Simple math problem from Europen Ch

[bluecalm]

[hv=pc=n&s=s98haj97dq72cq752&n=skjthkqtdakt4ct98&d=w&v=n&b=12&a=p1np2cp2dp3nppp]266|200[/hv]

 

2♠ 8♠ 7♠ J♠

 

You note 7 which is UD count which combined with their CC as to first lead makes 5-3 spade split most likely.

You now play 4 rounds of hearts and two round of diamonds. Both opponents followed 3 times to hearts and discarded small clubs to 4th. Both followed to diamonds.

 

Finesse or on top ?

 

Bocchi, Bessis, Brink and Fritsche got it right

Helness and Sylvan got it wrong.

 

What about you ?

What are approximate %'s here ?

[paulg]

I saw the hand, but I would ask whether the lead is fourth, 3/5 or attitude.

[bluecalm]

Yeah, let' assume spades are 5-3. If anything count of 2nd defender is quite reliable and it was 5th on some tables for sure (Fantoni - Nunes lead low from Hxxxx).

[ArtK78]

It seems to me that the odds are 7-5 in favor of the finesse.

 

Both opponents are known to hold 3 hearts. Opening leader is assumed to hold 5 spades, thus 5 other cards. Partner of the opening leader is presumed to hold 3 spades, thus 7 other cards.

 

The fact that they both followed to 2 rounds of diamonds does not alter the odds.

[Fluffy]

 

The fact that they both followed to 2 rounds of diamonds does not alter the odds.

 

it tells you no 5-1 diamond break, so should probably be worth something

[han]

So if diamonds are 3-3 the distributions are 3-3-3-4 with west and 5-3-3-2 with east, while if diamonds are 4-2 then the distributions are 3-3-4-3 with west and 5-3-2-3 with east. Both distributions are clearly exactly as likely, but given that on some hands with a doubleton diamond the jack of diamonds would have appeared, it seems that the percentage play is to play for the drop.

 

I ignored the fact that small clubs were discarded on the fourth round of hearts. The impact of this on the percentages cannot be computed.

[bluecalm]

I am already happy I made this thread because I see there is a flow in either my reasoning or Han's.

Experience so far suggests I am going to learn something here...

[ArtK78]

According to a calculator that takes into account vacant spaces, the odds of Jxxx onside are 57.144% compared to the odds of Jxx offside.

 

That is close to 7-5.

[bluecalm]

Yeah, I calculated this manually to arrive at same conclusion (not exact % because I only compared layouts which makes a difference).

I guess what Han is saying is that we should take spades into account too and not treat them as "known cards". I have some problems with that because possible combinations in spades doesn't affect affect chances of dealing either Jxx or xx of diamonds (in both cases there are exactly 5 spades).

I will post my exact argument later to allow yet unbiased people to post their thoughts :)

[han]

Maybe I was talking nonsense, let me think about it for a bit.

[han]

Yeah where I went wrong is that some of the 3-3 splits need to be ruled out as well: Jxx with west. Since there are more Jxx-xxx splits than xxxx-Jx splits one should finesse.

 

Of course the vacant places argument is much clearer.

[wyman]

After dealing out 8 major cards to LHO and 6 major cards to RHO, there are 12 minor cards to be distributed: 5 to LHO and 7 to RHO.

 

The cases we can be in are when LHO gets:

 

Jxx / xx - 5C2 * 6C2 = 150 cases

xx / xxx - 5C2 * 6C3 = 200 cases

 

In the former, we should drop. In the latter, we should finesse. This looks to me like 4:3 in favor of the finesse (or 57.14%).

 

Two notes:

 

1) I don't think the club discards should be taken into account, so I didn't account for them, except that

 

2) If anything, LHO is super-unlikely to discard a club from Kx, so perhaps we should give even a bit more weight to the finesse.

[bluecalm]

2) If anything, LHO is super-unlikely to discard a club from Kx, so perhaps we should give even a bit more weight to the finesse.

 

He shouldn't mind discarding from Kx but he doesn't have AK (because we saw a small one) and Ax (because then he would be endplayed after taking a trick with Jd.

Nice play would be discarding Ac from Axx while holding xx of diamonds :)

[Phil]

given that on some hands with a doubleton diamond the jack of diamonds would have appeared, it seems that the percentage play is to play for the drop.

 

In addition to that, we'd also see Jxx onside.

 

According to a calculator that takes into account vacant spaces, the odds of Jxxx onside are 57.144% compared to the odds of Jxx offside.

 

That is close to 7-5.

 

Yes, but we've seen the 3rd diamond from RHO, so aren't the odds 6:4?

[Phil]

By the way, LOVE the title.

[han]

Actually, I don't think that the vacant spaces argument to get the 7:5 odds is correct. It says that before seeing any diamond cards, the chance that west had the diamond jack was 7:5. But we can't ignore the fact that west has followed with 3 small diamonds and east with 2 small diamonds. Also, 7:5 is really not equal to 57.144%. so clearly there is something wrong.

 

It looks like the first part of my faulty argument still holds, namely that diamonds 4-2 (4 with west) is exactly equally likely as diamonds 3-3. Then again, we should rule out the Jx with east or Jxx with west cases, which is one third (5 out of 15) of the 4-2 splits but half (10 out of 20) of the 3-3 splits.

 

So we should compare the remaining cases, that gives 4/6 of the 4-2 splits vs 3/6 of the 3-3 splits, or in other words, 4-3 in favor of finessing. Note that 4/7 = 0.571428571, which is very close (but surprisingly not exactly equal) to what the calculator of Art gave.

[han]

In addition to that, we'd also see Jxx onside.

 

 

 

Yes, but we've seen the 3rd diamond from RHO, so aren't the odds 6:4?

 

Yes and no, the argument is correct but your numbers are wrong. We should subtract 3 from west's 7 and 2 from east's 5 to obtain 4:3..

[bluecalm]

It looks like the first part of my faulty argument still holds, namely that diamonds 4-2 (4 with west) is exactly equally likely as diamonds 3-3

 

We could deal every combo of 3-3 diamonds in 15 ways (along with 2 out of 6 clubs) and every combo of 4-2 diamonds in 20 ways (along with 3 out 6 clubs).

So assuming all clubs are equal (which is not exactly correct because as noted E wouldn't discard a club from AK or Ax) there are :

 

The cases we can be in are when LHO gets:

 

Jxx / xx - 5C2 * 6C2 = 150 cases

xx / xxx - 5C2 * 6C3 = 200 cases

 

ways to deal significant diamond layouts. 200/350 = 57.14% :)

Isn't it correct ? :)

 

By the way, LOVE the title.

 

The math is simple... once you know what math applies for the situation :)

 

EDIT: I see now, while every combination of 4-2 is more probable that every combination of 3-3 there are 20 combinations of 3-3 and 15 of 4-2 so it's evens out. So yeah, first quote is true :)

[han]

Not sure what you are saying there Bluecalm but I take it you quoted the only sentence that I wrote and you agreed with. ;)

[han]

I definitely agree that it is simple, even though it apparently still is easy to get confused, which is disappointing to me.

 

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

[bluecalm]

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.

 

"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...

[Phil]

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

 

It is?

[bluecalm]

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.

[han]

It is?

 

Yes?

[Phil]

Yes?

 

Perhaps I meant in isolation instead of this particular hand.