Given to me by a much better player

[gnasher]

Talking of card-reading, what clubs did West have? And do we think he would have bid over a strong 1♣ with something like xx xx KJxxx J10xx ? I know I would (at most vulnerabilities), but in this respect I'm not necessarily mainstream.

 

Edit: Who had ♦9 and 8? KJ98x is a more attractive overcall than KJ732.

Edited April 6, 2010 by gnasher

[dake50]

I like 4xh(drop 2xD), SKQA (see SJ drop or who has it), CAK, 5thH. Hope D10 +S10 +4thC scares someone.

[xcurt]

* The other 5 cards are either {SJ, DHHxx}, {SJ, DHxxx}, or {DHHxxx}.

There are only two cards left so west can not have started with ♠J ♦KJxx as then a diamond honour would have been played.

In this position (west has HHxx) the DJ is a "small" diamond -- West's choice isn't restricted.

[Mbodell]

* The other 5 cards are either {SJ, DHHxx}, {SJ, DHxxx}, or {DHHxxx}.

There are only two cards left so west can not have started with ♠J ♦KJxx as then a diamond honour would have been played.

In this position (west has HHxx) the DJ is a "small" diamond -- West's choice isn't restricted.

Yes, but West has only 2 cards at this point so he can't still hold ♠J and ♦KJ was the point. East holds at least 1 of these 3 cards when we start T12.

 

If we consider what we know about East's possible starting shape we get 3=4=4=2 or 4=4=3=2. A priori, knowing nothing else about anyone's hands, obviously these are equally likely. But West's possible starting shape is 3=2=4=4 or 2=2=5=4 respectively. A priori, the first is more likely. That means the drop will be favored all else equal.

 

The 4=4=3=2 opposite 2=2=5=4 is 60% as likely as 3=2=4=4 opposite 3=4=4=2 just considering the shapes (a 3-3 split to a 4-2 is 75% and a 4-4 to a 5-3 is 80% and the product is 60%; confirm as there are 189,000 East 4=4=3=2 compared to 315,000 East 3=4=4=2).

 

So this would make the drop 62.5% likely.

 

But now we have to figure out is this influenced by other things like choice of lead and discards, and were both the diamond K and J "special" cards that couldn't be played.

 

If West was 3=2=4=4 then they have the ♠J and one of the diamond honors in which case they might well have known they needed to keep Jxx of spades (in case declare finesses in spades) and thus pitched low diamonds. So this is ok and I don't think is cause for discounting. Similarly East would have 3 small spades and a diamond honor, and given the diamond pitches on the run of the hearts I think this would be consistent also.

 

If West was 2=2=5=4 then again I think they can work out to pitch the diamonds and keep the spades to not give up a marked finesse given the play. And here obviously east needs to keep all their spades from Jxxx so again I think this is no need for discount.

 

If we count the diamond honors and spade J as special in our count of the hands though and ask how many ways are there for West to hold Jxx xx Hxxx xxxx we get 90,000. And for West to hold xx xx HHxxx xxxx we get 45,000 (basically the x's are all the same so because the H in Hxxx could be Kxxx or Jxxx there are twice as many Hxxx as HHxxx hands and there are exactly as many Jxx hands as xx once you are placing the JHH and shuffling the rest). So in this analysis the drop is favored 2:1.

 

Since probability is hard and it is sometimes hard to know what you can take as useful fixed information versus just uninteresting revealed information (think restricted choice and monty hall unintuitive results) I'm not 100% sure which of the above numbers gives the "right" answer (2:1 in favor of drop or 5:3 in favor of the drop, I think it is the 5:3 number), but I think drop is clearly the right answer.

[Mbodell]

The 4=4=3=2 opposite 2=2=5=4 is 60% as likely as 3=2=4=4 opposite 3=4=4=2 just considering the shapes  (a 3-3 split to a 4-2 is 75% and a 4-4 to a 5-3 is 80% and the product is 60%; confirm as there are 189,000 East 4=4=3=2 compared to 315,000 East 3=4=4=2).

 

So this would make the drop 62.5% likely.

I guess even considering the ♠J in these we know that the J can't be in the 2 card in the 4-2 split case so the 189,000 is multiplied by 2/3 to get 126,000. And in the 3-3 case we know the J is in the West hand so that eliminates half the combinations in that case to give 157,500 of these. So the drop is only favored at 5:4 once we remember to adjust for the J. So the drop wins ~56% of the time.

 

I'm still playing for the drop.

[Cascade]

* The other 5 cards are either {SJ, DHHxx}, {SJ, DHxxx}, or {DHHxxx}.

There are only two cards left so west can not have started with ♠J ♦KJxx as then a diamond honour would have been played.

In this position (west has HHxx) the DJ is a "small" diamond -- West's choice isn't restricted.

If west started KJxx(x) then after the queen has gone the king and the jack are equals.

 

You can discard one of them but not both.

 

This is not the case with small cards so the jack is not equivalent to a small card.

 

I am not sure what this means for counting the possible distributions.

[Mbodell]

So I ran 2 simulations of 1,000,000 matching hands to get the numbers for what is best. If you assume that you haven't seen either of the diamond honors then the drop wins in around 52.42% of the time [this is pretty close to the 11:10 number that xcurt calculated]. If you assume nothing about the diamond honors (which is obviously fine as both will not have dropped no matter how you arrange them so it isn't like not making this assumption is unrealistic as in none of these is the ♦T high) then the drop wins around 55.59% of the time [this is pretty close to the 5:4 number that I got once I corrected for the spade jack in my last post].

 

Does anyone want to tell us what was the full hand at the table?

 

I'm also including my deal code in case anyone sees an obvious mistake.

 

Not caring about the diamonds (~55.6%):

south is "KQT2 Q2 AQ2 AQ32"
north is "A43 AKJT3 T3 K54" [space]
sdev drop [space]
main { [space] 
[space] [space]set cw [clubs west] [space] 
[space] [space]set dw [diamonds west] [space] 
[space] [space]set hw [hearts west] [space] 
[space] [space]set sw [spades west]
[space] [space]if {$cw!=4} { reject }
[space] [space]if {$sw < 2 || $sw > 3} { reject }
[space] [space]if {$hw != 2} { reject }
[space] [space]if {$sw == 2 && [west has JS]} { reject }
[space] [space]if {$sw == 3 && [east has JS]} { reject }
[space] [space]accept 
} [space]

proc write_deal {} { [space] 
[space] [space]variable drop
[space] [space]if {[west has JS]} {
[space] [space] [space] [space]drop add 1 [space] 
[space] [space]} else {
[space] [space] [space] [space]drop add 0 
[space] [space]} 
# [space]formatter::write_deal 
# [space]puts stdout "================================="
} [space]

deal_finished { [space] 
[space] [space]puts "drop works [drop average] percent of the time" 
}

 

 

Caring about the diamonds (~52.4%):

south is "KQT2 Q2 AQ2 AQ32"
north is "A43 AKJT3 T3 K54" [space]
sdev drop [space]
main { [space] 
[space] [space]set cw [clubs west] [space] 
[space] [space]set dw [diamonds west] [space] 
[space] [space]set hw [hearts west] [space] 
[space] [space]set sw [spades west]
[space] [space]if {$cw!=4} { reject }
[space] [space]if {$sw < 2 || $sw > 3} { reject }
[space] [space]if {$hw != 2} { reject }
[space] [space]if {$sw == 2 && [west has JS]} { reject }
[space] [space]if {$sw == 3 && [east has JS]} { reject }
[space] [space]if {$dw == 4 && [west has KD] && [west has JD]} { reject } 
[space] [space]if {$dw == 5 && [east has KD] && [east has JD]} { reject }
[space] [space]accept 
} [space]

proc write_deal {} { [space] 
[space] [space]variable drop
[space] [space]if {[west has JS]} {
[space] [space] [space] [space]drop add 1 [space] 
[space] [space]} else {
[space] [space] [space] [space]drop add 0 
[space] [space]} 
# [space]formatter::write_deal 
# [space]puts stdout "================================="
} [space]

deal_finished { [space] 
[space] [space]puts "drop works [drop average] percent of the time" 
}

[gnasher]

If west started KJxx(x) then after the queen has gone the king and the jack are equals.

 

You can discard one of them but not both.

 

This is not the case with small cards so the jack is not equivalent to a small card.

 

I am not sure what this means for counting the possible distributions.

If West started with ♠Jxx and ♦K, can he infer from the play that South doesn't have ♦J, and therefore throw a deceptive king to look like a man with ♠xx ♦KJ? I think this is too hard for most of us, so West will always throw a low diamond from this holding.

 

Therefore, from ♠Jxx ♦J West should throw the jack, to make it indistinguishable from ♠xx ♦KJ.

 

Similar arguments apply to East: with ♠xxx and ♦K, he will hold onto ♦K. Hence with ♠xxx and ♦KJ he should throw the jack not the king, and with ♠xxx and ♦Jxx he should also throw the jack.

[Cascade]

So I ran 2 simulations of 1,000,000 matching hands to get the numbers for what is best. If you assume that you haven't seen either of the diamond honors then the drop wins in around 52.42% of the time [this is pretty close to the 11:10 number that xcurt calculated]. If you assume nothing about the diamond honors (which is obviously fine as both will not have dropped no matter how you arrange them so it isn't like not making this assumption is unrealistic as in none of these is the ♦T high) then the drop wins around 55.59% of the time [this is pretty close to the 5:4 number that I got once I corrected for the spade jack in my last post].

 

Does anyone want to tell us what was the full hand at the table?

 

I'm also including my deal code in case anyone sees an obvious mistake.

 

Not caring about the diamonds (~55.6%):

south is "KQT2 Q2 AQ2 AQ32"
north is "A43 AKJT3 T3 K54" [space]
sdev drop [space]
main { [space] 
[space] [space]set cw [clubs west] [space] 
[space] [space]set dw [diamonds west] [space] 
[space] [space]set hw [hearts west] [space] 
[space] [space]set sw [spades west]
[space] [space]if {$cw!=4} { reject }
[space] [space]if {$sw < 2 || $sw > 3} { reject }
[space] [space]if {$hw != 2} { reject }
[space] [space]if {$sw == 2 && [west has JS]} { reject }
[space] [space]if {$sw == 3 && [east has JS]} { reject }
[space] [space]accept 
} [space]

proc write_deal {} { [space] 
[space] [space]variable drop
[space] [space]if {[west has JS]} {
[space] [space] [space] [space]drop add 1 [space] 
[space] [space]} else {
[space] [space] [space] [space]drop add 0 
[space] [space]} 
# [space]formatter::write_deal 
# [space]puts stdout "================================="
} [space]

deal_finished { [space] 
[space] [space]puts "drop works [drop average] percent of the time" 
}

 

 

Caring about the diamonds (~52.4%):

south is "KQT2 Q2 AQ2 AQ32"
north is "A43 AKJT3 T3 K54" [space]
sdev drop [space]
main { [space] 
[space] [space]set cw [clubs west] [space] 
[space] [space]set dw [diamonds west] [space] 
[space] [space]set hw [hearts west] [space] 
[space] [space]set sw [spades west]
[space] [space]if {$cw!=4} { reject }
[space] [space]if {$sw < 2 || $sw > 3} { reject }
[space] [space]if {$hw != 2} { reject }
[space] [space]if {$sw == 2 && [west has JS]} { reject }
[space] [space]if {$sw == 3 && [east has JS]} { reject }
[space] [space]if {$dw == 4 && [west has KD] && [west has JD]} { reject } 
[space] [space]if {$dw == 5 && [east has KD] && [east has JD]} { reject }
[space] [space]accept 
} [space]

proc write_deal {} { [space] 
[space] [space]variable drop
[space] [space]if {[west has JS]} {
[space] [space] [space] [space]drop add 1 [space] 
[space] [space]} else {
[space] [space] [space] [space]drop add 0 
[space] [space]} 
# [space]formatter::write_deal 
# [space]puts stdout "================================="
} [space]

deal_finished { [space] 
[space] [space]puts "drop works [drop average] percent of the time" 
}

Don't you need this:

 

if {$dw == 5 && [east has KD] && [east has JD]} { reject }

 

in the first case.

 

Since if west has 5 diamonds and east has both honours the ♦10 is good.

 

And don't you need

 

if {$dw == 5 && ([east has KD] || [east has JD])} { reject }

 

in the second case.

 

Since if west has five diamonds then east will be forced to throw an honour to guard the spades.

 

The first case gave me 58.3%

 

The second case gave me 73.2%

[fred]

Does anyone want to tell us what was the full hand at the table?

West held:

 

Jxx

xx

Kxxx

J10xx

 

Fred Gitelman

Bridge Base Inc.

www.bridgebase.com

[aguahombre]

Life is not fair. As the cards lie, any shluck who decides to play for either:

 

-diamond king on side,

-3-3 spades,

-the spade Jack drop

-righty has any 5+ in spades

 

+thinking because of the 4-2 heart split he has to forget about the extra chance of clubs splitting and pitches one on the fourth heart; knowing that squeezes give him a headache ---would bring it home.

 

(Edit) Actually declarer might accidentally fall into a pop squeeze if righty has four spades and the diamond king was stiff in lefty's hand.

[xcurt]

I worked through most of the game theory of this, but it's too much to compute the exact Nash equilibrium because it's dependent on the falsecarding (unnecessarily pitching an honor) strategy of both East and West -- and they cannot both falsecard at the same time lest the DT become good. Nonetheless, the conclusion that emerged was that against most mortal defenders who probably don't falsecard enough, if no honors appear before T12 the finesse is probably the better play at that point.

 

So I now think Fred's teammate made the "right" play even though it didn't work in practice.

 

I know this sounds weird since if declarer closed his eyes and played for the drop he would get favorable odds, but that's the Monty Hall problem in a nutshell -- two people can get different odds calling for the same door, if they have different information at the decision point.

[Mbodell]

I worked through most of the game theory of this, but it's too much to compute the exact Nash equilibrium because it's dependent on the falsecarding (unnecessarily pitching an honor) strategy of both East and West -- and they cannot both falsecard at the same time lest the DT become good. Nonetheless, the conclusion that emerged was that against most mortal defenders who probably don't falsecard enough, if no honors appear before T12 the finesse is probably the better play at that point.

I know this sounds weird since if declarer closed his eyes and played for the drop he would get favorable odds, but that's the Monty Hall problem in a nutshell -- two people can get different odds calling for the same door, if they have different information at the decision point.

That seems wrong to me. The numbers where you don't know which diamonds they play favor the drop. The numbers where you know they didn't play an honor favor the drop even more strongly. Thus I think you have to play for the drop unless you are thinking they false card a lot (and even then it may still be right to drop).

[xcurt]

I worked through most of the game theory of this, but it's too much to compute the exact Nash equilibrium because it's dependent on the falsecarding (unnecessarily pitching an honor) strategy of both East and West -- and they cannot both falsecard at the same time lest the DT become good.  Nonetheless, the conclusion that emerged was that against most mortal defenders who probably don't falsecard enough, if no honors appear before T12 the finesse is probably the better play at that point.

I know this sounds weird since if declarer closed his eyes and played for the drop he would get favorable odds, but that's the Monty Hall problem in a nutshell -- two people can get different odds calling for the same door, if they have different information at the decision point.

That seems wrong to me. The numbers where you don't know which diamonds they play favor the drop. The numbers where you know they didn't play an honor favor the drop even more strongly. Thus I think you have to play for the drop unless you are thinking they false card a lot (and even then it may still be right to drop).

A priori there are Choose[9,5] (126) ways to give West 5 of the {8 diamonds and the SJ}. He can't have 5 small diamonds (6 cases, the hand plays itself on a showup line), and he can't have {SJ, Dxxxx} (15 cases, East would have been forced to play an honor). The remaining 105 cases are 55:50 for the drop.

 

As someone pointed out, if West always truecards (meaning plays an honor only if forced), and he plays a spot, that eliminates the 15 cases where West has SJ, DHHxx, making it 40:50 for the drop (ie 5:4 for the finesse).

 

A true-carding East playing a spot eliminates the possibility that West held SJ, Dxxxx (15 cases), also making it 5:4 for the finesse.

[Cascade]

I worked through most of the game theory of this, but it's too much to compute the exact Nash equilibrium because it's dependent on the falsecarding (unnecessarily pitching an honor) strategy of both East and West -- and they cannot both falsecard at the same time lest the DT become good. Nonetheless, the conclusion that emerged was that against most mortal defenders who probably don't falsecard enough, if no honors appear before T12 the finesse is probably the better play at that point.

This conclusion seems wrong to me. That is the conclusion about the finesse and not the conclusion about the false card.

 

Either

 

1. West is 2=2=5=4 with East 4=4=3=2

 

In this case east is going to have to pitch all three diamonds in the two card ending therefore:

 

i/ the ♠J has dropped doubleton

 

ii/ east has ♦ KJx and is squeezed

 

iii/ east has ♦ Kxx or ♦ Jxx and an honour has appeared

 

iv/ east has ♦ xxx and west has ♦ KJxxx in which case west has an easy false card. I think this false card is mandatory. After the ♦ Q has gone it is straightforward for west to realize the diamond honours are equals and therefore to pitch one giving the illusion of fewer diamonds and more spades than actually held.

 

or

2. West is 3=2=4=4 with East 3=4=4=2

 

The relevant diamond holdings are

 

i/ east ♦ KJxx and an honour would have appeared

 

ii/ east ♦ Jxxx and east has the option of false carding the jack giving the illusion of holding ♦ Jxx and hence long spades - this false card is easy since partner is known to have the ♦ K when declarer throws the queen from hand.

 

iii/ east ♦ Kxxx and in theory has the option of false carding the king giving the illusion of ♦ Kxx. This false card is easy too when ten is in dummy. It is virtually impossible that south has the ♦ J and cashed the ace and pitched small and queen from AQJx.

 

iv/ east has ♦ xxxx and west would have produced an honour.

 

The false cards ii/ and iii/ also appear mandatory to me in that they can't lose except against insane declarer play (cashing ace from AQJx) but they might gain.

 

Since no honour has appeared we either have west with ♦ KJxxx or east with ♦ Kxxx or east with ♦ Jxxx with the player in the named position having failed to see the false card potential.

 

Kxxx or Jxxx with east seems more likely than KJxxx with west so that the drop is favoured.

 

Perhaps in some situations declarer can judge which defender is most likely to have failed to false card.

[Cascade]

I worked through most of the game theory of this, but it's too much to compute the exact Nash equilibrium because it's dependent on the falsecarding (unnecessarily pitching an honor) strategy of both East and West -- and they cannot both falsecard at the same time lest the DT become good.  Nonetheless, the conclusion that emerged was that against most mortal defenders who probably don't falsecard enough, if no honors appear before T12 the finesse is probably the better play at that point.

I know this sounds weird since if declarer closed his eyes and played for the drop he would get favorable odds, but that's the Monty Hall problem in a nutshell -- two people can get different odds calling for the same door, if they have different information at the decision point.

That seems wrong to me. The numbers where you don't know which diamonds they play favor the drop. The numbers where you know they didn't play an honor favor the drop even more strongly. Thus I think you have to play for the drop unless you are thinking they false card a lot (and even then it may still be right to drop).

A priori there are Choose[9,5] (126) ways to give West 5 of the {8 diamonds and the SJ}. He can't have 5 small diamonds (6 cases, the hand plays itself on a showup line), and he can't have {SJ, Dxxxx} (15 cases, East would have been forced to play an honor). The remaining 105 cases are 55:50 for the drop.

9C5 = 126

 

West's holdings

 

♦ xxxxx = 6 cases - squeeze

 

♦ Hxxxx = 2 (honours) x 6C4 = 30 cases - an honour showed from east with Hxx

 

♦ HHxxx = 6C3 = 20 cases (west might have false carded)

 

♠ J ♦ xxxx = 6C4 = 15 cases - east must have played an honour

 

♠ J ♦ Hxxx = 2 (honours) x 6C3 = 40 cases (east might have false carded)

 

♠ J ♦ HHxx = 6C2 = 15 cases - west must have played an honour.

 

Without any information the drop is:

 

70:50 on (eliminating the 6 cases where we made on a show up squeeze)

 

If they never false card then there are only two possibilities west HHxxx or west J Hxxx

 

and the drop is:

 

40:20 on

[Cascade]

So I ran 2 simulations of 1,000,000 matching hands to get the numbers for what is best.  If you assume that you haven't seen either of the diamond honors then the drop wins in around 52.42% of the time [this is pretty close to the 11:10 number that xcurt calculated].  If you assume nothing about the diamond honors (which is obviously fine as both will not have dropped no matter how you arrange them so it isn't like not making this assumption is unrealistic as in none of these is the ♦T high) then the drop wins around 55.59% of the time [this is pretty close to the 5:4 number that I got once I corrected for the spade jack in my last post].

 

Does anyone want to tell us what was the full hand at the table?

 

I'm also including my deal code in case anyone sees an obvious mistake.

 

Not caring about the diamonds (~55.6%):

south is "KQT2 Q2 AQ2 AQ32"
north is "A43 AKJT3 T3 K54" [space]
sdev drop [space]
main { [space] 
[space] [space]set cw [clubs west] [space] 
[space] [space]set dw [diamonds west] [space] 
[space] [space]set hw [hearts west] [space] 
[space] [space]set sw [spades west]
[space] [space]if {$cw!=4} { reject }
[space] [space]if {$sw < 2 || $sw > 3} { reject }
[space] [space]if {$hw != 2} { reject }
[space] [space]if {$sw == 2 && [west has JS]} { reject }
[space] [space]if {$sw == 3 && [east has JS]} { reject }
[space] [space]accept 
} [space]

proc write_deal {} { [space] 
[space] [space]variable drop
[space] [space]if {[west has JS]} {
[space] [space] [space] [space]drop add 1 [space] 
[space] [space]} else {
[space] [space] [space] [space]drop add 0 
[space] [space]} 
# [space]formatter::write_deal 
# [space]puts stdout "================================="
} [space]

deal_finished { [space] 
[space] [space]puts "drop works [drop average] percent of the time" 
}

 

 

Caring about the diamonds (~52.4%):

south is "KQT2 Q2 AQ2 AQ32"
north is "A43 AKJT3 T3 K54" [space]
sdev drop [space]
main { [space] 
[space] [space]set cw [clubs west] [space] 
[space] [space]set dw [diamonds west] [space] 
[space] [space]set hw [hearts west] [space] 
[space] [space]set sw [spades west]
[space] [space]if {$cw!=4} { reject }
[space] [space]if {$sw < 2 || $sw > 3} { reject }
[space] [space]if {$hw != 2} { reject }
[space] [space]if {$sw == 2 && [west has JS]} { reject }
[space] [space]if {$sw == 3 && [east has JS]} { reject }
[space] [space]if {$dw == 4 && [west has KD] && [west has JD]} { reject } 
[space] [space]if {$dw == 5 && [east has KD] && [east has JD]} { reject }
[space] [space]accept 
} [space]

proc write_deal {} { [space] 
[space] [space]variable drop
[space] [space]if {[west has JS]} {
[space] [space] [space] [space]drop add 1 [space] 
[space] [space]} else {
[space] [space] [space] [space]drop add 0 
[space] [space]} 
# [space]formatter::write_deal 
# [space]puts stdout "================================="
} [space]

deal_finished { [space] 
[space] [space]puts "drop works [drop average] percent of the time" 
}

Don't you need this:

 

if {$dw == 5 && [east has KD] && [east has JD]} { reject }

 

in the first case.

 

Since if west has 5 diamonds and east has both honours the ♦10 is good.

 

And don't you need

 

if {$dw == 5 && ([east has KD] || [east has JD])} { reject }

 

in the second case.

 

Since if west has five diamonds then east will be forced to throw an honour to guard the spades.

 

The first case gave me 58.3%

 

The second case gave me 73.2%

Whoops I just noticed that in the second case we also need:

 

if {$dw == 4 && [east has KD] && [east has JD]} { reject }

 

Since both west with HHxx and west with xxxx will lead to a diamond honour being played.

 

Now in the first case

 

58.3% is close enough to 70/120 and

 

after correcting my simulation the second case gives

 

66.7% which is close enough to 40/60

[nige1]

Thank you, Cascade. Please would you explain how to run your code?

[Cascade]

Thank you, Cascade. Please would you explain how to run your code?

That code is not mine.

 

It is using a program called Deal I think - you can google it.

 

I usually do my simulations with dealer.exe which is the same program that BBO uses for hand selection in the web version.

[gnasher]

It's available from here:

Thomas Andrew's Deal