Best Play, A, Q or T

[inquiry]

[hv=n=saqtxx&w=s&e=s&s=sxxxxx]399|300|You need to play for no losers in this trump suit, when you lead a small one, WEST plays the remaining x (leaving KJ out). East has a known 7 card or longer side suit.

 

What is the best percentage play for no loser?

 

Your side had a total of 2 spades... sorry about that. [/hv]

[cherdanno]

I think the question is more fun when we know the spade suit divides 4=6 (with East having 6)...

[inquiry]

I think the question is more fun when we know the spade suit divides 4=6 (with East having 6)...

corrction to problem made...sorry,... too lazy to post full hand caused the slip of giving spade count

[MFA]

Go deep.

 

T and Q handle one specific trump distribution each, but the one that will make the side suits more evenly distributed will also be more likely in total when we consider the whole hand.

 

If trumps are 3-0, the two remaining side suits will be divided 6-6 in total.

If trumps are 2-1, the two remaining side suits will be divided 7-5 in total.

(Given spades 4-7)

 

So we play for 3-0 trumps.

[MFA]

Btw, I thnk that cherdanno tried to hint that if spades were 4-6 instead of 4-7, it would be a complete toss-up to play Ace, Queen or Ten!

Apart from the fact that we should not play the ace, because a cunning defender might falsecard J from Jx, so when the J doesn't appear the likelihood of Jx should be discounted.

[manudude03]

12:6 left in vacant spaces, 2 left in the suit. The average break from here is 1.33:0.67. Ignoring the possibility of an original 1-2 for now, this means that it was breaking 2-1 2/3 of the time and 3-0 1/3 of the time. However the possibility of 1:2 would lower the average and so that means that 1.33:0.67 is too low (for LHS) for what we have to do. This would now mean that the 3-0 is more likely than either of the 2-1s.

 

Therefore I think percentage is to play the 10.

 

Edit: This assumes that OP did mean East had 7 spades or LONGER (ie. we know nothing about the West hand)

[G_R__E_G]

According to Richard Pavlicek's suit calculator, if we give East 6 vacant spaces and give West 13 vacant spaces (which pushes the odds as far as you can to a 3=0 break) then the numbers are 48.3% for a 2=1 break and 29.5% for a 3=0 break (0=3 break is now impossible and a 1=2 break can be ignored). Even if we eliminiate 1/3 of the 2=1 breaks since KJ/x is no longer possible, that still leaves the 2=1 break as being slightly more likely. Am I missing something?

[rogerclee]

According to Richard Pavlicek's suit calculator, if we give East 6 vacant spaces and give West 13 vacant spaces (which pushes the odds as far as you can to a 3=0 break) then the numbers are 48.3% for a 2=1 break and 29.5% for a 3=0 break (0=3 break is now impossible and a 1=2 break can be ignored). Even if we eliminiate 1/3 of the 2=1 breaks since KJ/x is no longer possible, that still leaves the 2=1 break as being slightly more likely. Am I missing something?

There you go

[jdonn]

Greg wins. But wait.

 

48.3% for a 2-1 break.

 

There is restricted choice on Jx (they are equals here really), so the following will summarize:

 

16.1% west has Kx and plays x.

16.1% west has KJ and plays J.

8.05% west has Jx and plays x.

8.05% west has Jx and plays J.

 

So KJx is a 29.5% to 16.1% favorite over Kx, no?

 

(previous nonsense deleted)

[manudude03]

According to Richard Pavlicek's suit calculator, if we give East 6 vacant spaces and give West 13 vacant spaces (which pushes the odds as far as you can to a 3=0 break) then the numbers are 48.3% for a 2=1 break and 29.5% for a 3=0 break (0=3 break is now impossible and a 1=2 break can be ignored).  Even if we eliminiate 1/3 of the 2=1 breaks since KJ/x is no longer possible, that still leaves the 2=1 break as being slightly more likely.  Am I missing something?

Unless I'm missing something, you still have to guess whether to drop the stiff K or stiff J offside.

[mgoetze]

Actually West has 12 empty spaces, as we already know he has x in the trump suit. East has 6 empty spaces. The number of spades we hold is irrelevant.

 

For the moment, I'll assume West will always play x if he has it. Then we have

 

W    E    Prob             Play
KJx  -    (12*11)/(18*17)  10
Jx   K    (6*12)/(18*17)   A
Kx   J    (12*6)/(18*17)   Q
x    KJ   (6*5)/(18*17)    none

 

i.e. the 10 will be right in 11/23 cases where it is possible to pick up the suit for no losers, and the A and Q get 6/23 each.

 

Under Josh's theory that West will play J 50% of the time if he has both J and x, Kx and x become twice as likely, therefore Q is right in 12/29 cases versus 11/29 for 10 and 6/29 for A.

 

If West plays the J from Jx just a little under 50% of the time, it becomes a total tossup between 10 and Q.

 

That said, I'm going to vote for the 10 anyway. B)

[inquiry]

This hand is from a recent ACBL Speedball, Matchpoints, on BBO.

 

I was reviewing the hands of a person who wanted advice from me about the strength and weakness of their game. Here they played low to the queen and lost a trump trick when west showed up with KJx.

 

I reviewed the play of others, and found only two people got the trump suit right (assuming low to the TEN is right, which I am as certain as I can be that it is). One of the people who got it right was a very poor player based on their average imp and matchpoint scores alone (I don't know this person). The other who got it right is a player known to me to be very good. For what it is worth, at least two people who I know have gold stars got this hand wrong.

 

Since a whole lot of people got this position wrong (it was matchpoints, and you had to get this right to make slam, but even at game, you would want to get it right). Due to the "poor" performance (at least imho) on this hand, I thought I would post it. Here, we are doing sooooo much better than those at the table, but that was Speedball and not really presented as a standalone problem like here.

[G_R__E_G]

According to Richard Pavlicek's suit calculator, if we give East 6 vacant spaces and give West 13 vacant spaces (which pushes the odds as far as you can to a 3=0 break) then the numbers are 48.3% for a 2=1 break and 29.5% for a 3=0 break (0=3 break is now impossible and a 1=2 break can be ignored).  Even if we eliminiate 1/3 of the 2=1 breaks since KJ/x is no longer possible, that still leaves the 2=1 break as being slightly more likely.  Am I missing something?

Unless I'm missing something, you still have to guess whether to drop the stiff K or stiff J offside.

That's a very good point (although my first thought was to say "Well obviously").

 

Forgetting about the restricted choice ramifications for now, even if you just look at the 2=1 or 3=0 break situation, you will always play the right card when you play for the 3=0 break but you'll only play the right card half the time if you play for a 2=1 break. So the 2=1 break would have to be twice as likely for it to be the correct play. It's actually pretty interesting to look at it if you assume that the opposing spades are 4=7. This would leave West with 8 vacant spaces (he's known to have 1 heart and 4 spades) and East with 6 vacant spaces. This would make the odds of a 2=1 split 46.15%. If you eliminate the KJ/x case then divide by two you get 15.38%. As it turns out, the odds of a 3=0 break with the above parameters is also 15.38%. So, assuming West would ever play the jack from Jx then it tips the scales in favour of playing the ten. And of course, any chance of East having more than 7 spades makes playing the ten even better.

[mgoetze]

(assuming low to the TEN is right, which I am as certain as I can be that it is).

Even if jdonn is sitting west? ;)

[jdonn]

(assuming low to the TEN is right, which I am as certain as I can be that it is).

Even if jdonn is sitting west? ;)

Um, especially if jdonn is sitting west!

[MFA]

It seems that everyone is making the same fatal mistake.

 

We can't just give east 7 spades and then count vacant spaces from there. We need to have a suit fully clarified before we can count it in. *)

 

Essentially you are calculating as if we were told that a suit is breaking 0-7. But what we actually know is that a suit is 4-7 or worse, which is a very different thing.

 

The method of vacant spaces can't handle such kind of imprecise information, so we have to split it up in subcases and look at each spade division at a time.

 

I did that by starting with spades 4-7. I found that spades 4-7 is significant enough for TEN to be right, so there was no need to proceed to spades 3-8 etc., since in those cases the TEN would obviously just be even better.

 

*)

One thing is allowed. If we know everything about a part of a suit, we can count that part in, provided that no player had the free choice to follow to a trick with a card from outside that part.

Typically, we can count in the x's of a suit if (1) we have seen them all and (2) the opps couldn't afford to play an honour instead of an x. If either (1) or (2) is not satisfied the calculation breaks down as 'restricted choice' joins the stage.

[mgoetze]

So apparently, by your analysis, the mistake isn't actually fatal. But actually it's not even a mistake, so that result isn't very surprising.

 

How do you calculate the conditional probability of a 4-7 break, given that East has at least 7? Well, it's quite simple, West has 13 empty spaces and East has 6, and then you distribute 4 spade cards...

 

Do you have any more substantial arguments than "no you can't do that!"?

[cherdanno]

So apparently, by your analysis, the mistake isn't actually fatal. But actually it's not even a mistake, so that result isn't very surprising.

 

How do you calculate the conditional probability of a 4-7 break, given that East has at least 7? Well, it's quite simple, West has 13 empty spaces and East has 6, and then you distribute 4 spade cards...

 

Do you have any more substantial arguments than "no you can't do that!"?

Uuuh Michael, MFA is actually right. It's the reason why dealing with constraints is so tricky unless you do it by brute force.

[fred]

The BBO web-client has a nice facility for exploring how information about the distribution of the hand can impact the odds of how a given suit is divided. Here is how it works:

 

1. Log in

2. Click "My BBO" and then "Hands and results"

3. Click "Hand editor"

4. Enter the suit you are interested in exploring

5. Click the "Options" button below the deal diagram, then "Analyze suit", then the appropriate suit

6. A window will appear. Use the controls in the "Known side suit cards" box to change the information known about the unseen hands and watch the odds change.

7. Try clicking the "More" button associated with a given way the suit could break in order to see the various specfic cases and their odds.

 

The "analyze suit" facility is also available through other areas of the web-client.

 

Fred Gitelman

Bridge Base Inc.

www.bridgebase.com

[MFA]

So apparently, by your analysis, the mistake isn't actually fatal. But actually it's not even a mistake, so that result isn't very surprising.

Cherdanno came to rescue, but I would like to answer your questions anyway.

 

The calculations are flawed and the percentage results presented in this thread are pretty far off. That's why I called it fatal. It is just a coincidence that we reach the same conclusion about the right play.

 

How do you calculate the conditional probability of a 4-7 break, given that East has at least 7? ...

Usually we would examine the bidding and make a subjective estimate. That would be best. But if it is to be a theoretical exercise we could use the a priori numbers for an 11-card suit. My book here has these percentages for the relevant distibutions:

 

4-7 15,88

3-8 4,77

2-9 0,72

1-10 0,05

0-11 0,001

 

We rescale so it adds to 100:

 

4-7 74,14

3-8 22,27

2-9 3,36

1-10 0,23

0-11 0.005

 

... Well, it's quite simple, West has 13 empty spaces and East has 6, and then you distribute 4 spade cards...

(...)

Do you have any more substantial arguments than "no you can't do that!"?

The vacant spaces is a simple model. It assumes that we know about certain cards and then it distributes the rest completely on random. When we just know "something" about a suit, then the model is in big trouble, because that "something" might be hard to fit in correctly.

 

Using the above percentages, we can see why it doesn't work just to 'give' east seven spades to start with and then calculate ahead. When the last four spades are distributed by the model, yes, surely west will get more than two of those four on average, but typically those four remaining spades will be divided 2-2 or 3-1 rather than 4-0.

 

The result is that the vacant spaces-model (as you apply it) is very far from getting the right theoretical spade distribution. There will be way-way too many 3-8 or 2-9 breaks, while 4-7 will be more like an exception rather than the 3/4 occurance it should be.

 

Therefore the output in the end will be wrong.

[cherdanno]

Cherdanno came to rescue, but I would like to answer your questions anyway.

Actually I think of it as coming to mgoetze's rescue (who is a friend from another world^H^H^Hgame), without having the time to write a full explanation.

[jdeegan]

:D The ten is better than the queen by 54-46. East has 8 vacant spaces - 4♠ and one trump . West has 6 vacant spaces - 7 ♠ are known. There are two missing cards - K & J of trump. The odds are 52 to 31 that RHO has a trump, but you will guess correctly only half the time - so you win only 26 times while the ten wins 31 times.

 

Curiously, if you assume that RHO would NEVER preempt with a stiff K of ♠, the odds are the same. Now, LHO has 6 known cards - 4♠ and Kx of trump, or 7 vacant spaces. RHO has the same 6 vacant spaces. Now there is only ONE missing card, the trump J. The odds are the same as before: 54 to 46 in favor of the ten.

 

If you assume that RHO will TEND not to preempt with the stiff K of trump, it favors playing the Q rather than the ace, BUT the ten is still the right play based on a Baysian-type analysis that imposes fractional vacant spaces on East and West.

 

Drs. Friedman and Savage (who posited that humans as a group inuitively select the choice with the best odds (from two choices) as long as the true odds are between 90-10 and 10-90) would be gratified with the results of your poll in which real human beings intuitively reject the worst case (trump ace) and favor the slightly better choice of the ten by 2 to 1. By comparison with their research, your problem is complex with three choices and unstated conditional probabilities (the stiff K issue).

 

James F. Deegan, Ph.D. Econometrician