Simple math problem from Europen Ch

[ArtK78]

The only facts that we have are:

 

Opening leader has 5 spades and his partner has 3 spades (assumed).

Both opponents followed to 3 rounds of hearts.

Opening leader has followed to 2 rounds of diamonds and his partner has followed to 3 rounds of diamonds, and the ♦J has not appeared.

 

The only two relevant cases are Jxxx-xx (with the Jxxx onside) and xxx-Jxx (with the Jxx offside), as we have seen all of the x's.

 

Given these facts, the vacant space suit distribution calculator that I found at http://www.automaton.gr/tt/en/OddsTbl.htmstates states that the odds of the Jxxx being onside are 57.144%.

 

I know that this is essentially the same thing that I stated above, but I wanted to provide the link so that others could check it out.

[bluecalm]

I used BBO build in calculater (web version -> hands and results -> hand editor ->analyze suit) and it gives 57.14% answer. (you need to make division yourself).

[inquiry]

Now that it has been worked out that the odds are 3:4 to play for the finesse. What if you decide spades were divided 4=4 to begin with? I think the implication of such a spade division will lead to an interesting reversal of the odds.

[JLOGIC]

I wanted to do this before I read the answers to see if I have it right at the table.

 

In general in spots like this when you must guess mid hand, I thought knowing 2 extra cards (eg 5-3 spades) meant that hooking was correct. If you know 1 extra card, it is generally about 50-50 (so on a hand like this if you knew spades were say 5-4 with opening leader, then def play for the drop to save an undertrick), unless they led a 4 card suit in which case you have extra inferences (depending on hand). Is this "rule of thumb" almost always correct, because it is what I always use at the table if I can't get anything better.

[ArtK78]

Now that it has been worked out that the odds are 3:4 to play for the finesse. What if you decide spades were divided 4=4 to begin with? I think the implication of such a spade division will lead to an interesting reversal of the odds.

According to the calculator, if the spades are 4-4, the odds are that the diamonds will be 3-3 57.142% of the time.

 

Also, in Justin's case, where the spades are divided 5-4 with the opening leader having 5, the odds are exactly 50-50.

[bluecalm]

Perhaps I meant in isolation instead of this particular hand.

 

Well, in isolation every 3-3 division has 10 cards out of 20 remaining on every side. You can choose it 20C10 ways and 4-2 splits have 9 and 11 cards out of 20 so 20C11 or 20C9 ways (those numbers are equal).

What is a ratio between 20C10 and 20C9 ? Let's see:

 

20C10 is:

20*19*18*17*16*15*14*13*12*11/10*9*8*7*6*5*4*3*2 and 20C9 is:

20*19*18*17*16*15*14*13*12/9*8*7*6*5*4*3*2

 

Luckily a lot of this simplifies once you put fraction line between them and it becomes:

11/10 so every specific 3-3 split is more likely than specific 4-2 split by 11 to 10 ratio (or 55% to 45%).

If we add to this that there are 20 3-3 splits and 15 4-2 splits we have 11*20 to 10*15 or 220 to 150.

 

 

Let's see how our calculations stack up against ultimate source of the truth, the Wikipedia: http://en.wikipedia.org/wiki/Bridge_probabilities

 

Here we see that 3-3 split has 0.36 of total probability and 4-2 0.48. Right, but so far we were concerned with 4-2 but there is also 2-4 so we shall multiple our 150 number by two arriving at 220 vs 300.

220/300 is 0.73348472336911643270024772914946 while wikipedia numbers are 0.36/0.48 = 0.75

 

Someone is wrong here and that's surely wikipedia. Let's try other sources: http://www.durangobill.com/BrSplitStats.html

They have our 0.733 number, let's try BBO hand calculator:

http://imgur.com/koOOZ

 

In fact our numbers are 48.45% and 35.53%. Let's see: 35.53/48.45 = 0.733 after all, so we were right and wikipedia was wrong. Oh well...

 

in which case you have extra inferences (depending on hand). Is this "rule of thumb" almost always correct, because it is what I always use at the table if I can't get anything better.

 

Yeah that seems to work in many cases.

[gnasher]

Why can't I withdraw an upvote? I glanced at the hand and thought "finesse, obvoiously". Then I read Han's first post and thought, "Oh yes, he's right". If someone changes my mind about something I usually give them an upvote, so I did. Then I read a few more posts, and found the one where Han refuted his earlier argument. What do I do now? In theory I should give him a a second upvote for changing my mind again, but that would be wrong because it just rewards him for making a mistake in the first place. What I want to do is withdraw my upvote from Han, and give myself a downvote for the erroneous upvote, but the forum software doesn't seem to offer this level of sophistication.

[wyman]

Just up vote every other post - past and future. That should compensate.

[inquiry]

Now that it has been worked out that the odds are 3:4 to play for the finesse. What if you decide spades were divided 4=4 to begin with? I think the implication of such a spade division will lead to an interesting reversal of the odds.

According to the calculator, if the spades are 4-4, the odds are that the diamonds will be 3-3 57.142% of the time.

 

Also, in Justin's case, where the spades are divided 5-4 with the opening leader having 5, the odds are exactly 50-50.

 

The 57.142 odds if spades were 4=4 is what I meant by "an interesting reversal of the odds". The reason being if ♠ are 4=4, both had 3 ♥, and once all the ♦ had been played but the ♦J you can count all the spades, hearts, and diamonds in the vacant space calculation. The hand that just followed to the diamond had 3♦, 4♠ and 3♥, leaving 3 vacant spaces. The other hand (yet to play) has the following known cards: 4♠, 3♥, 2♦ leaving 4 vacant spaces...

 

So with spades 5=3, when you get here, it is 4:3 for the finessee

with spades 4=5 when get here it is 4:3 for drop.

 

Justin's idea of 5=4 spades, all things equal would be...

leader 5♠. 3♥, 2♦ = 3 vacant spaces

3rd hand 4♠, 3♥, 3♦ = 3 vacant spaces, thus 50=50, either way,

 

According to Jeff Rubens (and others), you can only count diamonds as known cards when all but the key card (this case, the jack) has been played.

 

So the key to this hand, is HOW MUCH YOU TRUST your understanding of their carding and how truthful they are early in the hand (most are always truthful early).

[lalldonn]

The odds are 4/6 that a world class declarer would get this right.

[Tomi2]

The odds are 4/6 that a world class declarer would get this right.

 

what would you think is the reason that more open players went down here compared to seniors and la... women?

 

when I watched the board as it was played in vug, around half of the tables finished it with kinda 50% made for the guys and 75% made for the other competitions.

 

of course it might be, that "worse defenders" misdefended, but actually ALL tables played it from north and ALL got the spade lead, and I can't imagine any non-open had more chance to "misdefend" (e.g. throwing a dia) here. Or am I wrong?

[lalldonn]

My conclusion was based on the following scientific analysis:

Bocchi, Bessis, Brink and Fritsche got it right

Helness and Sylvan got it wrong.

[perko90]

The only facts that we have are:

 

Opening leader has 5 spades and his partner has 3 spades (assumed).

Both opponents followed to 3 rounds of hearts.

Opening leader has followed to 2 rounds of diamonds and his partner has followed to 3 rounds of diamonds, and the ♦J has not appeared.

We also know a lot about the club suit!

Let's take the 5-3 spade break as a given. Opening leader has at least 2 clubs and 3rd hand has at least 3. Each hand has only 1 vacant space! Opening leader can only be 5=3=3=2 or 5=3=2=3 and 3rd hand can only be 3=3=3=4 or 3=3=4=3. It would be a toss up, except as han pointed out, we know Jxx is not onside and Jx is not offside. In a 3-3 situation, Jxx offside is a 50% chance and for the now equally likely 4-2 split the Jxxx onside is a 66.7% chance (a priori suit split chances no longer factor in). So the finesse is a 4:3 favorite (⁴/₆ / ³/₆).

So, I believe han's 1st post + his follow-up correction tells the story.

[bluecalm's mention that opening leader can't have AK tight of clubs probably nudges the finesse to be even a tiny bit more of a favorite.]

[Fluffy]

Now for something more exciting, if East has all of ♣AK♠A the contract is cold even if diamonds are missguessed, is he more likelly to hold all those cards with 2 diamonds than 3?, if he has stiff ♣A at this moment the contract also makes. This might favour finese over drop.

[gnasher]

Now for something more exciting, if East has all of ♣AK♠A the contract is cold even if diamonds are missguessed, is he more likelly to hold all those cards with 2 diamonds than 3?, if he has stiff ♣A at this moment the contract also makes. This might favour finese over drop.

AQxxx xxx xx AKx is more likely than AQxxx xxx Jxx AK, which would be a reason to play for the drop.

 

Is it enough reason? With clubs 3-3, East will have ♣AK 1/4 of the time. With clubs 4=2, East will have ♣AK 1/4 of the time 1/15 of the time. So the new ratio is

 

finesse : drop

= 4/7 + 3/7 * 1/15 : 3/7 + 4/7 * 1/4

= 4/7 + 1/35 : 4/7

 

Hence the finesse is still better than the drop.

 

In addition, as Fluffy says, there is the chance that clubs are KJxx-Ax., which means that they misdenfended, but also argues for the finesse.

 

(Edited after I'd reread Fluffy's post and actually understood it.)

Edited June 22, 2012 by gnasher

[han]

Andy, following common claiming tradition on BBO, the next time I'm about to make a post that is likely to get an upvote, I won't make the post in order to "give it back".

[MrAce]

Why can't I withdraw an upvote? I glanced at the hand and thought "finesse, obvoiously". Then I read Han's first post and thought, "Oh yes, he's right". If someone changes my mind about something I usually give them an upvote, so I did. Then I read a few more posts, and found the one where Han refuted his earlier argument. What do I do now? In theory I should give him a a second upvote for changing my mind again, but that would be wrong because it just rewards him for making a mistake in the first place. What I want to do is withdraw my upvote from Han, and give myself a downvote for the erroneous upvote, but the forum software doesn't seem to offer this level of sophistication.

 

Its ok Andy. Calm down. It is just an upvote ffs :P

 

I am sure there had been a post or two of Han you forgot to upvote in the past :)

[PhilKing]

At the critical point you know that East has one of two distributions:

 

5323

5332

 

Both are equally likely, surely. After all you are missing six cards in each minor.

 

The "vacant spaces" argument is a red herring, since you could equally well apply it to the club suit with an assertion such as: "East has 5 vacant spaces compared to West's 7, so is likely to be shorter in clubs. Therefore he is more likely to have the diamond." Doesn't wash, really, does it.

 

Another way of looking at it is this: West either has four diamonds or he has four clubs. We know he is either 3343 or 3334. Explain to me why he is more likely to have four diamonds than four clubs.

 

East has an average of 2.5 cards in each minor. Period.

 

Finessing and going 2 down loses an extra imp. Assuming they play it differently in the other room you will lose 12 or gain 13 by playing for the drop.

[bluecalm]

5323

5332

 

Both are equally likely, surely.

 

Nope.

You forgot that we didn't see J♦ and that was the card they couldn't play.

So 5-3-xx-xxx ( assuming all clubs are equal for a while) could be dealt in 5C2 * 6C3 = 10*20 = 200 ways and 5-3-Jxx-xx could be dealt in 5C2 * 6C2 = 10*15 = 150 ways.

As every hand is equally probable the probability of getting 5-3-2-3 is 200/350.

What you say would be true if they had only equal cards in both clubs and diamonds.

 

I tested this ending out using JackBridge, which really is good at this sort of thing

 

Maybe you are not that good at testing then ? :)

[PhilKing]

 

Maybe you are not that good at testing then ? :)

 

This is certainly true. I tried it again with a One Spade overcall over One Club, just so Jack know the lead was not off a 3-carder and it came up with some absurd conclusions.

 

Back to the card combibations, have you factored down the 4-2 diamond breaks to reflect that the diamond jack has not dropped (a priori 5 combinations)?

 

There are, after all 10 ways for East to be dealt Jxx and 10 ways to be dealt xx, once the Jx's are removed.

 

Since East's average holding in diamonds is 2.5 when he holds 8 cards in the major, it seems there is no need to weigh the holding one way or tother.

[bluecalm]

Back to the card combibations, have you factored down the 4-2 diamond breaks to reflect that the diamond jack has not dropped (a priori 5 combinations)?

 

There are, after all 10 ways for East to be dealt Jxx and 10 ways to be dealt xx, once the Jx's are removed.

 

This is correct.

Notice though that with xx of diamonds there are 3 clubs to deal and with Jxx of diamonds there 2 clubs to deal.

There are more ways to deal 3 clubs out of 6 than to deal 2 clubs out of 6, or in other words there are more hands with xx of diamonds.

 

no need to weigh

 

I am not weighing anything, I just count hands and then apply fundamental theorem of bridge (every hand has the same probability of being dealt).

[PhilKing]

I decided it was time to "shut up and deal."(Well I managed the first part)

 

I dealt 100 hands using the "vacant spaces" method. The 12 cards outside hearts and spades were dealt into two piles of seven and five cards to reflect the known 2-card disparity. Obviously, on a large number of the cases the diamond position is known before declarer plays to the the third diamond (the most common being when Jxx showed up onside which represents the silver bullet for the 7/5 theory) and these are represented with a dash. Here are the results:

 

D = Drop, F = finesse, - = irrelevant

 

-D-D--FF-D--D-FFF-FD-D-F--DF------DFDD-D----DF-F-D-D---FFDF--D---FDD-D-----DFFF---F---D-FF--F-D-FD--F

 

Well done the finesse for an impressive late rush.

 

Ok, that's not a huge sample, but it's now penalties in Spain Portugal.

 

Anyway, that's 24/23 in favour of the finesse for 24 swings in of 12 imps and 23 losses of 13 imps against a "dropper".

[PhilKing]

You can deal any specific 4-2 in 6C3 = 20 ways and every specifi 3-3 in 6C2 - 15 ways. 6C3 and 6C2 are ways to deal various club holding to go along 5 spades and 3hearts we already know.

 

You can turn this on its head:

 

Each specific 3 card club holding with the opening leader can be dealt 10 ways (ie the 10 xx combinations in diamonds).

 

Each specific 2-card holding in clubs in the leader's hand can also be dealt ten ways (the 10 Jxx combinations).

 

Anyway, on the off chance that this argument is gibberish, here's another:

 

Things have changed from the time we knew it was 7-5 in favour of the non-leader having the diamond jack. 16 combinations have been eliminated from contention for the 7/5 guy having the diamond jack (jxx - 10, Jx - 5, J - 1) but ony six cases of the 5/7 guy holding that card can be discounted (Jx and J).

[bluecalm]

Each specific 3 card club holding with the opening leader can be dealt 10 ways (ie the 10 xx combinations in diamonds).

 

Each specific 2-card holding in clubs in the leader's hand can also be dealt ten ways (the 10 Jxx combinations).

 

This is correct. Why did you feel the need to point this fact out I do not know.

It just tells you that every specific club holding is equally probable. Let's see where it leads us:

 

1)5-3-2-3: there are 6C3 = 20 ways to deal 3 clubs.

2)5-3-3-2: there are 6C2 = 15 ways to deal 2 clubs.

 

As you noted all the ways are equally probable so again 1) is more probable than 2) by a factor of 20/15. Same conclusion as before "turning this argument on its head".

[PhilKing]

 

1)5-3-2-3: there are 6C3 = 20 ways to deal 3 clubs.

2)5-3-3-2: there are 6C2 = 15 ways to deal 2 clubs.

 

As you noted all the ways are equally probable so again 1) is more probable than 2) by a factor of 20/15.

 

This sounds pretty tight.

 

I think I'd better run another simulation. :(