A restricted choice situation

[cnszsun]

AK9x

Qxx

Let's suppose no other relevant information. After A and Q, RHO played 10 on 2nd round.

I know the odds favor finesse now , but what's the ratio of finese to drop? Thanks in advance.

[kenberg]

I get 2 to 1, the same as in the more familiar situation where a quack comes down on the first round of the suit. Call a Jack or Ten a Tack, and assume that a player holding both Tacks will play them randomly. So the play goes A, small to Q, small toward dummy, spots being plaued up to the point of decision. All spots are equal and there were four of them. So we have 8 possibilities of Tack-x and 4 possibilities for JTx.

 

Or, if we look at the spot we have seen on the right as specified, we have 2 Tack-x possibilities and 1 JTx possibility.

 

This all assumes we treat the play of the ten or the Jack as the play of an unidentified Tack. Correct, if our opponent plays either Tack indiscriminately.

 

These things can be tricky but at least at first thought this seems right.

[Stephen Tu]

I get 2 to 1, the same as in the more familiar situation where a quack comes down on the first round of the suit. Call a Jack or Ten a Tack, and assume that a player holding both Tacks will play them randomly. So the play goes A, small to Q, small toward dummy, spots being plaued up to the point of decision. All spots are equal and there were four of them. So we have 8 possibilities of Tack-x and 4 possibilities for JTx.

 

It's very close, but slightly less than 2 to 1, because an individual 3-3 break possibility is slightly more likely than an individual 4-2 possibility. So it's ~64.5% to 35.5%.

[han]

Another way to say it is that LHO has 10 empty spaces and RHO has 11. So the odds are 20-11.

[lalldonn]

Another way to say it is that LHO has 10 empty spaces and RHO has 11. So the odds are 20-11.

Let's suppose dummy had AK8x instead and on the first two rounds RHO played two of the J-T-9. Now I know the odds favor the finesse much more than in the original example but the empty spaces are still 10-11. So how can that logic be right? On the other hand you are han so I know it's right, but I don't understand why.

[han]

In the original example there are two holdings where finessing works and one where dropping works. Assign 10 to each 4-2 split and 11 to each 3-3 split and you get 20-11.

 

In your example there are three holdings (J10, J9, 109) where finessing works and one (J109) where dropping works. So we get 30-11.

[lalldonn]

You are better than most teachers I had!

[JLOGIC]

This is pretty lol but it worked for me against a national champion at a recent tournament. At trick 2 in 1N on a blind auction, RHO played the CK and dummy on my left had Axx. I had Jx and played the jack. As luck would have it, he had KQ9x and had to guess whether I had JTx or Jx, and he assumed the J was JTx (I also play ud carding) and played for the drop.

 

It took him a while to convince him that the J would never blow a trick from Jx and that he was a fish. Anyways, it will probably work pretty frequently if you're not known as a tricky player or if the declarer is bad or if the declarer is good and thinks you're bad. Ofc usually they won't have KQ9x and it won't matter.

 

Another similar spot is when they have Ax opp Q9xxxx and they lead the ace and you have KJx onside and play the jack then low. This one so far has only worked once for me in my life but you gotta keep trying. It is probably more effective when Ax is in the dummy since you can't see the Q9xxxx and you are just hoping they have that.

[PhilKing]

 

It took him a while to convince him that the J would never blow a trick from Jx and that he was a fish.

 

 

He sounds like the bridge equivalent of Phil Hellmuth.

 

:ph34r:

[kenberg]

It's very close, but slightly less than 2 to 1, because an individual 3-3 break possibility is slightly more likely than an individual 4-2 possibility. So it's ~64.5% to 35.5%.

 

Yes, right. I took the question to mean: Is it essentially the same as the standard restricted choice situation? And the answer is yes. Both are, approximately, 2:1. By standard situation I mean A234 opposite KT567, the play goes A-8-5-J then 2-9-?. I suppose the usually quoted 2:1 in favor of the finesse also needs a modest adjustment.

 

Dropping from 66.7 to 64.5 is more than I expected though. Nor had I thought though using empty spaces as Han did to make the adjustment. His 20-11 translates to a probability of 20/31= 0.64516

 

 

Anyway, yes, somewhat under 2:1. Mea culpa, sort of. I didn't take the question as asking for that level of precision.

[billw55]

... or if the declarer is good and thinks you're bad.

Does this happen to you often? :o

[JLOGIC]

I wish. Although the reverse has it's upside also, sometimes you do something stupid or you just do something normal and they assume you're good so you're doin something good.

[y66]

Yes, right. I took the question to mean: Is it essentially the same as the standard restricted choice situation? And the answer is yes. Both are, approximately, 2:1. By standard situation I mean A234 opposite KT567, the play goes A-8-5-J then 2-9-?. I suppose the usually quoted 2:1 in favor of the finesse also needs a modest adjustment.

 

Dropping from 66.7 to 64.5 is more than I expected though. Nor had I thought though using empty spaces as Han did to make the adjustment. His 20-11 translates to a probability of 20/31= 0.64516 (approximately)

 

 

Anyway, yes, somewhat under 2:1. Mea culpa, sort of. I didn't take the question as asking for that level of precision.

 

fyp

[kenberg]

fyp

 

Thanks. :)

[Zelandakh]

This is pretty lol but it worked for me against a national champion at a recent tournament.

You need to add these to the falsecarding thread Justin!

[Fluffy]

I wish. Although the reverse has it's upside also, sometimes you do something stupid or you just do something normal and they assume you're good so you're doin something good.

 

This happens to me often :)

 

Anywa the KJx position, just insta playing the jack at second trick often works for me.