[dburn]

With:

54

AKQ3

you cash three rounds. West follows with the six, seven and ten; East with the two, eight and nine. If you had to bet your life on which opponent had the jack, whom would you choose?

The answer is East, because you will survive four times out of seven. Unless you understand this, please don't try to answer the next questions.

With:

65

AKQ4

you cash three rounds. West follows with the two, seven and ten; East with the three, eight and nine. On which opponent will you now place your bet? What if West had followed with the two, three and ten, and East with the seven, eight and nine?

If your answer is "it depends on the bidding, the opening lead, the opponents' signalling methods, the competence of East and West in the field of game theory, the weather, the..." then you are a very fine bridge player and should turn your mind to many of the real problems on this forum. If on the other hand you have some time to help me with a baffling question, please contribute - especially if you are Jeff Rubens.

[Little Kid]

50-50 for first, West for second

[JanM]

My mathematical expert says you didn't provide enough information . Is the doubleton or the AKQx in dummy? Which of the opponents is playing first to each trick?

[pooltuna]

JanM, on Dec 16 2009, 07:44 PM, said:

My mathematical expert says you didn't provide enough information . Is the doubleton or the AKQx in dummy? Which of the opponents is playing first to each trick?

I think the layout implies that North holds the doubleton with South realing off the winners. I would be glad to help if I could but the psychologic aspects of some the card play is an area of weakness so I will be watching with interest.

[dburn]

JanM, on Dec 16 2009, 07:44 PM, said:

My mathematical expert says you didn't provide enough information . Is the doubleton or the AKQx in dummy? Which of the opponents is playing first to each trick?

I know what your expert means Assume for the sake of argument that South is the dummy, and that after round one of the suit both East and West know that North has a doubleton.

The actual case cited by Rubens had these pips:

A643 (dummy)

KJ (declarer)

For reasons best known to himself, declarer chose to play the king, then the jack to the ace, then ruff the third round.

Rubens says only that when each opponent has followed to three rounds, the fact that each of them has three cards in the suit may be regarded as "fixed" for the purposes of vacant space calculations. This is important, because it contradicts received wisdom that only suits whose complete distribution is known may be included in such calculations.

But Rubens implies, if I have read him aright, that the location of the remaining card in the suit ("The Last One Left") is arbitrary - that is: assuming nothing else is known about the distribution, West is as likely as East to have the twelfth and master card in the suit.

I know (or at least I think I know) that if dummy's holding were A543, the location of The Last One Left is not arbitrary at all - the holder of the two is a 4/3 favorite to have the queen-equivalent. But in Rubens's actual example, the holder of the five must also play it before the fourth round, so that... well, so that for the moment I am inclined to accept Little Kid's view of the matter. But I don't know. That's why I asked.

[mrdct]

In both cases the 10 by LHO is akin to restricted choice so I guess I'll play RHO for the J.

[jdonn]

mrdct, on Dec 16 2009, 11:30 PM, said:

In both cases the 10 by LHO is akin to restricted choice so I guess I'll play RHO for the J.

But once the ten was played if east had J9 remaining that became restricted choice. So your point is a wash.

[MarkDean]

I the suit breaks 4-3 there are 35 holdings for the person with four cards to have.

5 with neither the 2 or 3 (basically the three card holding can pick one of the five non-2,3 cards to have)
10 with the 2 and not the 3 (pick 3 of the five non-2,3 cards)
10 with the 3 and not the 2
10 with both the 2 and 3 (pick 2 of the five non-2,3 cards)

If the 2 and 3 are in the opposite hands, you know it is one of the two middle cases, and it is a wash.
If the 2 and 3 are in the same hane, you know it is the first or fourth case, so there is a 2/3 chance the hand with the 2 and 3 have the last card.

[quiddity]

Mark, why can't you make exactly the same argument about the 7 and the 8?

[ulven]

Anders Wirgren has written an article about exactly this which will be published in The Bridge World. The article was submitted quite a while back, inspired by deal from the 2008 Europeans in Pau. Should appear soon.

I have a copy of the article but can't distribute or discuss it.

[hanp]

Interesting topic.

Quiditty, all cards can be played in any order the defenders want except for the cards lower than our spots. Those cards must be played in the first three rounds and are therefore special and we know we will always see them in the first three rounds (as long as the suit splits 4-3, which is now a given).

As dburn politely explained, in the first example the person who holds the four card suit is 4/3 times as likely to hold the the 2 as the person who holds 3 cards in the suit. In other words, the person who has the 2 is 4/3 as likely to hold 4 cards as the person who did not hold the 2.

The situation is even more extreme if you hold 32 with AKQ7 in dummy, and LHO follows with 8, 9, 10 while RHO follows with 4, 5, 6. In this case the chance that RHO holds the remaining card is 4/5.

[gnasher]

Very interesting. Why hasn't anyone written a book about this?

Another way to look at it is to treat each group of equally ranked cards as a separate suit.

With AKQ3 opposite 54, the opponents have 6 high cards and one low card. When the low cards break 0=1, the vacant places ratio is 13:12.
The chance of a specific 3=3 break is equal to the chance of a specific 4=2 break. (Once I've put 3 on the left and 2 on the right, the vacant places are equal, so it's evens where the last card goes.)
There are 20 3=3 breaks and 15 4=2 breaks, so the 3=3 break is more likely by a ratio of 4:3.

With AKQ4 opposite 65, they have 5 high cards and two low cards.
If the low cards are 1=1, it's evens whether the high cards are 3=2 or 2=3.
If the low cards are 2=0 the odds of a specific 1=4 are the same as the odds for a specific 2=3. (Put three on the right and one on the left to equalise the vacant places, then it's evens where the last one goes.) There are twice as many 2=3 breaks than 1=4 breaks, so it's 2:1 that they're 2=3.

[Hanoi5]

Can someone explain this topic as if I were a 4-year-old boy?

[Free]

Hanoi5, on Dec 18 2009, 10:11 AM, said:

Can someone explain this topic as if I were a 4-year-old boy?

We'll teach you when you grow older

[Codo]

So the idea is that the owner of the 2 has the 4 card suit 4 out of seven times because this is the only "small card" which is not needed to beat the 4. card from declarer?
Sorry, but how can this be true?

We define that the distribution of the suit is 4432 around the table. How can a 6 or 7 can ever take a trick in this layout? So these cards are not significant at all. They are equal to the two. The most high cards (given that dummy has two low) that may be played in three rounds are AKQ + JT9, so there cannot be a relevant card lower then the 8. (If higher cards are in dummy, this can change, but in all examples there weren't.)

Or can you construct any distribution where the small pipes (smaller the 8) can win the 4. trick?

So my question is: There are 3 and 4 in dummy, one partner shows the 2, the other the 5 and 6 and you have AKQ7. Who has the four card suit?

[ulven]

This is actually also discussed in a BOLS tips many years ago.

http://www.haroldsch...m/MaxRebatu.htm

[gnasher]

Codo, on Dec 18 2009, 12:30 PM, said:

So the idea is that the owner of the 2 has the 4 card suit 4 out of seven times because this is the only "small card" which is not needed to beat the 4. card from declarer?
Sorry, but how can this be true?

We define that the distribution of the suit is 4432 around the table. How can a 6 or 7 can ever take a trick in this layout? So these cards are not significant at all. They are equal to the two. The most high cards (given that dummy has two low) that may be played in three rounds are AKQ + JT9, so there cannot be a relevant card lower then the 8. (If higher cards are in dummy, this can change, but in all examples there weren't.)

Or can you construct any distribution where the small pipes (smaller the 8) can win the 4. trick?

To understand the effect that Burn is talking about, assume that the defenders are playing double-dummy, so they know that the 6 and the jack are equivalent, and they can play them in any order. The 6 can win a trick if the defenders choose to play the J, 10, 9, 8 and 7 on the first three tricks. The 7 can win a trick if the defenders choose to play the J, 10, 9, 7 and 6 on the first three tricks.

However, the 2 isn't equivalent to the 6, because the player who holds the 2 must play it on one of the first three rounds.

Quote

So my question is: There are 3 and 4 in dummy, one partner shows the 2, the other the 5 and 6 and you have AKQ7. Who has the four card suit?

Assuming double-dummy defence, the 2, 5 and 6 are equivalent, so there are three small cards and four high ones.
If the small cards are 2=1, the chance of a specific 1=3 break is equal to the chance of a specific 2=2 break. There are six possible 2=2 breaks and four possible 1=3 breaks, so 60% of the time the remaining high card will be with the two small cards.

[Codo]

So at least we have a solution for following proplem:

We are declarer and want to judge which defender holds a length in a critical suit. We own 6 cards in a 4-2 beak. We have no opportunity to get a picture of the breaks by elimination or cause of the bidding (or lack of it). In the meanwhile, the defenders do not just know how the suit breaks, they know that all cards from jack down to the six are equal.

I am very happy to see this problem solved. It was an everyday nightmare for me how to react in these kind of situations. I guess my overall results will be much better in 2010 then in 2009 now.

[vuroth]

Cool thread, thanks for sharing. Always nice to learn something.

[james876]

Regarding articles on the subject of this discussion:
In the March 2010 issue of The Bridge World (info at www.bridgeworld.com), there will be two articles (one by Anders Wirgren) titled X Truly Marks a Spot; these pieces are relevant to the idea underlying analyzing this type of combination.
Jeff Rubens
Editor, The Bridge World