[Jlall]

So say you have this suit:

AT9xx opp
Jxxx

You lead low to the 9, and RHO wins an honor. Now RHO doesn't return his partners suit, so you know he has a stiff spade (this suit is clubs).

Should you now play for the drop, based on RHO being far more likely to be 1-2 than 1-1, or should you play restricted choice?

I figured the way to analyze this would be to look up the odds of 1-2 vs 1-1 and it looked like about at least 3.5:1 more likely (I just compared 6511 vs 5521, 7411 vs 6421 is even more), which obviously trumps the 2:1 of restricted choice by a lot.

However, when Roger asked me what the right play was I told him hooking was clearly right imo. I then asked my dad and he insta hooked. I feel like almost all good players would hook (maybe I'm wrong).

This means that probably I'm going about analyzing this problem wrong, since the a priori shape odds would indicate that it's almost twice as likely to be right to drop than hook.

The other possibility is that I and my dad and probably others would just be very far off in our answer, and I'm analyzing this well. Obv this is a possibility, and people in general have a bad feel for math/stats/probability. I still feel like I'm doing something wrong in the analysis though, and my intuition is correct.

So anyone good at math, what am I doing wrong here?

Also, please ignore bidding clues/etc etc I'm considering this purely a math problem at this point (RHOs first chance to bid was at a very high level and he's weak so the bidding clues aren't relevant).

Edit: I guess we can say rule out RHO having an 8+ card suit from the bidding if that helps at all.

[bluecalm]

My intuition tells me to play for a drop.
I would be very happy if there is a way to analyze this kind of problem as I encountered similar problem several times and I came to the conclusion that without computer simulations it's probably not doable (but I am not very good at math).
I think you should compare all possible distributions though (for which we need to know number of cards we have in all other suits). Not only compare some with 1-1 to some with 2-1. Is there any tables with a priori probability of various distributions ?

[gnasher]

You seem not to have hold us how many spades they opponents have. I'll assume six, so the suit is 5=1 and the vacant places ratio is 8:12.

The chance of a specific 3=1 club break is proportional to 8 x 7 x 6 x 12
The chance of a specific 2=2 club break is proportional to 8 x 7 x 12 x 11

Specific 3=1 : Specific 2=2 = 6:11

Hxx-H : xx=HH = 12:11

So, with these conditions the finesse is sligtly bettern than the drop. If you make the spades 6=1, the odds become 10:11, and the drop becomes correct.


I expect that when I get up tomorrow mornng I'll find about 10 posts refuting this analysis.

[dburn]

Not in the least - looks reasonably fine to me. We can't answer the original question without knowing how many spades the opponents have between them, but as soon as we do know, we can give an answer based on gnasher's reasoning above. Of course, the vacant places analysis will need to take into account the "low-club" suit, on which we have a complete count, but that is a minor detail.

[Cascade]

how many spades are out?

[Jlall]

There are 7 spades out, so 6-1. Sorry for not including that!

[Jlall]

So I was wrong but it's close, that I find a lot more believable. Thanks.

[bluecalm]

Wow, this reasoning really seems to be correct. I am amazed Thanks !

[MFA]

I don't quite follow gnasher's math, but I get the same result, so ...

There is a short cut. We should just count vacant spaces and play for drop if east has more than twice the number of vacant spaces as west.

With spades 5-1.
At the kill point west has 6 vacant spaces (5♠+2♣) while east has 11. So the odds are 2*6:11, 12:11 as gnasher also finds. We should finesse.

With spades 6-1.
At the kill point west has 5 vacant spaces (6♠+2♣) while east has 11. So the odds are 2*5:11, which is 10:11. We should play for the drop.

[fred]

There is an odds calculator in the web-client that makes it easy to solve this sort of problem.

I used it to verify that Gnasher's (apparently correct) contention that the drop is a 11:10 favorite given that a side suit is 6-1.

Fred Gitelman
Bridge Base Inc.
www.bridgebase.com

[fred]

Apparently if a side suit is 7-4 this becomes a complete guess.

Fred Gitelman
Bridge Base Inc.
www.bridgebase.com

[kenrexford]

I'm not sure that the 1-1 vs. 2-1 analysis is the way I'm thinking about this right now. It seems to me that the line of thought I'm seeing from others is to focus on the expectations of this suit splitting when we know pattern elsewhere.

This could be in the nature of one known suit, like spades being 6-1. That leaves LHO with 7-2=5 cards elsewhere and RHO with 12-1=10 cards elsewhere. If RHO have 2:1 on cards elsewhere, this seems to make a 3:1 split less likely than expected without that knowledge.

But, the existence of a stiff is not the key. If spades were known to be 6-2, for instance, then we still end up with a 5:9 ratio. If some other suit were known to be split 2:5, then we end up with a 3:4 ratio.

It just seems like analysis of likely split with residue holdings, however they may be known.

It would seem to be worthwhile, IMO, to generate a table along lines reflecting this.

For example, a 3:1 split exists x% of the time when...

1. Number of cards remaining
2. Ratio of knowns (and hence ratio of unknowns)

In other words, maybe stuff like:

a. a 3:1 split is x% when 10 unknowns split 4:6
b. a 3:1 split is x% when 10 unknowns split 3:7
c. a 3:1 split is x% when 10 unknowns split 2:8

If you did this for all possible knowns, I bet you could figure out where the odds change, probably on some sort of curve or line on a chart. There is probably even a set of ratios where the 3:1 becomes less likely than the 2:2. Just as an example, maybe that point is something like:

11:7
11:6
11:5
10:5
10:4
9:4
9:3
8:3
8:2
7:2
7:1
6:1
5:1

(Random numbers, just for illustration)

[MFA]

fred, on Apr 11 2010, 01:40 AM, said:

Apparently if a side suit is 7-4 this becomes a complete guess.

Fred Gitelman
Bridge Base Inc.
www.bridgebase.com

Yes.

West will have 4 vacant spaces at the kill point, and east will have 8.
2*4 = 8, so it's a toss up.

[Jlall]

Yeah this was really easy I'm not sure why I got confused lol. Clee tricked me!

[gnasher]

dburn, on Apr 11 2010, 12:21 AM, said:

Of course, the vacant places analysis will need to take into account the "low-club" suit, on which we have a complete count, but that is a minor detail.

Treating the low clubs and high clubs as different suits just means that we remove the "8 x 7" from both of my calculations.

[FrancesHinden]

kenrexford, on Apr 10 2010, 11:41 PM, said:

It just seems like analysis of likely split with residue holdings, however they may be known.

It would seem to be worthwhile, IMO, to generate a table along lines reflecting this.

Try reading Jeff Rubens' latest book.

[barryallen]

MFA, on Apr 10 2010, 06:34 PM, said:

I don't quite follow gnasher's math, but I get the same result, so ...

There is a short cut. We should just count vacant spaces and play for drop if east has more than twice the number of vacant spaces as west.

With spades 5-1.
At the kill point west has 6 vacant spaces (5♠+2♣) while east has 11. So the odds are 2*6:11, 12:11 as gnasher also finds. We should finesse.

With spades 6-1.
At the kill point west has 5 vacant spaces (6♠+2♣) while east has 11. So the odds are 2*5:11, which is 10:11. We should play for the drop.

Not sure I am comparing apples with apples, but for the point we are now at. Having finessed once and lost to an honour, we now assume that East only started with one spade. Now assuming you have the option of being able to take the second finesse we are now left with the following possibilities in clubs.

1) - v Hx
2) x v H
3) H v x
4)Hx v -

Leading towards North, options 1 and 3 are an irrelevance. You can do nothing about 1, and 3 being obvious which ever method. That just leaves the individual odds for 2 and 4, with 2 being close to a 5/2 favourite for the drop?

[dburn]

FrancesHinden, on Apr 11 2010, 04:22 AM, said:

kenrexford, on Apr 10 2010, 11:41 PM, said:

It just seems like analysis of likely split with residue holdings, however they may be known.

It would seem to be worthwhile, IMO, to generate a table along lines reflecting this.

Try reading Jeff Rubens' latest book.

Or even Jean-Marc Roudinesco's Dictionary of Suit Combinations, where such tables are explicitly shown.

The reason they are not of much practical use, though, may be seen if we consider the actual calculations a player would need to perform. In the original example, it was said that at East's first turn to bid the auction was at such a level that he would never bid whatever his hand. This is rare, and if one places the original example within a context where East might bid, this is what you have to do:

Call the chance that playing for the drop works D; call the chance that taking the finesse works F (of course, F = 1-D since one or the other line must work).

Now, the relative probability of D is 11/21 (the chance that East has ♣KQ) multiplied by the chance (call it C10) that East would not bid with that club holding and ten red cards. The relative probability of F is 10/21 (the chance that East has a singleton club honour) multiplied by the chance (call it C11) that East would not bid with that club honour and eleven red cards. To recalculate the absolute probabilities of D and F, divide each by the sum of the newly calculated relative probabilities.

Of course, no table exists in which you can look up the chance that any given East will enter the auction with any given hand - you must assess C10 and C11 subjectively. In general, players are more likely to bid with eleven cards in two suits than with ten; if you think it likely that East would bid with 60% of his possible 1=X=[11-X]=1 hands but only 50% of his possible 1=X=[10-X]=2 hands, you should finesse. In fact, and fairly obviously, you should finesse whenever you consider that C11/(C11+C10) > 11/21, because then F*C11 > D*C10.

In the actual example C11=C10=1 (East would never bid whatever his hand), so there is no need to perform the above calculation and, as gnasher and Fred's odds checker say, you should play for the drop. That is, you should play for the drop unless you consider that West is less likely to have done whatever he did in the auction with six spades and two low clubs than with six spades and three clubs to an honour. How much less likely? Well, we will leave that as an exercise for the reader, whilst observing that unless the reader is unusually proficient at both mental arithmetic and assessing his opponents, he should incline to rely more on intuition than on mathematics.