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Simple math problem from Europen Ch

Poll: Simple math problem from Europen Ch (24 member(s) have cast votes)

Your play ?

  1. Finesse (20 votes [83.33%])

    Percentage of vote: 83.33%

  2. Drop (3 votes [12.50%])

    Percentage of vote: 12.50%

  3. Dunno (1 votes [4.17%])

    Percentage of vote: 4.17%

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#21 User is offline   bluecalm 

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Posted 2012-June-21, 10:32

I was disagreeing with you at first (because every single 3-3 is less probably than every single 4-2) but then I realized that despite that getting a 3-3 is equally likely to getting 4-2.

Quote

"west has 4 vacant spaces, east has 3 vacant spaces, so the odds are 4:3" is very simple.


I am still not capable of intuitively grasping vacant spaces argument. I need to count combinations to convince myself. I guess I will get there...
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#22 User is offline   Phil 

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Posted 2012-June-21, 10:56

View Postbluecalm, on 2012-June-21, 10:32, said:

every single 3-3 is less probabley than every single 4-2


It is?
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#23 User is offline   bluecalm 

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Posted 2012-June-21, 10:59

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.
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#24 User is offline   han 

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Posted 2012-June-21, 11:14

View PostPhil, on 2012-June-21, 10:56, said:

It is?


Yes?
Please note: I am interested in boring, bog standard, 2/1.

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#25 User is offline   Phil 

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Posted 2012-June-21, 11:15

View Posthan, on 2012-June-21, 11:14, said:

Yes?


Perhaps I meant in isolation instead of this particular hand.
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#26 User is offline   ArtK78 

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Posted 2012-June-21, 11:26

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.
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#27 User is offline   bluecalm 

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Posted 2012-June-21, 11:37

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).
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#28 User is offline   inquiry 

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Posted 2012-June-21, 13:13

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.
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#29 User is offline   JLOGIC 

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Posted 2012-June-21, 13:31

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.
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#30 User is offline   ArtK78 

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Posted 2012-June-21, 13:36

View Postinquiry, on 2012-June-21, 13:13, said:

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.
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#31 User is offline   bluecalm 

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Posted 2012-June-21, 14:05

Quote

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....e_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.durangobi...SplitStats.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...

Quote

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.
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#32 User is offline   gnasher 

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Posted 2012-June-21, 15:51

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.
... that would still not be conclusive proof, before someone wants to explain that to me as well as if I was a 5 year-old. - gwnn
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#33 User is offline   wyman 

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Posted 2012-June-21, 16:00

Just up vote every other post - past and future. That should compensate.
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#34 User is offline   inquiry 

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Posted 2012-June-21, 17:22

Quote

View Postinquiry, on 2012-June-21, 13:13, said:

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).
--Ben--

#35 User is offline   lalldonn 

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Posted 2012-June-21, 17:38

The odds are 4/6 that a world class declarer would get this right.
"What's the big rebid problem? After 1♦ - 1♠, I can rebid 1NT, 2♠, or 2♦."
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#36 User is offline   Tomi2 

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Posted 2012-June-21, 18:10

View Postlalldonn, on 2012-June-21, 17:38, said:

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?
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#37 User is offline   lalldonn 

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Posted 2012-June-21, 18:25

My conclusion was based on the following scientific analysis:

View Postbluecalm, on 2012-June-21, 08:57, said:

Bocchi, Bessis, Brink and Fritsche got it right
Helness and Sylvan got it wrong.

"What's the big rebid problem? After 1♦ - 1♠, I can rebid 1NT, 2♠, or 2♦."
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#38 User is offline   perko90 

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Posted 2012-June-21, 22:22

View PostArtK78, on 2012-June-21, 11:26, said:

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.]
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#39 User is offline   Fluffy 

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Posted 2012-June-22, 00:55

Now for something more exciting, if East has all of AKA 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.
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#40 User is offline   gnasher 

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Posted 2012-June-22, 01:47

View PostFluffy, on 2012-June-22, 00:55, said:

Now for something more exciting, if East has all of AKA 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.)

This post has been edited by gnasher: 2012-June-22, 01:54

... that would still not be conclusive proof, before someone wants to explain that to me as well as if I was a 5 year-old. - gwnn
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