[mgoetze]

AKQ8

765

RHO contributes the Jack on the first round and the Ten on the second round of the suit.

(1) Assume your opponents play perfectly. Calculate the probability that (a) the finesse, (b) the drop will succeed.

(2) Do you believe that the probability that an intermediate opponent will play J then T from JT9 is greater, equal to or less than 1/6?

(3) Do you believe that the probability that an advanced opponent will play J then T from JT9 is greater, equal to or less than 1/6?

(Assume that there is no reason for the opponents to give suit preference.)

[Quantumcat]

When deciding what to do with AKQ8 opposite 765, the only holdings where it matters what you do (in all other, either the play is obvious, or there is nothing you can do), are:

xxx JT9
9xxx JT
Txxx J9
Jxxx T9

A priori, a 3-3 break is 35.5% and a 2-4 break is 24.25%. If we assume it has to be one or the other, adjusting so they remain in the same proportion but add to 100%, we have 3-3 break 59.4% and 2-4 break 40.6%.

So we have
xxx JT9 = 59.4%
9xxx JT = 13.5%
Txxx J9 = 13.5%
Jxxx T9 = 13.5%.

Edit: there are some rounding issues so this only adds to 99.9%. Add (0.01/3)% to each of the 13.5% if you are concerned.

We divide the 2-4 break in thirds because as soon as it doesn't split in any of these ways, we would be ceasing this analysis, we'd either know how to play it or wouldn't be able to.

Now we work out the percentage chance of RHO playing any combination of his cards.

First round
J (JT9) 19.8%
T (JT9) 19.8%
9 (JT9) 19.8%
J (JT) 6.75%
T (JT) 6.75%
J (J9) 0% ---> cards are not equal so he does not have free choice
9 (J9) 13.5%
T (T9) 6.75%
9 (T9) 6.75%

Second round
J then 9 (JT9) 9.9%
J then T (JT9) 9.9%
T then 9 (JT9) 9.9%
T then J (JT9) 9.9%
9 then J (JT9) 9.9%
9 then T (JT9) 9.9%

All the doubletons remain the same percentage as the 1st round, since they have no choice what to play next:

J then T (JT) 6.75%
T then J (JT) 6.75%
9 then J (J9) 13.5%
T then 9 (T9) 6.75%
9 then T (T9) 6.75%

Now we are in a position to determine RHO's most likely holdings given his play.

He has played T then J: the ratio JT9:JT = 9.9:6.75, or around 3:2

So, it is better to play for 3-3, assuming your opponents don't religiously give count and don't give suit preference.

More interesting is 9 then J. Then the ratio JT9:J9 is 9.9:13.5 or around 3:4, so you would finesse.

I worked all this out for myself, and don't know of any standard approach to this particular problem, so if there is a mistake in my analysis, or the standard approach is to finesse, please let me know.

[nate_m]

The following are my own calculations, but:

1.
a. 73.17%
b. 26.83%


P(A|B)=P(B|A)P(A)/P(B)= P(B|A)P(A)/[P(B|A)P(A)+P(B|~A)P(~A)]=P(B|A)P(A)/[P(B|A)P(A)+P(B|C)P©]=.7317

P(C|B)=P(B|C)P©/P(B)= P(B|C)P©/[P(B|C)P©+P(B|~C)P(~C)]=P(B|C)P©/[P(B|C)P©+P(B|A)P(A)]=.2683

Where A is RHO having J10, and B is that I see the J10 drop on the first 2 rounds of the suit. Call C the event RHO has J109.
P(B|C) is C(2,2)/C(3,2)=1/3. Assuming random plays from equal holdings, the order of the drops doesn't matter. (I assume that's what's meant by "perfect" play). We also make the assumption that RHO wouldn't dump the J10 from J10x, i.e. we assume that once he drops the J10, he holds either J10 doubleton or J109. If he's dropping out of 4 cards or longer, no play matters so we don't include that in our calculation.
P(B|A)=C(2,2)=1. This shouldn't shock, if RHO has J10 doubleton he must play the J10.
P(A)=[C(2,2)/C(6,2)][C(6,2)C(18,11)/C(13,26)]=.01615
The bracketed parts are the odds of pulling specifically the J10 out of any 2-4 split and the odds of a suit splitting 2-4 respectively.
P©=[C(3,3)/C(6,3)][C(6,3)C(18,10)/C(13,26)]=.01777
The bracketed parts are the odds of pulling specifically the J109 out of any 3-3 split and the odds of a suit splitting 3-3 respectively.

We obtain:
P(A|B)=(1)(.01615)/[(1)(.01615)+(1/3)(.01777)]=.7317
P(C|B)=(1/3)(.01777)/[(1)(.01615)+(1/3)(.01777)]=.2683

B.
C.
Neither opponent will vary their play. It does not pay to alter your play from equals. The reason "restricted choice" is valuable is that it is not a statement of bridge psychology, it is a statement for probability. To be willing to alter your play, you would have to be convinced that a given opponent would play J10 out of J109, keeping the 9, a bit over 90% of the time. (That is the break even point for the two probabilities.) Nobody does this, and even if they did, you could not be sure of it. It can be demonstrated, using a more general calculation, that it CANNOT pay to do so. That is, any dramatic alteration of strategy to always play J10 out of J109 carries a cost, because every time you DO drop a 9, the odds of J109 being your holding are minimal.

Restricted choice is NOT a statement of tendency. It is a fact of probability. Its power can be seen more clearly if you ask the more useful general question:
"If RHO drops ANY 2 equal honors, what are the odds of him holding that PARTICULAR doubleton or the specific holding J109. After all, there is nothing terribly unique about the play of the J10 as opposed to the J9 or 910 etc. You are ALWAYS better off hooking.

[JLOGIC]

From an at the table point of view, thinking of it as 3:1 is fine (JT, J9 T9 vs JT9). It is slightly worse since a 3-3 combo is more likely than a 4-2 combo, just like the classic AKT9x opp xxxx when they drop an honor on the first one is not quite 2:1 since a 2-2 is more likely than a 3-1.

I guess to answer your question I think good RHOs will play jack and then 9 basically never with JT9, so I think all of the other possible permutations are more likely than usual and would be about equal.

[JLOGIC]

Quantumcat, if you are letting their choice of cards affect your play then you are allowing them to exploit you. Surely you can see that it would be a poor overall strategy to allow your play to be different whether they play 9 then jack or jack then ten or whatever.

If you want to look at the exact cards played, then in the case of say, jack then ten they could have JT9 or JT. With JT they could play ten then jack, or jack then ten. With JT9 they could play jack then ten, jack then 9, ten then jack, ten then 9, 9 then ten, 9 then jack. it is roughly 6:2 (or roughly 3:1) that they have JT rather than JT9.

Also, your assumption about not being able to play jack from J9 is faulty, J9 is restricted choice. Even if you might have Tx and might choose to hook later in the context of the hand (which is very unlikely/impossible, if you needed 4 tricks you would have just taken a double finesse), it is a good play since you might convince them that you have JT9 and to go for the drop when otherwise they would certainly hook. I agree as a practical matter, bad players are never playing jack then 9 from J9 doubleton but against reasonable opps this does not apply.

[nate_m]

Quantumcat, on 2012-September-18, 19:43, said:

When deciding what to do with AKQ8 opposite 765, the only holdings where it matters what you do (in all other, either the play is obvious, or there is nothing you can do), are:

xxx JT9
9xxx JT
Txxx J9
Jxxx T9

A priori, a 3-3 break is 35.5% and a 2-4 break is 24.25%. If we assume it has to be one or the other, adjusting so they remain in the same proportion but add to 100%, we have 3-3 break 59.4% and 2-4 break 40.6%.

So we have
xxx JT9 = 59.4%
9xxx JT = 13.5%
Txxx J9 = 13.5%
Jxxx T9 = 13.5%.

I don't quite follow your reasoning. Firstly, in the PARTICULAR problem we are dealing with, any holding EXCEPT J10 doubleton or J109 is irrelevant, because if he had a doubleton containing the 9, he'd have been forced to drop it already, and he played the J10.

However, assuming that we are dealing with the general question of how do we play it after they drop 2 of the J109, I believe your scaling is incorrect.

It is the case that if I scale the odds of suit breaks, the odds of 3-3 is around 60% and 2-4 40%. However, having decided I only am going to consider ANY doubleton honor (J10, 109, J9) and the specific tripleton J109, the odds change again. The J109 combination is only one twentieth of the possible 3-3 splits, and the splits that contain J10, 109, or J9 are only one fifth of the possible 2-4 splits. I must take that into account if I am going to use a scaling the odds approach.

I assumed that order of play would not matter from doubletons, as you pointed out, it may matter in the case of J9 (though I think many would card randomly from J9 as well) Regardless, restricted choice is so with the odds that I don't think it affects the overall result terribly much.

[JLOGIC]

Even if they never played the J from J9, they could simply also never play J then 9 from JT9, and I'm sure that's what many do. Of course, those people will probably forget to play 9 then jack twice as often to compensate. I will admit I never play the J from JT9 and play the J very often from J9 vs weak opps, but this is simply to try and exploit weak players who see jack then 9 and assume JT9 since they've never seen it before. Against good players I just never play jack then 9.

This is all academic of course, in reality even if you have some amazing read on them that they are unbalanced in how they follow suit, you are never going to overcome 3:1, even if you read them well and figure out that it is only 2:1 to hook because of this, you're still gonna hook lol. I guess it might matter in some bizarre combining chances spot, but as a practical matter you will never play enough boards against someone to learn their exact tendencies and even get it down from 3:1 to 2:1

Basically, I agree with nate_m on this:

Quote

Regardless, restricted choice is so with the odds that I don't think it affects the overall result terribly much.

[mgoetze]

Solid first post, Nate, welcome to the forums!

When this came up at the table yesterday, I instictively hooked and it turned out to be correct. I'm glad to see my intuition about the math on this was correct as I was too tired to work it out myself.

Just one point though: just because it can be demonstrated that it doesn't pay to e.g. play JT from JT9 >90% of the time, doesn't mean there isn't a player in the world who does it anyway.

[bluecalm]

Now I want to see a hand when it's close because of revealed holding in a side suit...
What about this one:

---

The lead is Q♠. You draw trumps in two rounds then play AK of diamonds and two honors drops on the right !
Does restricted choice matter now ? What about if the trumps were 3-1 with 3 in opener's hand ? Your play ?

[JLOGIC]

bluecalm, on 2012-September-19, 03:09, said:

Now I want to see a hand when it's close because of revealed holding in a side suit...
What about this one:

---

The lead is Q♠. You draw trumps in two rounds then play AK of diamonds and two honors drops on the right !
Does restricted choice matter now ? What about if the trumps were 3-1 with 3 in opener's hand ? Your play ?

If LHO had one heart then I would hook, playing him for 6142 and RHO for 3325. This would be much more likely than 6232 because of restricted choice, and also because LHO would probably need to the CA if he was 6232 to open 2S (might depend on opps, I would assume most people are not opening 2S w/w on QJT9xx xx xxx xx)

If LHO has 2 hearts, it is interesting. Obv it seems right to go for the drop since with 6241 LHO would lead his stiff club. That is true unless LHO had the stiff A of clubs. However, if we decide that LHO would never preempt with 6232 without the CA, then we are playing him for the CA either way. On top of that, with 6232 and Ax of clubs, he might have chosen to lead his Ax of clubs.

Anyways, I think enough people would open 2S with 6232 and no CA that it is right to play for the drop if LHO has 2 hearts. If I was sure lho could have 6232 and just the QJ of spades, it would be more interesting

If LHO had 3 hearts, my head would definitely explode. Again with 6331 he would lead his stiff with no CA. So it is 6340, or 6331 with stiff CA. However, if LHO is 6340, that gives RHO Kxx x xx Axxxxx which would certainl have bid 4S. There is some problem that LHO might have led the SK from KQ because he had a club void trying to ensure his ruff but I still think generally RHO would bid 4S with 7321 and a stiff heart. Accordingly, I would play for LHO to have stiff A of clubs and thus the drop.