[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]

gnasher, on 2012-June-21, 15:51, said:

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

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]

Quote

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.

Quote

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]

bluecalm, on 2012-June-27, 13:58, said:

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]

Quote

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.

Quote

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]

bluecalm, on 2012-June-21, 10:59, said:

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]

Quote

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]

bluecalm, on 2012-June-27, 16:39, said:

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.

[Statto]

bluecalm, on 2012-June-27, 16:39, said:

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.

You already pointed out that all combinations where ♦J drops have been eliminated, and as Phil K correctly stated, that leaves just 10 combinations for each.

If we assume the ♠ lead to be from a 5-card suit, East has shown 5 ♠, 3 ♥ and 2 ♦, and has 3 spaces left for the remaining ♦. West has shown the same number of cards in red suits, but only 3 ♠, so has 5 vacant spaces for the last 2 ♦. This makes it much more likely the finesse will succeed. A quick check at http://www.automaton.../en/OddsTbl.htm makes the finesse 63% and the drop 53%.

Of course it's different if the lead was from a 3-card suit, now just 37% finesse, still 53% drop. But enough more often than not it will be from a 5-card suit to favour the finesse, I think.

Edit: if you think the lead will be from a 3 card suit rather than 5 card suit more than 40% of the time, you should play for the drop...

[PhilKing]

Statto, on 2012-June-27, 21:37, said:

A quick check at http://www.automaton.../en/OddsTbl.htm makes the finesse 63% and the drop 53%.

Of course it's different if the lead was from a 3-card suit, now just 37% finesse, still 53% drop. But enough more often than not it will be from a 5-card suit to favour the finesse, I think.

Any tool that can reach a total chance of 116% for two almost mutually exclusive lines has my admiration. I don't think the chance of AQxxx xxx Jxx AK fully accounts for this, somehow. How it can then drop to a cumulative 90% shows a worrying drop in class.

Did you program in that the non-leader followed to three diamonds?

Anyway, I will do an extra simulation later today, fully expecting a result of 4/3 against, although it will not be for some time due to having just finished a monster poker session.

[Statto]

PhilKing, on 2012-June-27, 23:39, said:

Any tool that can reach a total chance of 116% for two almost mutually exclusive lines has my admiration.

They're not mutually exclusive at all. You always make if it started with Jxx onside (27%, assuming lead was 5th), and never make if it started with Jxxx offside (11%). Finesse succeeds if it started with Jxxx onside (36%), and drop if it started with Jxx offside (27%). 27% + 36% ~= 63%. 27% + 27% ~= 53% (subject to rounding errors).

[helene_t]

sorry nonsense argument

[PhilKing]

Statto, on 2012-June-28, 02:35, said:

They're not mutually exclusive at all. You always make if it started with Jxx onside (27%, assuming lead was 5th), and never make if it started with Jxxx offside (11%). Finesse succeeds if it started with Jxxx onside (36%), and drop if it started with Jxx offside (27%). 27% + 36% ~= 63%. 27% + 27% ~= 53% (subject to rounding errors).

But the question is whether to finesse or play for the drop when second hand plays low on the third round - not whether to win the trick when he plays the jack.

[bluecalm]

Quote

You already pointed out that all combinations where ♦J drops have been eliminated, and as Phil K correctly stated, that leaves just 10 combinations for each.

Paraphrasing previous poster: I don't know why you quoted me and then wrote this comment so I take it as you quoted a part you agree with...

Quote

You always make if it started with Jxx onside (27%, assuming lead was 5th), and never make if it started with Jxxx offside (11%). Finesse succeeds if it started with Jxxx onside (36%), and drop if it started with Jxx offside (27%). 27% + 36% ~= 63%. 27% + 27% ~= 53% (subject to rounding errors).

All those percentages are wrong.
You can try build in BBO calculator to see this, it will show you exact combinations with % or you can calculate it by hand which should be easy if you follow my posts in this thread.
EDIT: unless you mean after playing exactly two round of diamonds but not 3rd. Then the percentages are almost correct (it should be 10.71%, 26,78%, 35,7%) but we really should eliminate Jxx onside if we talk about what decision to make here.

[PhilKing]

Simulation 2 was way in favour of the finesse, running off 6 anaswered winners at one point. Added to sim one the results are pretty close to the 57% (4/3) expected.

[bidule5]

No need for simulation, it is really simple math.
I am with wyman with different notation:
West 5332 with DJ = C(2,5)*C(2,6)=10*15
West 5323 without DJ =C(2,5)*C(3,6)=10*20
so finesse =4/7

[Statto]

bluecalm, on 2012-June-28, 07:35, said:

EDIT: unless you mean after playing exactly two round of diamonds but not 3rd. Then the percentages are almost correct (it should be 10.71%, 26,78%, 35,7%) but we really should eliminate Jxx onside if we talk about what decision to make here.

Sorry all, confusion, yes I did mean before embarking on the 3rd round of ♦. And yes those numbers are very familiar, I just rounded them.

But it's far simpler. There are 4 vacant spaces against 3 after the half round of ♦, assuming lead from 5. If the ♠ lead is from 3 it's 5:2 against. Happily, 26.78:35.7 is also 3:4, so we're all right. Apart from those who were wrong.