Suit combo

[pclayton]

Learned a new one today:

 

KQ98532

T7

 

You lead 10, 4, K, A...

 

You get back in and lead the 7, 6, and...?

 

(I think its the same ratio as another better-known combo)

[ArtK78]

All other things being equal (which they rarely are) my initial thought is to play for the opponents' cards in the suit to be divided 2-2.

 

I suppose your argument may be that RHO's play of the Ace on the first round is restricted. This would be true if his Ace were singleton or if he had AJ doubleton. We already know that those are the only two choices available.

 

A priori the chance of the suit being divided 3-1 is roughly 50%. 3-1 with the singleton in RHO's hand is 25%. The singleton being the Ace is 6.25%.

 

A priori the chance of the suit being divided 2-2 is roughly 40%. There are three different combinations of 2 card holdings (AJ, A6 and A4), and each such combination can exist in both RHO's hand or LHO's hand, leading to 6 different combinations. So, a priori, each two-card combination in each hand has roughly a 6.67% chance.

 

Therefore, comparing only the singleton Ace in RHO's hand (6.25% chance a priori) and the doubleton AJ in RHO's hand (6.67% chance a priori) the chance of AJ doubleton being a small favorite over the chance of singleton A.

 

So, unless there is something wrong with my analysis (and I am sure everyone will jump right on it) the correct play is to play for the suit to be divided 2-2 (all other things being equal).

 

Needless to say, with such a small difference in probabilities, any information about the rest of the hand could change the relative likelihood of a 2-2 division of the suit vs. a 3-1 division of the suit. Also, entry considerations may change matters (why is RHO taking the first round of the suit?).

[lexlogan]

With 4 cards outstanding, the suit is more likely to break 3-1 than 2-2 BUT, as with trying to find the Queen with 4 out, once you reach this point the odds slightly favor the drop IF you have absolutely no other clue about the hand and no reason to keep one or the other opponent off the lead. It's about 52% to 50%. In general, with an even number of cards outstanding, the odds of dropping a specific card are slightly better than a finesse. Similar cases include AKQ10 opposite xxx (6 cards out) and AJ10xxx opposite Qxxxx (2 cards out.)

 

A player who always finessed in these cases except when there were clues pointing toward LHO having the missing honor OR he could afford to lose the lead to LHO but not RHO would do far better than one who slavishly followed the "eight ever, nine never" nursery rhyme.

[Trumpace]

With 4 cards outstanding, the suit is more likely to break 3-1 than 2-2 BUT, as with trying to find the Queen with 4 out, once you reach this point the odds slightly favor the drop IF you have absolutely no other clue about the hand and no reason to keep one or the other opponent off the lead. It's about 52% to 50%.

Once LHO follows to the second round, the odd becomes ~52 vs 48 for drop vs finesse.

 

In terms of the vacant spaces argument, one card left (the Jack): 11 spaces with LHO, 12 spaces with RHO.

 

Might be wrong though.

 

Also, I am not able to think of a way to bring Bayes(Restricted choice) into the picture... RHO didn't have a choice and it seems we cannot read anything into the way in which LHO played his spots.

[helene_t]

I would play RHO for the Jack because LHO might have covered the 10 if he held the Jack. This may not be true against strong opps but then we are back to Trumpace's 52-48 anyway. I can't see a reason for RHO to hold up from AJ.

[kenrexford]

Also, I am not able to think of a way to bring Bayes(Restricted choice) into the picture... RHO didn't have a choice and it seems we cannot read anything into the way in which LHO played his spots.

This is interesting.

 

LHO either started with J64 or just 64. If he is a thinking defender, he should play the 6 and 4 randomly both times, it seems. However, it seems like people in practice usually just play up the line with two small. So, there may be a "restricted attention" issue here, where 4-6 suggests doubleton.

 

This is, of course, assuming that this is trumps. In a side suit, some people have a tendency to give count rather frequently, especially if the entry back is not 100% clear. That might change the odds somewhat the other way.

[fred]

The same honor combination with a 6-2 fit is, not surprisingly, much more complicated. It is also very interesting IMO - a computer program I wrote a long time ago "discovered" the answer for me:

 

KQ9832

 

107

 

Fred Gitelman

Bridge Base Inc.

www.bridgebase.com

[ArtK78]

My analysis - 6.67% vs. 6.25% - is also roughly 52/48 as others have noted.

[rbforster]

LHO either started with J64 or just 64. If he is a thinking defender, he should play the 6 and 4 randomly both times, it seems. However, it seems like people in practice usually just play up the line with two small. So, there may be a "restricted attention" issue here, where 4-6 suggests doubleton.

But if the LOL is playing their lowest card each time, you'll see 4...6 from either 64 or J64 so I'm not sure what that tells you. Might be more interesting in a case with 5 missing cards instead of only 4.

[ArtK78]

The same honor combination with a 6-2 fit is, not surprisingly, much more complicated. It is also very interesting IMO - a computer program I wrote a long time ago "discovered" the answer for me:

 

KQ9832

 

107

 

Fred Gitelman

Bridge Base Inc.

www.bridgebase.com

According to an odds table I found on the internet (http://www.automaton.gr/tt/en/OddsTbl.htm) the odds of bringing this suit in for 5 tricks are as follows:

 

Play towards the K, then, regardless of the result, play towards the Q.

 

53.130%

 

Play towards the K, then, regardless of the result, run the 10 on the second round.

 

39.565%

 

Run the 10 on the first round.

 

48.043%

 

This result, which is solely a factor of the likelihood of the lie of the cards, does not take into account whether LHO or RHO ducks the ace on the first round (assuming that they can).

 

By the way, this same odds calculator says that, for the original problem in this thread, the relevant splits of the suit are 6.783% for 2-2 with AJ in RHO's hand, and 6.217% for 3-1 with singleton A in RHO's hand. This is slightly more than 52-48.

[TimG]

The same honor combination with a 6-2 fit is, not surprisingly, much more complicated. It is also very interesting IMO - a computer program I wrote a long time ago "discovered" the answer for me:

 

KQ9832

 

107

Charlie Coon once told me that when both defenders duck the first trick, you should play the better player to hold the Ace.

 

It seems that suitplay and Roudinesco disagree about this holding. I'm not going to embarrass myself by taking sides.

[Guest Jlall]

Didn't read the other answers but play for the drop, AJ tight is more likely than stiff A.

[Guest Jlall]

The same honor combination with a 6-2 fit is, not surprisingly, much more complicated. It is also very interesting IMO - a computer program I wrote a long time ago "discovered" the answer for me:

 

KQ9832

 

107

 

Fred Gitelman

Bridge Base Inc.

www.bridgebase.com

2 possible lines:

 

(A ) low to the king, low to the queen

(B ) run the ten

 

Assuming the objective is 5 tricks, 1 loser then:

 

A beats B on: Axxx, xxx, Axx(2), Ax(2)

B beats A on: AJxx(2), Jxx(2), Jxxx

 

So, A is better by a full combination.

 

Btw, some might wonder why low to the queen then run the ten next is not the correct line. This is because you cannot pick up AJ tight offside (if you plan on playing for the drop if they win the ace, then they can occassionally win the A from Ax, and you are no longer picking up Ax offside always).

[SoTired]

The same honor combination with a 6-2 fit is, not surprisingly, much more complicated. It is also very interesting IMO - a computer program I wrote a long time ago "discovered" the answer for me:

 

KQ9832

 

107

 

Fred Gitelman

Bridge Base Inc.

www.bridgebase.com

I looked at this once a while ago.

Low to K. If this wins, then low to Q. If the K loses to the A, then low to Q spurning finesse. This is counter-intuitive.

[pclayton]

A few weeks ago I won from Ax with KQ98xx in dummy and later lost a trick to pard's Jxx.

 

The declarer told me, I'd thought you'd duck with Ax.

[kgr]

Did I understand this correct?:

1)

KQ9832

 

107

 

2)

KQ983

 

107x

 

In one we play twice to KQ and in 2 we play to K and finesse the 9 the 2nd trick.

The difference is because in 1 we cannot pick up AJxx with West.

do we also finesse 2nd round in 1 if the suit is trump in 6S and we see good possibility to run all side suit winners so that K9 in dummy can be left as last (and West AJ if he started with AJxx)

[TimG]

In case 1, you also can't pickup Jxxx/A (once you've played the K or Q on the first round), you can in case 2.

[Guest Jlall]

A few weeks ago I won from Ax with KQ98xx in dummy and later lost a trick to pard's Jxx.

 

The declarer told me, I'd thought you'd duck with Ax.

gotta know your opps...

[dburn]

By the way, this same odds calculator says that, for the original problem in this thread, the relevant splits of the suit are 6.783% for 2-2 with AJ in RHO's hand, and 6.217% for 3-1 with singleton A in RHO's hand.  This is slightly more than 52-48.

Of course it is. The chance that a suit will be 1-1 as opposed to 2-0 or 0-2 is exactly 52 to 48, because the chance that East will be dealt the second card after West has been dealt the first is exactly 13 (the number of vacant spaces East has) to 12 (the number of vacant spaces West has).

 

The chance that West will have two cards of four missing cards is somewhat harder to calculate in this way, because it can happen in a number of different ways (six, to be precise). West can receive cards 1 and 2, East cards 3 and 4; or West can receive 1 and 3, East 2 and 4; and so forth. If we look at the likelihood of each of these in terms of vacant spaces, we soon see that each is equally likely:

 

West gets card 1 happens with probability 13/26

West gets card 2 then happens with probability 12/25

East gets card 3 then happens with probability 13/24

East gets card 4 then happens with probability 12/23

 

West gets card 1 happens with probability 13/26

East gets card 2 then happens with probability 13/25

West gets card 3 then happens with probability 12/24

East gets card 4 then happens with probability 12/23.

 

In short, each of the six ways in which West can receive exactly two of the missing cards happens with probability (13*13*12*12)/(26*25*24*23), or 6.7826087%. Multiplying this by six gives the actual probability of a 2-2 break, which is 40.69565217% (give or take a pip or two).

 

The same sort of calculation can of course be performed to arrive at the probability that West has three cards and East one: each case has a probability of (13*12*11*13)/(26*25*24*23), or 6.217391304%, which we multiply by four to give 24.86956522%. Not surprisingly after all this, the ratio of the chance that West has two cards to the chance that he has three is the ratio of:

 

13*13*12*12 to 13*13*12*11, or exactly 12/11. This is not 52 to 48, but nobody ever said it was - or if he did, he erred.

[Echognome]

Dburn long post

 

I'm just wondering, isn't it equivalent after we've seen 3 cards played from opponents that the odds are just 12 to 11 based on vacant spaces? Isn't that much easier than all of the combinatorics above?

[dburn]

I'm just wondering, isn't it equivalent after we've seen 3 cards played from opponents that the odds are just 12 to 11 based on vacant spaces?  Isn't that much easier than all of the combinatorics above?

Very much so. The exercise in futility above was an attempt to show how distributional probabilities are calculated in terms of vacant places. Or to put it another way, I was bored, and I didn't see why I should be the only one.