[Flame]

Echognome, on Oct 26 2005, 11:05 PM, said:

For best play in theory, we only need to consider 4 cases.

--W----E---Ex-Ante%---Conditional %
--T--Q9xx---2.83---------21.7
-T9---Qxx---3.39---------26.1
-QT---9xx---3.39---------26.1
QT9----xx---3.39---------26.1


Although I had to look up the percentages, the idea without the exact numbers is straightforward enough. You have seen 3 cards and that only leaves the holdings above as possibilities. Finessing the 8 gains on the first layout. Finessing the J gains on the second layout. Playing the K gains on the third layout. Nothing helps you on the last layout. The question you ask next is, would West play the T from each of those holdings. With the first holding, he has no choice. From the second holding he plays the T half the time and 9 half the time. From the 3rd holding he'd play the T most of the time. From the last holding he'd probably play the T or 9 equally as likely, but it's not really relevant since 5 tricks just aren't there. So my thinking is that since playing for either case 2 or case 3 is theoretically better than case 1 AND in case 2 West will falsecard more often, then I will play for case 3 and play for the drop.

For me this response is accurate and anything beyond it is philosophical.

[Blofeld]

cherdano, on Oct 26 2005, 06:36 PM, said:

Blofeld, on Oct 27 2005, 01:33 AM, said:

Ben, that was my initial thought, but LHO can't have Tx, as RHO has played two x's already.

Arend, if declarer holds the 9 then he won't be applying restricted choice, so I'm not sure that it's so suicidal to play the Q from QT. "Not sure", mind - I could still be convinced either way on this. Need to think about it a little more

When declarer has AKJ9x and the queen drops, he can just safely cash the king and win against both Q singleton (LHO shows out, so lead low from dummy and finesse the 9 on the 3rd round) or any 3-2 split (both follow, so just cash the J).

Arend

I shouldn't have been posting last night.

My point was that if declarer has the 9, and the T drops, then if you never play Q from QT he's going to play the K next, so you won't make any tricks anyway, and it doesn't cost to drop the Q. It doesn't gain anything either, but it does potentially gain when declarer doesn't have the 9, so it becomes a valid strategy again.

Does that make any sense at all?

[Fluffy]

Flame, on Oct 27 2005, 12:06 PM, said:

Echognome, on Oct 26 2005, 11:05 PM, said:

For best play in theory, we only need to consider 4 cases.

--W----E---Ex-Ante%---Conditional %
--T--Q9xx---2.83---------21.7
-T9---Qxx---3.39---------26.1
-QT---9xx---3.39---------26.1
QT9----xx---3.39---------26.1


Although I had to look up the percentages, the idea without the exact numbers is straightforward enough.  You have seen 3 cards and that only leaves the holdings above as possibilities.  Finessing the 8 gains on the first layout.  Finessing the J gains on the second layout.  Playing the K gains on the third layout.  Nothing helps you on the last layout.  The question you ask next is, would West play the T from each of those holdings.  With the first holding, he has no choice.  From the second holding he plays the T half the time and 9 half the time. From the 3rd holding he'd play the T most of the time.  From the last holding he'd probably play the T or 9 equally as likely, but it's not really relevant since 5 tricks just aren't there.  So my thinking is that since playing for either case 2 or case 3 is theoretically better than case 1 AND in case 2 West will falsecard more often, then I will play for case 3 and play for the drop.

For me this response is accurate and anything beyond it is philosophical.

Well matt restrricted choice says that you should multiply stiff T % by 2 because the 9 is equally probable and they are equivalen't, same thign happesn to QT and Q9. am I wrong?.


Adn to this point I start to understand what was justin talking about

[Fluffy]

Jlall, on Oct 27 2005, 01:54 PM, said:

Why is that? If I had AKJ9xx opp xxx and I cashed the ace and the T dropped, I would finesse always next. The defender could have stiff T or Tx vs just QT.

I think 10x is not an option when you see the last x appear on your right.

[Flame]

Fluffy, on Oct 27 2005, 08:56 AM, said:

Flame, on Oct 27 2005, 12:06 PM, said:

Echognome, on Oct 26 2005, 11:05 PM, said:

For best play in theory, we only need to consider 4 cases.

--W----E---Ex-Ante%---Conditional %
--T--Q9xx---2.83---------21.7
-T9---Qxx---3.39---------26.1
-QT---9xx---3.39---------26.1
QT9----xx---3.39---------26.1


Although I had to look up the percentages, the idea without the exact numbers is straightforward enough.  You have seen 3 cards and that only leaves the holdings above as possibilities.  Finessing the 8 gains on the first layout.  Finessing the J gains on the second layout.  Playing the K gains on the third layout.  Nothing helps you on the last layout.  The question you ask next is, would West play the T from each of those holdings.  With the first holding, he has no choice.  From the second holding he plays the T half the time and 9 half the time. From the 3rd holding he'd play the T most of the time.  From the last holding he'd probably play the T or 9 equally as likely, but it's not really relevant since 5 tricks just aren't there.  So my thinking is that since playing for either case 2 or case 3 is theoretically better than case 1 AND in case 2 West will falsecard more often, then I will play for case 3 and play for the drop.

For me this response is accurate and anything beyond it is philosophical.

Well matt restrricted choice says that you should multiply stiff T % by 2 because the 9 is equally probable and they are equivalen't, same thign happesn to QT and Q9. am I wrong?.


Adn to this point I start to understand what was justin talking about

No your not wrong, and 26.1*2 is better then 26.1 or 21.7*2.

[Trpltrbl]

Hannie, on Oct 26 2005, 04:37 PM, said:

---

When you cash the ace, west drops the 10. You cross to dummy and play a low one up, east plays low.

What is the best play in theory? Does this differ from your play in practice?

Beginner/intermediate please hide your answer.

In theory the best is the 8.
Will most play the 8, I doubt it.

When playing in real the situation is different, because of maybe bidding, already played cards maybe even having almost complete count on the hand.

For example, without knowing anything you have a 39% chance of playing for 5 tricks.
Let's say you know half the deck in opps hand. I am putting 7 "known" cards in each hand.
Now it becomes about 45%.
If I have even more info let's say I know 10 cards in each hand, the chance becomes 50%.
And of course, it really really depends on who you play against.

Is it the person, who just started reading the book, the one that has been stuck at the 3rd chapter for 30 years or the guy that wrote it

GBB

[Echognome]

I do not disagree with Justin that false-carding changes the odds significantly. I am not so sure that this is a case where restricted choice has such a strong bearing. The odds I put up were ex-ante odds (updated without account for false-carding). However, they still are the ONLY possibilities. That being said, if your opponents look like they couldn't possibly produce a falsecard to save their lives, then clearly finessing the J or playing the K are better than finessing the 8. If the opponents falsecard "optimally", then it becomes quite complicated. At a first cut, let me think of the problem as a conditional probability.

Let Pr(W10) be the probability that West plays the 10. Then given the limited spaces, we can write this out as:

Pr(W10) = P(W10|T)Pr(T)+P(W10|T9)Pr(T9) + P(W10|QT) + P(W10|QT9)

Where the conditional arguments are on west's actual holding and the updated unconditional probabilities from the table above (respectively 21% and 26% for the others). Since West always plays the T from T singleton, P(W10|T) = 1. Suppose that West randomizes T from T9 evenly. So then P(W10|T9) = 0.5. From the third holding, let's say that West almost always plays the T, but sometimes plays the Q (could be a mistake or falsecard). So say P(W10|QT) = 0.1. Finally from QT9, suppose again that west randomizes between the T and the 9. So Pr(W10|QT9) = 0.5

Then we get Pr(W10) = 1(0.21) + 0.5(0.26) + (0.9)(0.26) + (0.5)(0.26) = roughly 70%

Now I'm going to condition back on this:

P(T|W10) = Pr(T and W10)/P(W10) = 0.21/0.7 = approx 30%
P(T9|W10) = Pr(T9 and W10)/P(W10) = (0.5)*(0.26)/0.7 = approx 18.5%
P(QT|W10) = Pr(QT and W10)/P(W10) = (0.9)*(0.26)/0.7 = approx 33%
P(QT9|W10) = Pr(QT9 and W10)/P(W10) = (0.5)*(0.26)/0.7 = approx 18.5%

So in this model, which accounts for false-carding, I'd say that playing for the drop is slightly superior. Of course, this all depends on what assumptions you make about falsecarding.

[Echognome]

I should add one final note. Although I think the model above takes into account all of our views on West's actions, it doesn't take into account our view on East's. Will have to think about that a bit in terms of joint probabilities. As Justin points out, we need to consider probability that EAST plays 2 small from his possible holdings as well. I'm sure it can be easily updated. But as I'm off to the a Pairs League Match I'll wait until later tonight. Maybe one of the more mathematically or statistically inclined can update it.

[Echognome]

Got back quickly from work, so had a little extra time. I'm having way too much fun with probabilities.

Basically, the analysis comes down to the following questions:

With QT, how often do you think West will play the T?
With QT9, and T9, how often do you think West will play the T?
With 9xx, how often do you think West will play xx?
With Q9xx, how often do you think West will play xx?

For the 1st question, I interpreted Justin's "will false from QT/Q9 some of the time." as say 10%.

For the 2nd question, I used the assumption of playing the T 50% of the time.

For the 3rd and 4th question, I again relied on Justin who said "(with 9xx) good defenders will very often (some more than 50% of the time) play the T or 9 on the second round. Let's say they'll do it about 50% of the time." And "..with QTxx or Q9xx to start with, he will always play low and then low." So that is 100%.

Skipping all of the math (but you can check easily enough) I find then that:

Prob(T, Q9xx) = 36.4%
Prob(T9, Qxx) = 21.9%
Prob(QT, 9xx) = 19.7%
Prob(QT9, xx) = 21.9%

Which tends to suggest that the finesse gives the highest odds. But wait! There's more. With QTxx or Q9xx, you actually have a false carding situation. For starters you can never make more than 1 trick and you might make zero tricks. So I will argue that you should play the T or the 9 some of the time. If you assume as much as 50% of the time, then the odds change. They actually become:

Prob(T, Q9xx) = 22.3%
Prob(T9, Qxx) = 26.8%
Prob(QT, 9xx) = 24.1%
Prob(QT9, xx) = 26.8%

And now you want to finesse the J! If I assume you falsecard from Q9xx only about 25% of the time, then it's back to finessing the 8. If you falsecard about 40% of the time, the odds are then split almost exactly even over the 4 possibilities.

It's a fun problem. I can just say that whatever you choose for the questions above will affect the optimal play. This is your subjective belief about your opponents so is of course subject to your knowledge of them. I am only trying to make you consistent with your own beliefs.

[jchiu]

From a preprint of a recent article of mine ...
And you thought Matthew's post was complicated ...

There are four ways to distribute the missing queen
and ten between the defenders, namely 10 opposite Q975, Q10 opposite
975, 109 opposite Q75, and Q109 opposite 75. Clearly, the last case
is hopeless. The trick here is the observation that RHO plays the
nine in the second holding with probability $0 \leq p \leq 1$. Note
that by bridge logic, no RHO will ever play the nine from the first
holding, almost no LHO will ever play the queen from the second
holding. By restricted choice, the probability that the third case
occurs is halved. Assume also that the probability ``lost'' by
playing the nine in the second case is distributed proportionally to
the first and third case. By the law of total spaces, the {\textit {a
priori}} probability of the first case is $\gamma = 0.0283$ and the
{\textit {a priori}} probability of the remaining cases is $\delta =
0.0339$. Then, the probability that playing the king wins (the second
case) is $(1-p) \delta$, the jack (third case) is $\frac{\delta}{2} +
p \delta \frac{\delta}{\gamma+\delta}, and the eight (first case) is
$\gamma + p \delta \frac{\gamma}{\gamma+\delta}$. Solving the
appropriate inequalities, the king is better than the jack whenever $p
\leq \frac{\delta^2 + \gamma \delta}{2 \delta^2 + \gamma \delta} =
0.6472$, the king is better than the eight whenever $p \leq
\frac{\delta^2 - \gamma^2}{\delta^2 + 2 \gamma \delta} = 0.1135$, and
the jack is better than the eight whenever $p \geq \frac{\delta^2 -
\gamma \delta - 2 \gamma^2}{2 \gamma \delta - 2 \delta^2} = 3.719$
(never). Hence, play the eight whenever RHO falsecards occasionally,
but otherwise play the king. As a corollary, the optimal strategy
with thse assumptions requires RHO to play the nine from 976 with
probability $0.1135$, and this minimizes the success rate at $0.3794$.