[phil_20686]

JLOGIC, on 2012-November-16, 14:56, said:

A simpler way of looking at it is that there are 3 clubs remaining, and your play only matters if LHO has a singleton remaining. 2 of the clubs are small and 1 is the ten, so it is 2:1 to hook.

This logic feels intuitive, but is actually wrong. Suppose the missing pips are the 432, and on the first round lho plays the 4. The relevant holdings are lho having 43, 42, T2 initially, but from either of the first two he would have played the 4 only half the time, so those two are exactly half as likely as the Tx, and its fifty fifty.

In essence, although you are right that restricted choice doesn't apply at trick two, it does apply at trick one, as lho's play is forced from Tx but not from xx, and rho's play of the K is always forced.


Again, the fool proof way of looking at these is to look at the a-priori holdings. Given that lho has a doublton and rho the K, there are 6 relevant holdings, Tx-Kxx three times, and xx-KTx three times, the card lho plays at trick one does not affect the a priori holdings, so it remains 50-50.

[phil_20686]

Phil, on 2012-November-16, 09:27, said:

Rubens calls the non-K clubs a 'suit within a suit' or a 'subsuit'. There are four non-king clubs, and they will split 2-2 more often than 3-1. We can't cope with a 3-1 remainder either way, so it more or less mirrors a two way finesse.

So this type of reasoning is fine, but it is implied that you do not use the a-priori odds for a 2-2/finesse. If all you know is that rho has the K, clearly 3-1 is more likely than 1-3 for example.

Again, in this particular case it is easy to work out the relative odds exactly: the drop is exactly as likely to succeed as the finesse.

[JLOGIC]

You are right obv not sure what I was thinking lol.

Basically once the king is known it is equally likely that either player has Tx vs xx obviously.

[MrAce]

phil_20686, on 2012-November-20, 13:31, said:

This logic feels intuitive, but is actually wrong. Suppose the missing pips are the 432, and on the first round lho plays the 4. The relevant holdings are lho having 43, 42, T2 initially, but from either of the first two he would have played the 4 only half the time, so those two are exactly half as likely as the Tx, and its fifty fifty.

In essence, although you are right that restricted choice doesn't apply at trick two, it does apply at trick one, as lho's play is forced from Tx but not from xx, and rho's play of the K is always forced.


Again, the fool proof way of looking at these is to look at the a-priori holdings. Given that lho has a doublton and rho the K, there are 6 relevant holdings, Tx-Kxx three times, and xx-KTx three times, the card lho plays at trick one does not affect the a priori holdings, so it remains 50-50.

I have to admit i also thought just like Justin, untill i read this reply.

[SteveMoe]

phil_20686, on 2012-November-20, 13:31, said:

This logic feels intuitive, but is actually wrong. Suppose the missing pips are the 432, and on the first round lho plays the 4. The relevant holdings are lho having 43, 42, T2 initially, but from either of the first two he would have played the 4 only half the time, so those two are exactly half as likely as the Tx, and its fifty fifty.

In essence, although you are right that restricted choice doesn't apply at trick two, it does apply at trick one, as lho's play is forced from Tx but not from xx, and rho's play of the K is always forced.


Again, the fool proof way of looking at these is to look at the a-priori holdings. Given that lho has a doublton and rho the K, there are 6 relevant holdings, Tx-Kxx three times, and xx-KTx three times, the card lho plays at trick one does not affect the a priori holdings, so it remains 50-50.

Echoing others, well said Phil.

[gszes]

Sometimes it is just plain easier to understand a concept by using
specific examples rather than just using generic cards. The opposing
missing clubs are KT764. We need only concern ourselves with the
best way to play the club suit WHERE WE HAVE A CHANCE TO GET
IT RIGHT. This is importand because there are many club holdings
where we have no chance no matter how we play the suit and those
holdings are IRRELEVANT to our decision making process here.

For us to have any chance at success rho must have started with
3 clubs (with the K) OR the KT doubleton. When we play the first
round of clubs the K/7 disappear from the opps club suit and all
that is left are the T64. When we play the second club and a small
card comes from rho the KT doubleton chance is gone and all that
is left is for the original club holding to have been:

T7 K64
76 KT4
74 KT6

Assuming both opps are capable of randomizing their spot card play
note that once the hands are reduced to 3 card possibilites the odds
favor the finesse by 2/1 since there are 2 possible holdings for rho
to hold the T and only 1 possible holding for LHO to hold the T.

That means the original guess to drop the T doubleton was based on
using the wrong logic at the right time. Lending further proof to the long
standing theory "I'd rather be lucky than good".

[phil_20686]

gszes, on 2012-November-26, 09:02, said:

For us to have any chance at success rho must have started with
3 clubs (with the K) OR the KT doubleton. When we play the first
round of clubs the K/7 disappear from the opps club suit and all
that is left are the T64. When we play the second club and a small
card comes from rho the KT doubleton chance is gone and all that
is left is for the original club holding to have been:

(a) T7 K64
(b) 76 KT4
© 74 KT6

Assuming both opps are capable of randomizing their spot card play
note that once the hands are reduced to 3 card possibilites the odds
favor the finesse by 2/1 since there are 2 possible holdings for rho
to hold the T and only 1 possible holding for LHO to hold the T.

That means the original guess to drop the T doubleton was based on
using the wrong logic at the right time. Lending further proof to the long
standing theory "I'd rather be lucky than good".

I thought I killed this zombie.

Holding (a) is twice as likely as (b) or ©, since on the first trick lho could have played either spot from (b) and ©, but not from (a), while rho had to play the K.

So its fifty fifty.

[aguahombre]

phil_20686, on 2012-November-26, 15:19, said:

I thought I killed this zombie.

Holding (a) is twice as likely as (b) or ©, since on the first trick lho could have played either spot from (b) and ©, but not from (a), while rho had to play the K.

So its fifty fifty.

Maybe he doesn't read other posts.

[PhilKing]

I go for the finesse.

Either it's 50/50, or the ever-dwindling band of restricted choice fluffers are right.

Alternatively, op will crack and tell us which spot card West played on the first round ...

[phil_20686]

PhilKing, on 2012-November-26, 17:12, said:

I go for the finesse.

Either it's 50/50, or the ever-dwindling band of restricted choice fluffers are right.

Alternatively, op will crack and tell us which spot card West played on the first round ...

In real life there is a strong bias towards playing your lowest spot in these situations, at least that is my experience. If LHO produced the seven I would say there is a strong bias in favour of the drop, as I don't think most players would randomise optimally from 7x.

[Phil]

I think this thread is very offensive to zombies.

[PhilKing]

phil_20686, on 2012-November-26, 15:19, said:

I thought I killed this zombie.

You can't kill a zombie. They are undead.

Phil, on 2012-November-26, 19:34, said:

I think this thread is very offensive to zombies.

And this is doubtful.

[y66]

PhilKing, on 2012-November-26, 19:54, said:

You can't kill a zombie. They are undead.

According to pop zombologist, Max Brooks, you can kill a zombie by obliterating its brain.

[wyman]

Phil, on 2012-November-26, 19:34, said:

I think this thread is very offensive to zombies.

...and fluffers

[gszes]

gszes, on 2012-November-26, 09:02, said:

T7 K64
76 KT4
74 KT6

Assuming both opps are capable of randomizing their spot card play
note that once the hands are reduced to 3 card possibilites the odds
favor the finesse by 2/1 since there are 2 possible holdings for rho
to hold the T and only 1 possible holding for LHO to hold the T.

I realize i am self correcting========
The above logic is fine as far as it goes but it fails to take into consideration
all of the latest information. The second round of clubs RHO follows with
a small card.
If it was the 6 then the original 3 card holdings are reduced to
74 KT6
T7 K64
If it was the 4 then the original 3 card holdings are reduced to
76 KT4
T7 K64

In both cases it has become a 5050 proposition. It has nothing to do with
vacant spaces and little if anything (except at the beginner levels) to do with
restricted choice. Cold hard deductive reasoning (apparently something i was
not capable of after a ten day bridge layoff sigh)

[Zelandakh]

gszes, on 2012-November-27, 15:56, said:

I realize i am self correcting========

...or you could have just given Phil a well-deserved +1.

[aguahombre]

Zelandakh, on 2012-November-28, 08:27, said:

...or you could have just given Phil a well-deserved +1.

The Scottish Phil. M's Clayton and King are involved in the movement for fair treatment of the undead --- probably an offshoot topic which actually does merit further exploration.

[mike777]

ya off topic but you raise the issue do robots or at least scottish robots deserve fair treatment if undead or unborn?

[Zelandakh]

mike777, on 2012-November-28, 10:04, said:

ya off topic but you raise the issue do robots or at least scottish robots deserve fair treatment if undead or unborn?

The Scots never deserve fair treatment, dead or alive or undead.

[wyman]

Quote

The Scots never deserve fair treatment, dead or alive or undead.