Expert Bridge Complicated

[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]

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]

My mathematical expert says you didn't provide enough information :D. 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]

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]

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.haroldschogger.com/MaxRebatu.htm

[gnasher]

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.

 

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

[kenberg]

I'm wrestling with this, may I think out loud a little? Certainly I agree with LittleKid that in the asked case it is even odds when one opp plays the 2 and the other the 3.

 

My thoughts:

 

First, about restricted choice and the openeing example:

With the AKQ3 in dummy the various restricted choice arguments don't see to work when applied to the 9,T,J. The AK are played producing spots and then the Q is played. The defender with two cards remaining can play either one, even if the two are the J9, since the ten is the only card he hasn't seen. He does not even care whether declarer or his partner has the ten, it's falling. So any restricted choice argument involving inference from the third round play of the ten or jack won't work because there is an equally strong argument for the fall of the nine. Whoever holds two ot the 9,T,J could play either, and the player who held only one has his choice restricted.

 

The deuce, however, is different from the 7. I realize others have said this, I'm just thinking it through in my way, Take the original hand: AKQ3 in dummy with the cards splitting 4-3 on the play of the tops. The deuce must be played. If it lies with the three card holding then the rules require it, while if it lies with the four card holding then the 3 in dummy requires the deuce to be played.

 

So the situation is equivalent to declarer being able to ask: Which of you guys holds the deuce? They must answer. With the 7, it's not the same. The 7 is more like the jack. After three rounds of play if the 7 is with the three card holding then it must be played, but if it is with the four card holding then that player may keep the 7 or one of his other cards higher than the deuce.

 

 

So this gets us to the 4-3 odds: It's 4-3 that the person holding the deuce is the one holding four cards. It's also true that the person holding the 7 holds four cards, but they have to tell you about the deuce and they don't have to tell you about the 7.

 

 

Now the asked about example.

Missing the 2 and the 3, They have to tell you the locatin of both. If one player has one and the other has the other, then the 2 and 3 are equal, so they provide equal inference and so it's even. When one player has both then he must show them both. Surely it is easier for a person with four cards to hold both than for the person with three cards to hold bot, so the odds are in favor of the person holding both to have four cards. This is, however, one of those "inversion of probabilities" things: It is easy enough to calculate the probabilty that the player holding four cards holds the 2 and the 3, but the question is the probability of X holding four cards given that everyone followed and one player had both the 2 and the 3. This is the sort of thing Baye's theorem was invented for, and I'll get back to you after some arithmetic. Technically we need Baye's theorem in the first case as well but I think that the situation is simple enough that we can bypass it. Not in this second case, however.

[kfay]

This is definitely interesting to me.

 

The idea that is nagging me is this: the deuce has become a significant card, essentially, right? Doesn't that put more vacant spaces in the holder's partner's hand? Thus equalizing the probability of who holds 4 cards?

 

I'm dealing the 7 remaining cards in the suit, give someone the deuce, and now deal the remaining cards in the pack at random... doesn't this really make it 4:7 minus some percent?

[gnasher]

This is definitely interesting to me.

 

The idea that is nagging me is this: the deuce has become a significant card, essentially, right? Doesn't that put more vacant spaces in the holder's partner's hand? Thus equalizing the probability of who holds 4 cards?

 

I'm dealing the 7 remaining cards in the suit, give someone the deuce, and now deal the remaining cards in the pack at random... doesn't this really make it 4:7 minus some percent?

There are 6 remaining cards, not 7.

 

Give the 2 to East. Now West has 13 vacant spaces and East has 12.

 

Take three specific high cards and give them to West. West now has 10 vacant spaces.

 

Take two specific high cards and give them to East. East now has 10 vacant spaces.

 

I have one high card left to deal. Because the vacant spaces are the same, each player has a 50% chance of getting the last high card.

 

So, a specific 3=3 break and a specific 4=2 break are equally likely.

 

There are 20 3=3 breaks and 15 4=2 breaks, so the 3=3 break occurs 20/35 or 4/7 of the time.

[kenberg]

My arithmetic sucks here, or did before I corrected it. See Cascade's/gnasher's correction below. And it's true that earlier arguments get the same answer w/o going through the complications I use. But I will leave it for whatever interest it may hold.

 

I like to think that math makes intuitive sense but when dealing with probabilities this often needs to be modified to "intuitive sense after you think about it right".

 

Suppose you have two random events A and B that may or may not occur. There is a certain probability either will occur, that both will occur, that neither will occur. There is also the probability that A will occur given that B occurred, and vice versa.

 

 

 

Now I am going to suggest that we can get away with saying that we need only consider hands where the suit turns out to split 4-3, one direction or the other. Everything in probability is a little risky, but I think this is kosher.

 

Also I want to concentrate on E being the guy with the 2 in the first example or the 23 in the second.

 

It may be a little easier to calculate the probability that E holds three cards. This will be 1 minus the probability that he holds 4.

 

I will take the second example.

 

The event A is E holds three cards

 

The event B is E holds the 2 and the 3.

 

 

Here is the key item: Let A & B mean that both A and B occur. Let P(A) be the probability that A occurs. Let P(A|B) be the probability that A will occur given that B occurs. This means you watch for all of the times when B holds the 2 and the 3, and see what percentage of the time he holds three cards (where we look only at the deals where he holds either 3 or 4).

 

 

Principle: P(A&B)=P(B) X P(A|B). This is not so crazy. Without worrying about the specifics of what A and B are, suppose that B occurs 2/3 of the time and of the cases where B occurs, A also occurs 1/4 of the time. The principle says that A and B occur together (2/3) X (1/4)= 1/6 of the time.

 

 

 

If you are with me so far, then you will probably also accept that P(A&B)=P(A) X P(B|A) .

 

So

 

P(B) X P(A|B)=P(A) X P(B|A)

 

 

We would like to calculate P(A|B), the probability that E holds three cards given that he holds the 2 and the 3. If we can calculate the other probabilities, then we have this one as well from the equation.

 

 

The others come from combinatorics. Let em do the simple case first, the one where the tops are played from AKQ3 and E shows the deuce. This time B is the event that E holds the deuce.

 

P(B) is the probability that E holds the deuce. We are talking of a priori probabilities here, or almost a priori meaning that we are assuming the suit splits 4-3 one way or the other but we are assuming nothing more. So P(B)=1/2. Either player is equally likely to hold the deuce.

 

 

 

Similarly, P(A)=1/2. One player holds 3, the other 4, a priori it's fifty-fifty which way. So P(A|B)=P(B|A), and P (B|A) is the probability that E holds the 2 given that he holds three cards. That's 3/7, as advertised

 

 

Now to the main event. B is the event that E holds the 2 and the 3.

 

P(A) is still 1/2, but the other tow probabilities require a little more effort.

 

 

P(B) is the probability that E holds the 2 and the 3. Again, we are assuming that the suit is split 4-3 one way or the other. The probability that E holds 3 is the same as the probability that he holds 4, and so the probabilty that E holds both the 2 and the 3 is the average of the prob of him holding the 2 and 3 when holding three cards and the prob of him holding the 2 and the 3 when holding 4 cards.

 

Correcting my previous post, agreeing with Cascade that 4+3 is actually 7, not 6, we get:

 

From combinatorics, P(E holds 2&3 | E holds three cards) is 5 choose 1 divided by 7 choose 3 (the number of three card holdings including the 2 and the 3 divided by the number of three card holdings). This is 5 divided by 35, or 1/7.

 

P(E holds 2&3 | E holds four cards) is 5 choose 2 divided by 7 choose 4. This is 10 divided by 35, or 2/7.

 

 

P(B) is the average of 1/7 and 2/7, namely 3/14 You still there? The hardest part is over.

 

 

 

P(B|A) is the probability of E holding the 2 and 3 given that E holds three cards. We did this above, the answer was 1/7.

 

 

 

 

 

So we have agreed that

 

P(B) X P(A|B)=P(A) X P(B|A)

 

from which (3/14) X P(A|B)=(1/2) X (1/7)=1/14

 

So P(A|B)=(1/14)/(3/14)=1/3.

 

When the suit splits 3-4 and one hand holds the 2 and 3, the probability that this hand has length three is 1/3, and so the probability it has length four is 2/3.

[gnasher]

[complicated stuff]

I confess that I haven't tried to follow this in detail, but I think you failed at the final hurdle:

P(A) is still 1/2

...

P(B) is the average of 1/5 and 2/5, namely 3/10

...

P(B|A) is the probability of holding the 2 and 3 given the holding of three cards. We did this above, the answer was 1/5.

 

So we have

 

P(B) X P(A|B)=P(A) X P(B|A)

 

from which (2/5) X P(A|B)=(1/2) X (3/10)

I think you have

  3/10 x P(A|B) = 1/2 x 1/5

so

  P(A|B) = 1/3

That is, the player who produces the 2 and the 3 will hold four cards 2 times out of 3.

 

Which is, of course, the same result as these:

there is a 2/3 chance the hand with the 2 and 3 have the last card.

it's 2:1 that they're 2=3.