxxxxx
xx xxx xxxx xxxx (3/7) xxxx hhx x xxxxx

Quote

If there was no club jack but the heart 9 was the 10 then we should compare xx + 6 clubs with Jxx + 5 clubs. In that case the odds give 3:2 in favor of the drop.

I'm still with you:
xx xxx xxxx xxxx (3/5) xxxx Txx x xxxxx
xx Txxx xxxx xxx (2/5) xxxx xx x xxxxxx

Quote

If we put the club jack back in, we notice that we should take the hands where LHO started with the club queen and 4 hearts out of the equation, in that case he would have been squeezed to part with a heart. On 3/9 of the hands where LHO started with 3 clubs, he has the club queen. So we should take out 1/3 of those hands, in effect multiplying the odds for the finesse with 2/3.

I think we are agreeing that the possible layouts are therefore
xx xxx xxxx Qxxx (1/2) xxxx Txx x xxxxx
xx Txxx xxxx xxx (1/2) xxxx xx x Qxxxxx
so the odds are 1:1.

Quote

That means that when dummy has the 9, hooking is 2/3 of 4:3, or 8:9. So play for the drop.

This seems a very complicated way to get to the answer. "That means that" suggests a linear line of reasoning, but 2/3 and 4:3 seem to be from thin air here. Anyway, I cannot follow.

The only layouts I see consistent with Fred's formulation are these two:
xx Hxxx xxxx xxx (2/3) xxxx Hx x Qxxxxx
xx xxx xxxx Qxxx (1/3) xxxx HHx x xxxxx
so I think the finesse is a pedestrian 2:1.

I can come up with 8:9 if I consider
xx Hxxx xxxx xxx (8/17) xxxx Hx x Qxxxxx
xx xxx xxxx Qxxx (4/17) xxxx HHx x xxxxx
xx xxx xxxx xxxx (5/17) xxxx HHx x Qxxxx
but the last is impossible when RHO has played five little clubs.

Quote

When dummy has the 10 the odds are 9:4 in favor of the drop, which seems intuitive.

Seems right:
xx xxx xxxx xxxx (5/13) xxxx Txx x Qxxxx
xx xxx xxxx Qxxx (4/13) xxxx Txx x xxxxx
xx Txxx xxxx xxx (4/13) xxxx xx x Qxxxxx

[hanp]

ceeb, on Aug 26 2010, 06:25 AM, said:

Quote

That means that when dummy has the 9, hooking is 2/3 of 4:3, or 8:9. So play for the drop.

This seems a very complicated way to get to the answer. "That means that" suggests a linear line of reasoning, but 2/3 and 4:3 seem to be from thin air here.

You are probably right that I used a complicated way to get the answer. I do think that I have a very good understanding of the words "that means that". I assure you 2/3 and 4:3 did not come from think air. Once you convince yourself that 8:9 is the right answer, perhaps you can read my post again

Quote

The only layouts I see consistent with Fred's formulation are these two:
  xx Hxxx xxxx xxx  (2/3)   xxxx Hx  x Qxxxxx
  xx xxx  xxxx Qxxx (1/3)   xxxx HHx x xxxxx

I can come up with 8:9 if I consider
  xx Hxxx xxxx xxx  (8/17)   xxxx Hx  x Qxxxxx
  xx xxx  xxxx Qxxx (4/17)   xxxx HHx x xxxxx
  xx xxx  xxxx xxxx (5/17)   xxxx HHx x Qxxxx
but the last is impossible when RHO has played five little clubs.
so I think the finesse is a pedestrian 2:1.

I tried to explain earlier why I think it is not correct to use that RHO played 5 little clubs. I am glad to see that the 8:9 also occurs when you compute the relative likelyhoods of ♥Hx ♣Qxxxxx and ♥J10x ♣?xxxx.

[ceeb]

fred, on Aug 25 2010, 04:29 PM, said:

hanp, on Aug 25 2010, 08:09 PM, said:

fred, on Aug 25 2010, 01:44 PM, said:

We can answer question 1 by comparing these numbers:

1) The number of ways that RHO could be dealt honor-doubleton of hearts and Qxxxxx of clubs

2) The number of ways that RHO could be dealt J10x of hearts and 5 small clubs

Do you agree with this?

I think this part of what you say is not correct. The fact that RHO played small on the 11th trick gives no information because at that time (when we had already played the club jack from the dummy) he could play the queen or small from Qxxxxx. To get the correct answer, we should also consider Qxxxx of clubs on our right.

This kind of argument can get a bit messy. Best is to ignore the card RHO plays. If RHO "falsecards" exactly the right amount of time from Qxxxxx there is no information to get from the play of the low club on our right.

Let me try to phrase it in terms of strategies, I think it is easier. If we play for the drop no matter whether RHO plays low or the queen, we win if RHO started with J10x of hearts and Qxxxx or xxxxx of clubs, but lose if RHO started with Jx or 10x of hearts and Qxxxxx of clubs.

♥J10x and ♣xxxxx is exactly equally likely as ♥Hx and ♣Qxxxxx, but ♥J10x and ♣Qxxxx is a bit more likely than ♥J10x and ♣xxxxx, and thus also more likely than ♥Hx and ♣Qxxxxx.

Therefore we should play for the drop.

Thanks very much, Han.

Yes, this makes sense and it is actually pretty similar to the theory I was afraid to post out of fear of further embarassing myself (though you definitely did a better job of explaining it to me than I did explaining it to myself!).

Indidentally, the reason I found this problem interesting was because I could not really decide if the Queen of clubs should be treated of as a pure x (as it is in question 3), a pure non-x (like all the known cards in spades and diamonds), or as something in between.

Fred Gitelman
Bridge Base Inc.
www.bridgebase.com

Of course something in between.

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.

Hanp has reformulated the original problem to say that RHO's last discard is "a club" instead of "a small club". Under that stipulation here are the layouts in question (posterior probabilities in parentheses):
xx Hxxx xxxx xxx (8/17) xxxx Hx x Qxxxxx
xx xxx xxxx Qxxx (4/17) xxxx HHx x xxxxx
xx xxx xxxx xxxx (5/17) xxxx HHx x Qxxxx
giving 8:9 for finesse:drop.

Assuming instead that you actually watch the spots, the 3rd possibility is eliminated but the 1st is counted pro-rata according to RHO's tendencies. If RHO last discards Q:x from the first case with ratio 5:4 in imitation of the Qxxxx:xxxxx prior probabilities, then yes 8:9 is correct. However in real life no one is that smart. At best they randomize 1:1 between the last two cards in which case the drop is 4:4 when the Q does not appear and 5:4 when it does. More likely you can judge RHO psychologically as either being a "clever" player who usually plays Q from Qx in order to "seem" longer in hearts -- in which case the drop approaches 4:0 when the Q does not appear and 5:8 when it does -- or a subtle (or oblivious) player who tends toward x from Qx -- in which case the drop approaches 4:8 when the Q does not appear and 5:0 when it does.

[ceeb]

hanp, on Aug 26 2010, 06:59 AM, said:

ceeb, on Aug 26 2010, 06:25 AM, said:

Quote

That means that when dummy has the 9, hooking is 2/3 of 4:3, or 8:9. So play for the drop.

This seems a very complicated way to get to the answer. "That means that" suggests a linear line of reasoning, but 2/3 and 4:3 seem to be from thin air here.

You are probably right that I used a complicated way to get the answer. I do think that I have a very good understanding of the words "that means that". I assure you 2/3 and 4:3 did not come from think air. Once you convince yourself that 8:9 is the right answer, perhaps you can read my post again

Quote

The only layouts I see consistent with Fred's formulation are these two:
  xx Hxxx xxxx xxx  (2/3)   xxxx Hx  x Qxxxxx
  xx xxx  xxxx Qxxx (1/3)   xxxx HHx x xxxxx

I can come up with 8:9 if I consider
  xx Hxxx xxxx xxx  (8/17)   xxxx Hx  x Qxxxxx
  xx xxx  xxxx Qxxx (4/17)   xxxx HHx x xxxxx
  xx xxx  xxxx xxxx (5/17)   xxxx HHx x Qxxxx
but the last is impossible when RHO has played five little clubs.
so I think the finesse is a pedestrian 2:1.

I tried to explain earlier why I think it is not correct to use that RHO played 5 little clubs. I am glad to see that the 8:9 also occurs when you compute the relative likelyhoods of ♥Hx ♣Qxxxxx and ♥J10x ♣?xxxx.

I know they are not from thin air but they did not come from the immediately previous either, so I was complaining that I could not following your thinking. I understand that "seem to be from thin air [here]" can sound like "are from thin air" but such was not my intent.

Yes, I followed your earlier argument and therefore tried to abort my post. Thought I had caught it too.

This post has been edited by ceeb: 2010-August-26, 06:27

[hanp]

Quote

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.

If you want to write a book about what to do against the beginner (finesse because they will never play the queen) or the advanced player (certainly drop because they always play the queen in an attempt to be an expert, but finesse when they play the queen) go ahead and make a fool of yourself. But please don't write in your book that I changed the problem, it is clear that Fred assumes the opponents defend properly when he gives this problem in the A/E forum.

[Fluffy]

hanp, on Aug 26 2010, 02:23 PM, said:

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]

Quote

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:

Quote

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

Quote

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

This as already explained by previous posts is nonsense.

Quote

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

Quote

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]

Quote

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:

Quote

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]

Fluffy, on Aug 26 2010, 10:05 AM, said:

hanp, on Aug 26 2010, 02:23 PM, said:

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.

keylime, on Aug 26 2010, 12:32 PM, said:

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]

hanp, on Aug 26 2010, 09:23 AM, said:

Quote

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.

Quote

... 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]

bluecalm, on Aug 26 2010, 03:21 PM, said:

Quote

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:

Quote

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

Quote

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

This as already explained by previous posts is nonsense.

Quote

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

Quote

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.

Quote

♠big ♥honor ♣Q
♠big ♥honor ♣x, or
♠big ♥-- ♣Qx

Quote

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:

Quote

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:

Quote

♠big ♥honor ♣x

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

and at the

Quote

♠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]

dburn, on Aug 27 2010, 10:21 PM, said:

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