[phil_20686]

benlessard, on 2012-February-01, 17:48, said:

It a simpler but equivalent solution to what i was thinking before. Before the run of the clubs West has 7 know cards and 6 vacant spaces & so do East. This IMO is the important starting point. We know that West has at least 3 idles cards (he cannot have Ks and the 3 remaining hearts) and that west has at least 2 idle cards (for him the K of S is an idle card since he can discard it easily after dummy discard his J of S). So the H distribution should be ---5 vacants spaces for West and 6 vacants spaces for East. West is going the have Jxx in hearts slightly more than 6% and East has twice the odds (more than 12%) to have the 3 remaining H.

As I understand vacant spaces, you should only include suits where the suit break is known completely, so diamonds cannot be included until the thirteenth is discarded on the run of the clubs.

[gnasher]

gnasher, on 2012-February-08, 07:43, said:

What's difficult about this is getting to the point where we realise that the answer is simple.

See what I mean?

[mikeh]

han, on 2012-February-01, 06:53, said:

I think the main idea that you can take out of this is that (against optimal defenders) the discards of the opponents (for example the fact that LHO pitched two diamonds) are completely irrelevant, but the fact that LHO cannot have 4 hearts and the spade king is relevant. This makes it more likely that RHO has 4 hearts.

For Bluecalm the following argument might be convincing. There are these possibilities:

A) hearts split 3-2.

B) LHO has the king of spades and Jxxx of hearts.

C) LHO has the king of spades and RHO has Jxxx of hearts.

D) RHO has the king of spades and Jxxx of hearts.

E) RHO has the king of spades and LHO has Jxxx of hearts.

Scenarios A and B are irrelevant since you always make it in these cases.

In the remaining 3 scenarios RHO has Jxxx of hearts twice (C and D) and LHO has Jxxx of hearts once (E). Since C and E are exactly equally likely, we should play RHO for the heart length.

To make this argument correct we have to convince ourselves that the discards are really irrelevant. See my previous post for how you can do that, it takes some work but it is not hard.

At the risk of merely emphasizing my ignorance in these matters, I take what you are saying to be that we have 3 scenarios that we need to consider: C, D and E. Since C and E cancel each other out, the fact that the 'un-cancelled' D gives the Jxx to RHO means we play him for it.

But isn't D known, on the original post, not to be present? By the time we make the decision, RHO is reduced to 3 cards, and he didn't pitch the spade K, therefore he didn't have D.

How then, when comparing which of C and E existed (the only two scenarios remaining as possible) do we give any weight to D?

Isn't what we are doing simply trying to assess the a priori chances that LHO was dealt a hand that resulted in C compared to a hand that resulted in D?

And that entails figuring out whether 5=1=5=2 is more or less probable than 2=4=5=2 as of the time of the actual deal. Thus, before anyone looked at their cards, LHO was probably some 4432 and almost as likely to be some 4333 and very unlikely to be either 2=4=5=2 or 5=1=5=2, but we've eliminated all but the last two shapes from consideration, and the ratio of their probabilities remains unchanged, assuming uninformative discards.

I suspect something is wrong with this, since I have earlier suggested that RHO should throw the spade K when he has it in order to make it look as if D existed.

if we assume that he will sometimes throw the spade K when he holds no hearts, the fact that in the actual case he didn't, makes E less likely than C. If he always throws the spade K when he has it then at the point of decision, RHO would have pitched the spade K and thus D would be back into the equation. Conversely, if RHO always pitches the spade K when he has it, the lack of the pitch takes E out. A blended strategy appears to be indicated.

My head hurts.

(ducking now in anticipation of a hopely polite explanation of my undoubted error)

[WellSpyder]

mikeh, on 2012-February-08, 11:58, said:

My head hurts.

I'm not surprised! How about this for a simple pragmatic reason for playing RHO for 4 hearts?

a) half the replies/analysis seems to suggest it makes no difference who you play for 4 hearts. In that case, you lose nothing by playing for it to be RHO
b) the other half argue that it is better to play RHO for 4 hearts.
c) as far as I can see, no-one thinks it is better to play LHO for 4 hearts.

So whatever (non-zero) odds you put on a) or b) being right you are better off playing RHO for 4 hearts! Simple, and you can rest your head because you don't even have to decide who is right

[fred]

It seems like some people are still having trouble (as I did) with the basic point of this problem: that the information regarding the diamond layout is not relevant. What follows is what I think is a good way to understand this point.

I am going to assume that the spades are xx opposite Ax in order to simplify things.

Consider the strategy that a defender (either one) with 5152 might adopt (and assume that declarer knows what that strategy is).

If a defender with 5152 never discards two diamonds then declarer will guess correctly 100% of the time that he sees two diamond discards (because he will know that the two diamond discards were forced plays from a defender with 2452).

If a defender with 5152 always discards two diamonds then declarer will guess correctly well over 50% of the time that he sees two diamond discards by playing the defender that discarded diamonds for 5152 (because that shape is considerably more likely than 2452).

So in the never case declarer can do better than a 50-50 guess by playing the diamond discarder for 2452 and in the always case declarer can do better than a 50-50 guess by playing the diamond discarder for 5152. I am not sure if I could prove it formally, but to me the implication is that there exists a certain amount of sometimes (which obviously falls between never and always) that a 5152 hand should discard diamonds that will leave declarer with a complete guess as to if he should play the diamond discarder for 5152 or 2452.

If the defenders adopt this strategy (discarding diamonds from 5152 the appropriate amount of time) then diamond discards are not relevant - declarer has a pure guess regardless of what the discards are.

Fred Gitelman
Bridge Base Inc.
www.bridgebase.com

[mikeh]

fred, on 2012-February-08, 12:48, said:

If a defender with 5152 always discards two diamonds then declarer will guess correctly well over 50% of the time that he sees two diamond discards by playing the defender that discarded diamonds for 5152 (because that shape is considerably more likely than 2452).

Fred Gitelman
Bridge Base Inc.
www.bridgebase.com

I would have thought, and a brief search online seems to confirm this, that a priori a 5422 shape is far more probable than a 5521 shape. Admittedly, these are with respect to any 5422 and any 5521, and we are now reduced to more specific possibilities, but is Fred right on his assertion that 5152 is 'considerably more likely than 2452?.

[omarsh10]

The D discards for LHO are the must. The holder of the K of spades would have tried to inform his p about it unless he has both 4-hearts and K of spades. LHO failure to discard spades first or second time points to RHO as likely K-of-s holder so I would play LHO for having 4 hearts

[MrAce]

omarsh10, on 2012-February-08, 13:16, said:

LHO failure to discard spades first or second time points to RHO as likely K-of-s holder so I would play LHO for having 4 hearts

LHO can not possibly discard ♠K because dummy's ♠ J is not discarded yet.

[fred]

mikeh, on 2012-February-08, 13:03, said:

I would have thought, and a brief search online seems to confirm this, that a priori a 5422 shape is far more probable than a 5521 shape. Admittedly, these are with respect to any 5422 and any 5521, and we are now reduced to more specific possibilities, but is Fred right on his assertion that 5152 is 'considerably more likely than 2452?.

One easy way to see this: if 2452 then spades are 2-7, but if 5152 then spades are 5-4.

Or maybe this works better for you: who is more likely to have 4 hearts? The hand with 5 diamonds (hence 2452) or the hand with 3 diamonds (hence 5152)?

A priori doesn't really matter since we know how diamonds and clubs break and since we are assuming hearts are 4-1.

Fred Gitelman
Bridge Base Inc.
www.bridgebase.com

[HighLow21]

gnasher, on 2012-February-08, 07:43, said:

I think that's an excellent way of looking at it. Another, equivalent, approach is to consider the situation after we have cashed the last club. Of the four layouts:
(1) LHO 4 hearts, RHO ♠K
(2) LHO ♠K, RHO 4 hearts
(3) LHO 4 hearts + ♠K
(4) RHO 4 hearts + ♠K
1 and 2 are equally likely; 3 is known to be impossible; 4 is still possible. Thefeore we should play RHO for four hearts.

What's difficult about this is getting to the point where we realise that the answer is simple. To do that, we have to understand that:
(a) The opponents' small spades and small diamonds are all equivalent, so we should ignore which ones they throw
(b) From RHO's point of view in the 3-card ending, ♠K is just another irrelevant small card, so we don't care whether he's thrown it or not.

(1) and (2) are unequivocally NOT equally likely. I know A/E players have developed a disdain for the concept of Vacant Spaces, but it is absolutely germane in this case. We KNOW LHO started with 2 more diamonds than RHO, so RHO is much more likely to hold the remaining heart stack. Whether Lefty gave us that info accidentally or on purpose, we have it. 5-to-1 is the calculation I explained earlier, and it has been corroborated by the analyses of others.

I realize that LHO might have been playing a deceptive game from 5152 shape to try to fool us, or from 2452 to try to fool us, but his cleverness (or lack thereof, depending on the true shape) cannot change the fact that he's shown us the 2 missing diamonds and therefore is much less likely to have also been longer in ♥.

Put simply, he started with 6 major suit cards originally, and his partner, 8. The original division of 2♠4♥ to the left and 7♠1♥ to the right is far less likely than 5♠1♥ at left and 4♠4♥ at right.

And by the way-->if he was anything other than 2452, his two diamond discards were a mistake, because a smart declarer who appreciates the importance of the diamond discards will then play ♥ correctly, much to his partner's chagrin. (And if Lefty WAS 2452, you should congratulate him on his brilliant defense!)

Yes, if he was 2452 originally, his diamond discards were forced, lest he give up the hand entirely, but that is far less likely than 5152, the only other relevant case.

[mikeh]

fred, on 2012-February-08, 14:33, said:

One easy way to see this: if 2452 then spades are 2-7, but if 5152 then spades are 5-4.

Or maybe this works better for you: who is more likely to have 4 hearts? The hand with 5 diamonds (hence 2452) or the hand with 3 diamonds (hence 5152)?

A priori doesn't really matter since we know how diamonds and clubs break and since we are assuming hearts are 4-1.

Fred Gitelman
Bridge Base Inc.
www.bridgebase.com

I have come up with another way of looking at it (I can hear the groans already).

My earlier reference to a priori was mistaken. We should instead look at the odds of either hand shape not on the deal (true a priori) but after we have looked at the hands held by declarer and responder.

The cards dealt to our opps are far more constrained than on the original deal.

When we are assigning 52 cards equally between 4 hands, 2452 is more likely than 5152.

But we know we're not doing that. We are assigning 9=5=8=4 cards equally between 2 hands.

We do that, and surely 5=1=5=2/4=4=3=2 is more likely than 2=4=5=2/7=1=3=2?

Each option contains one balanced/semi-balanced hand and one shapely hand.

if we compare 4432 to 5422, we'd expect 4432 to be more common. When we compare the unbalanced companion hands, we'd expect 5521 to be more common than 7321.

So when we compare the combination of 5152/4432 to 7132/2452, we find that both elements on the first are more common than their counterpart on the second.

So given the constraints arising from our hands being known, and given the elimination of all other shapes, it seems to me to be clear to hook rho.

I'm actually quite proud of this....even though it took me a long time (and a lot of help from others here) to get to this point.....so I really hope I don't shortly learn that I've screwed it up again.....like I said, my head hurts.

[benlessard]

Quote

If a defender with 5152 never discards two diamonds then declarer will guess correctly 100% of the time that he sees two diamond discards (because he will know that the two diamond discards were forced plays from a defender with 2452).

If a defender with 5152 always discards two diamonds then declarer will guess correctly well over 50% of the time that he sees two diamond discards by playing the defender that discarded diamonds for 5152 (because that shape is considerably more likely than 2452).

So in the never case declarer can do better than a 50-50 guess by playing the diamond discarder for 2452 and in the always case declarer can do better than a 50-50 guess by playing the diamond discarder for 5152. I am not sure if I could prove it formally, but to me the implication is that there exists a certain amount of sometimes (which obviously falls between never and always) that a 5152 hand should discard diamonds that will leave declarer with a complete guess as to if he should play the diamond discarder for 5152 or 2452.

If the defenders adopt this strategy (discarding diamonds from 5152 the appropriate amount of time) then diamond discards are not relevant - declarer has a pure guess regardless of what the discards are.

Fred Gitelman
Bridge Base Inc.
www.bridgebase.com

The theoretical anwser is that West should discard randomly from a pool of S and D. (and of course never discard a H or the K of S).

he has 4S/2D at this point and he has to make 3 discards. One of them is always a S. So he should discard SDD about 10%,SSS about 30% and SSD 60%. So this is why this problem is counter intuitive. After west discard 2 D we know that Spades are more likely to break a priori 54 than 27. However only 10% of the 5152 really count (because 90% of the times when west is 5152 hes NOT going to discard 2 diamonds). So that why if dummy had Ax in S and you play against a perfect defender you would have no clues on who to finesse because 5152--4432x10% equal the frequency of 2452-7132x100% (West with a 2452 is always going to discard 2D+1S)

[phil_20686]

benlessard, on 2012-February-09, 07:47, said:

The theoretical anwser is that West should discard randomly from a pool of S and D. (and of course never discard a H or the K of S).

he has 4S/2D at this point and he has to make 3 discards. One of them is always a S. So he should discard SDD about 10%,SSS about 30% and SSD 60%. So this is why this problem is counter intuitive. After west discard 2 D we know that Spades are more likely to break a priori 54 than 27. However only 10% of the 5152 really count (because 90% of the times when west is 5152 hes NOT going to discard 2 diamonds). So that why if dummy had Ax in S and you play against a perfect defender you would have no clues on who to finesse because 5152--4432x10% equal the frequency of 2452-7132x100% (West with a 2452 is always going to discard 2D+1S)

While there are cases where a mixed strategy by the defence lets them hide all info, surely this is not one of them, as the discards of the person who has 4 hearts is mandated.

My head hurts, going to bed.

[han]

I recommend searching back to a short post by gnasher which contains a 1-line explanation for why you should play RHO for the heart length, followed by some thinking about why this single line is correct.

(to be more clear, I recommend the searching followed by the thinking, I did not mean to suggest that gnasher's post contains thinking)

[phil_20686]

Suppose that the hands were 4432-5152 LHO has to discard three spades. if 3442 then he gets two spades and one diamonds, and if 2452 then two diamonds and one spade. So if LHO is any of the 5152 6142 or 7132, then on the third he must discard three spades, if the middle or the first I have some choices.

However, I need to use my mixed strategy not only to give equal probability to 5152 and 2452, but also to balance 4432 and 6142 and it is not clear to me that these aims are simultaneously acheivable. For example, 4432 could be enough more common than any other layout that the best you can for your overall strategy is to always try to look like 4432 whenever you do not have four hearts. In that case the best strategy overall would never involve discarding two diamonds from 5152.

Gnashers argument of treating them all like one suit would be true if all the distributions were close to equally likely, because in that case you can definitely achieve an optimal mixed strategy. I have no idea if a perfect mixed strategy is acheivable here. If it is then the odds are for RHO because lho was not squeezed, because you will be able to balance the relative likelyhood of diamonds being 5-3 in discards on other hands, so its irrelevant. If you cannot acheive a mixed strategy, or if your best mixed strategy involves mostly trying to look like diamonds are 4-4 then it would almost never involve discarding two diamonds, and so I should include that information.

[HighLow21]

phil_20686, on 2012-February-15, 12:30, said:

Suppose that the hands were 4432-5152 LHO has to discard three spades. if 3442 then he gets two spades and one diamonds, and if 2452 then two diamonds and one spade. So if LHO is any of the 5152 6142 or 7132, then on the third he must discard three spades, if the middle or the first I have some choices.

However, I need to use my mixed strategy not only to give equal probability to 5152 and 2452, but also to balance 4432 and 6142 and it is not clear to me that these aims are simultaneously acheivable. For example, 4432 could be enough more common than any other layout that the best you can for your overall strategy is to always try to look like 4432 whenever you do not have four hearts. In that case the best strategy overall would never involve discarding two diamonds from 5152.

Gnashers argument of treating them all like one suit would be true if all the distributions were close to equally likely, because in that case you can definitely achieve an optimal mixed strategy. I have no idea if a perfect mixed strategy is acheivable here. If it is then the odds are for RHO because lho was not squeezed, because you will be able to balance the relative likelyhood of diamonds being 5-3 in discards on other hands, so its irrelevant. If you cannot acheive a mixed strategy, or if your best mixed strategy involves mostly trying to look like diamonds are 4-4 then it would almost never involve discarding two diamonds, and so I should include that information.

I see you slept well thinking about this. ;-)

Hey Fred --> who had the J♥ guarded so we can all stop dreaming about making this grand?

[benlessard]

Quote

but also to balance 4432 and 6142 and it is not clear to me that these aims are simultaneously acheivable

The side that has 4H his discard are forced. The other side should discard S and D at random to give the illusion that hes the one with 4H. Its really a mixed strategies case.