[ron_ron]

ron_ron, on 2012-January-31, 23:27, said:

If east started with a singleton heart and the SK, he must play the SK some fraction of the time (4/7, I believe) to balance out the times he is forced to play the SK (when he started with 4 hearts). If he plays the SK more often or less often than 4/7 of the time, declarer has a strategy that does better than break even. Declarer should therefore play east for Jxxx when no SK appears. If east never plays the SK when holding stiff H, declarer has a straight guess, so he will never do worse than break even (no matter east's strategy), and will do better than break even when east plays optimally (or in fact if east plays the SK some nonzero percentage of the time when he started with a stiff heart).

Actually, I think part of the above is not quite right; east can play the K anywhere from 0 to 4/7 of the time. The conclusion is still correct, though, I think.

[mike777]

Let me back up is this a math question where the issue is who has the best PHD in Math or is this a bridge question for exp bridge players?



Actually bridge is a timed event but in this thread no one seems to be aware there are penalties for taking all this time.


--


If you want to prove the best math great but this is bridge not unlimited time but no one really seems to care about that.

In other words is this a bridge issue or a math issue where we can take hours?

[han]

My impression is that Fred intented this as a theoretical issue only. Looking at it again, it seems to me that I gave the right answer yesterday, although my numbers may have been off. If I find some time today I'll try to write a proof to show that no strategy can do better than the one I proposed (always playing RHO for the heart length no matter the discards), although I fear that such a proof will not be convincing to everybody.

[ron_ron]

Whoops. I miscounted the vacant spaces. Replace 4/7 by 3/6...

[bluecalm]

Quote

I'll try to write a proof to show that no strategy can do better than the one I proposed (always playing RHO for the heart length no matter the discards), although I fear that such a proof will not be convincing to everybody.

Please do.
I am still not convinced but more from not getting it than from having counter argument.

[han]

I haven't written it up but I've convinced myself that it is actually not as hard as I thought. In order to convince yourself that the correct solution is to play RHO for the heart jack no matter the discards, you only have to give one discard strategy for EW such that for any of the possible discards the chance that RHO has ♥Jxxx is larger than the chance that LHO has ♥Jxxx. This is not hard to do because there is a lot of room (by which I mean, the total chance that RHO has ♥Jxxx is so much larger than the total chance that LHO has Jxxx that there is a lot of freedom in how you choose the discards). I could try to write it out here but it would be long and nobody would read it, it's probably more useful if you try to work it out for yourself. Certainly more useful for me.

[benlessard]

Quote

If you want to prove the best math great but this is bridge not unlimited time but no one really seems to care about that.

Even if calculating this example has no real life benefit, the ideas behind the solution will be useful on others hands, of course trying to find the perfect solution or proof might have no application in real life, but in general there is always something not obvious to learn from these case and trying to find out by yourself is often a good start to understand these things IMO even if you know your tries are likely to be far from the solution.

[han]

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.

[bluecalm]

Ok, so it comes down to:
"Only honest (ie not volunteered by opponents) information is that they have 5 minor cards each; we know they both have 8 more cards. Since LHO would be squeezed if he had both Ks and Jxxx of hearts those layouts are out and all other are symmetrical leaving the ones with RHO having Ks and Jxxx unmatched, so there are less layouts with LHO having 4 hearts".

Now when I spelled that out it seems quiet obvious to me
Thanks, that was nice explanation, I wouldn't have thought of this.

[han]

It's not clear that that explanation is correct (what does it mean that the information was volunteered? If LHO has Jxxx of hearts his discards are forced), I'm convinced that the conclusion is correct.

Fred has posted similar threads in which the underlying question was similar: which of the plays by the opponents have impact on the odds, and which do not? Here the discards are not impacting the odds, but I think it requires some thought. Perhaps Fred has found a neat way to think about this though.

[benlessard]

Edited Sorry Han and others, Im a native french speaker and work night shift and its close to my sleeping time now.

IMO I don't think we should say the discards are irrelevant.

After cashing the fist 7 tricks we learn

---

Its symmetrical and if we make the asumption that H break 4-1 we can say for sure that E and W are equally likely to have 4H (in a no-bidding vacuum).

Here are the a priori west hands depending if west started with 4H or with only 1. Lets give the name "DS" for D or S idle cards.

with 4H
x
xxxx
xxx
xx
with 3 (DS) or with 2DS+Ks

with a stiff H
x
x
xxx
xx
with 6DS or with 5DS +Ks

on the run of the clubs west two first discards will be 2DS, he will never discard a H or the K of S. If its one diamond & one spade or 2 spades or 2 diamonds its irrelevant. He always have 2 free DS available for discarding, so these are non significant discards. However on the 5th clubs the 3rd discard by West he still cannot discard the K of S nor a heart so at this point if he got 4H & the K of S he wont have an idle DS available to discard. So in that position when he discard a DS here its a meaningful discard not an irrelevant one. So that change the odds. The fact that he knowned with 1 more free DS than EAST mean hes got one less vacant space in his hand. If we put West with 5 vacant space and 6 vacant space to east we get these H splits.

No-W---E--Probability--Times--Total
1--0---Jxx--12.121------1-----12.121
2--x---Jx---15.152-----2-----30.303
3--xx--J----12.121-----1-----12.121
4--J---xx---15.152-----1-----15.152
5--Jx--x----12.121-----2-----24.242
6--Jxx--0----6.061------1-----6.061

Bonus question what if East discard the Ks after north discard his J of S ?

[han]

Benlessard, your writing style is incomprehensible, do you honestly expect anybody to read that and understand it without spending a lot of time deciphering it?

[bluecalm]

Ok, here is my latest try:
we forced them to play 5 minor cards, one spade and one heart each. That leaves 12 cards in their hands out of which 8 are irrelevant and 4 are relevant: Jxx ♥ and K ♠
Possible layouts are:

a)LHO has 6 irrelevant cards; RHO has 2 irrelevant card and both Jxx♥ and K♠
b)LHO has 5 irrelevant cards and K♠ ; RHO has 3 irrelevant cards and Jxx♥
c)LHO has 3 irrelevant cards and Jxx♥ ; RHO has 5 irrelevant cards and K♠
d)LHO has 2 irrelevant card and both K♠ and Jxx♥ ; RHO has 6 irrelevant cards

We already know that a) and d) are not the case since we saw 3 irrelevant cards from both of them (played to clubs).
We also know that b) and c) are symmetrical and equally likely.
How does information about RHO being able to play K♠ come to play?

My understanding is that if they agree not to play K♠ ever then it's just a guess on actual layout but it wouldn't be a guess if layout a) occurred. RHO had to play K♠ then and we play him for Jxx of hearts too. RHO can try to some mixed strategy of throwing K♠ from layout c) but then every time he plays irrelevant cards that layout becomes less likely (because his strategy consists of throwing K♠ from it sometimes)

In other words:
If we always play RHO for Jxxx of hearts we succeed every time against layout a), c), d) and lose to layout b)
If we always play LHO for Jxxx of hearts we succeed against layouts b) and d).
So if we just close our eyes (despite looking for K♠) and play RHO for Jxxx ♥ we succeed more often so we should do so.
Opponent may play in such a way to give us exactly 50-50 flip on this exact hand (by refusing to play K♠ no matter what) but for using that strategy they would pay by giving us 100% in case a).

Are current bridge playing programs capable of recognizing such situations ?

[cherdano]

I think there is nothing to add to Han's and bluecalm's explanations, but let me try an intuitive reasoning:

You should always ignore information that the defenders volunteered.

Here we should ignore the information that LHO has 5 diamonds - he could also have thrown spades instead if he has a singleton heart. But we should also ignore the fact that we know RHO does not have ♠K ♥Jxxx - as Andy pointed out, if RHO has ♠K and ♥x he is free to choose to provide us with this information, or not to provide us with this information.

[fred]

han, on 2012-February-01, 07:27, said:

It's not clear that that explanation is correct (what does it mean that the information was volunteered? If LHO has Jxxx of hearts his discards are forced), I'm convinced that the conclusion is correct.

Fred has posted similar threads in which the underlying question was similar: which of the plays by the opponents have impact on the odds, and which do not? Here the discards are not impacting the odds, but I think it requires some thought. Perhaps Fred has found a neat way to think about this though.

The way I came to the same conclusion as Han involved no real math. To me it is qualifies as "neat", but I am not sure if the mathematicians out there would consider this reasoning to be valid. Valid or not, like Han I am also confident that the conclusion is correct.

The problem that I posted was the same as the one I was given, but let's simplify things by changing the spades to Ax opposite xx (this changes the relative likelyhood of which defender has Jxxx of hearts, but it doesn't change the basic reasoning or the conclusion).

Ignoring LHO's possible discarding strategies for a minute, it is clearly more likely that RHO has the Jxxx of hearts. The simplest way to see this involves no more than realizing that LHO is known to be longer than RHO in diamonds and all other information is neutral so RHO is more likely to be the defender with longer hearts.

Now let's stop ignoring LHO's possible discarding strategies and ask ourselves why he would discard his diamonds. If he was 2452 then he had no choice in the matter, but if he was 5152 he gave us information that he didn't need to (because he could have discarded spades instead). Why would he do this? One possible answer is that he didn't really think about what he was doing, but it is also possible that a strong LHO might discard this way intentionally from 5152 trying to make it look like he had 2452 instead.

Several posters basically said the same thing in their analysis - well done by them.

But let's go a step further. If LHO is a super-strong player (and gives declarer credit for being able to get this far), he might switch horses and discard 3 spades from 5152. Now declarer is left with a complete guess, but declarer might then ask himself "if my super-strong LHO is actually 5152, why didn't he discard his diamonds in order to make it look like he is 2452? Maybe he is really 4432 and had no choice about his discards". Of course that would leave RHO with 5152 and then the same the same question could be asked about him (if he is also super-strong).

This sort of spy versus spy game can go on forever - it is one of those "he knows that I know that he knows that I know..." type of situations. The only reasonable way for declarer to attempt to win this game is not to play it. He should not get involved in trying to figure out how many rounds of bluff and double bluff a super-strong LHO might have considered and just use the actual information that he has.

That means going back to square one and playing RHO for Jxxx of hearts.

Fred Gitelman
Bridge Base Inc.
www.bridgebase.com

[gnasher]

I'm going to stick my neck out and say that Fred's reasoning is invalid.

If this is a theoretical problem, LHO will treat all his diamonds and small spades as equivalent, and discard them randomly. Throwing ♦6, ♦8 and ♦10 is exactly equivalent to throwing ♠4, ♠9 and ♦7. LHO has three suits:
- Small pointed cards
- Hearts
- King of spades
The particular three small pointed cards that LHO chooses to throw have no bearing on his length in the other two suits.

If this is a real-world problem, we should try to outthink them, obviously.


[Edited because I very briefly posted some complete nonsense. I hope no one saw it.]

This post has been edited by gnasher: 2012-February-01, 10:31

[han]

gnasher, on 2012-February-01, 10:27, said:

If this is a real-world problem, we should try to outthink them, obviously.

I absolutely agree with this. Especially if I think that I'm better at this than my opponents I should be willing to play this spy vs spy game against them.

Your idea that the remaining low spades and diamonds should be considered as equivalent is a very nice way of thinking about the problem and makes the solution obvious. I regret I didn't think about it earlier.

[JLOGIC]

Very smart people in this thread. Great read, enjoyed it.

[bluecalm]

Quote

Your idea that the remaining low spades and diamonds should be considered as equivalent is a very nice way of thinking about the problem and makes the solution obvious. I regret I didn't think about it earlier.

Well I did think about it in my latest try post thanks exactly to your previous posts and before I read them I didn't see the problem at all

I think Fred's reasoning is invalid too. This:

Quote

The simplest way to see this involves no more than realizing that LHO is known to be longer than RHO in diamonds and all other information is neutral so RHO is more likely to be the defender with longer hearts.

Is incorrect. If we are that naive LHO will just discard diamonds from 2-4-5-2 but spades from 5-1-5-2 and we will be outplayed badly. Or in similar hands where diamonds where 4-4 only defender with Jxxx discards a diamond thus fooling us if use that information.
I think that if spades are xx to Ax it's pure 50-50 guess.

Quote

You should always ignore information that the defenders volunteered.

I tried this but while it seems to be good general rule I think there will often be problem interpreting what information was volunteered and which wasn't.

[fred]

bluecalm, on 2012-February-01, 13:36, said:

If we are that naive LHO will just discard diamonds from 2-4-5-2 but spades from 5-1-5-2 and we will be outplayed badly. Or in similar hands where diamonds where 4-4 only defender with Jxxx discards a diamond thus fooling us if use that information.
I think that if spades are xx to Ax it's pure 50-50 guess.

I am now (mostly) convinced that I had lost my mind (hopefully only on a temporary basis) in my earlier thinking about this problem. Thanks to those who helped to point this out with the above quote by Bluecalm really getting the job done for me.

Fred Gitelman
Bridge Base Inc.
www.bridgebase.com