[petterb]

76542

KJ93

What is the best way to play this suit combination when you need 4 tricks?

[Cyberyeti]

There are 16 possible holdings on your right, against void, 8, 10, Q, A, 108 nothing works, against AQ anything works so there are 9 where it makes a difference.

3 possible lines:

Play the lowest card which beats RHO's (9)
Play the J if RHO plays 10 or 8, K if he plays Q (J)
Play the K unless RHO plays A (K)
RHO :working play(s)
AQ108 : 9
AQ10 : 9 J
AQ8 : J
A108 : K
Q108 : 9
A8 : K
A10 : K
Q10 : 9 J
Q8 : J

9 works in 4 cases 1x 4-0 2x 3-1, 1x 2-2
J works in 4 cases 2x 3-1 2x 2-2
K works in 3 cases 1x 3-1 2x 2-2

Since each 2-2 break is a little more likely than each 3-1 which is a little more likely than each 4-0

J>9>K

[Fluffy]

Low to King only when you have only 1 entry.

[Cyberyeti]

Fluffy, on 2014-March-04, 16:17, said:

Low to King only when you have only 1 entry.

Isn't J just as good, J works v Q8/Q10, K works v A8/A10

[sfi]

Yes. Dropping the stiff queen offside only helps when you have the 10.

[gszes]

cyberyeti supplies a great chart and, even more importantly,
mentions the concept of imagining the cases where your choices
make a difference. I like this kind of instruction so I hate saying
this but when rho has the QT the 9 never works. Since you are using
the statistics from these cases to help make a choice the actual
results are J=4 K=3 9=3 which gives the J the clear edge over the
other two though (as seems to be the infuriating case in bridge) you
will still be wrong 60% or so of the time if you pick the best choice.

[Cyberyeti]

gszes, on 2014-March-05, 18:09, said:

cyberyeti supplies a great chart and, even more importantly,
mentions the concept of imagining the cases where your choices
make a difference. I like this kind of instruction so I hate saying
this but when rho has the QT the 9 never works. Since you are using
the statistics from these cases to help make a choice the actual
results are J=4 K=3 9=3 which gives the J the clear edge over the
other two though (as seems to be the infuriating case in bridge) you
will still be wrong 60% or so of the time if you pick the best choice.

You don't understand what I meant by 9. 9 indicates that my intention is to play the 9 unless RHO plays a higher card, in which case I'll beat it as cheaply as possible, so I actually play the J or K depending on which one of Q10 RHO plays, just as I won't play the K if RHO plays the A from AQ when K is my choice.

[mfa1010]

If they play T, we should play J to cater to QT or AQT, where K would cater to AT only.

If they play 8, it is in principle a pure guess between J and K. J handles Q8 and AQ8, and K handles A8 and AT8.
(Going 9 would be epsilon worse since it handles QT8 and AQT8 - two less balanced distributions.)

But a clever 2nd hand can play T from AT8 to force us to misguess to the stiff Q. If 2nd hand might find that play, then J>K when we see the eight, because AT8 becomes less likely.

[gszes]

Cyberyeti, on 2014-March-05, 18:16, said:

You don't understand what I meant by 9. 9 indicates that my intention is to play the 9 unless RHO plays a higher card, in which case I'll beat it as cheaply as possible, so I actually play the J or K depending on which one of Q10 RHO plays, just as I won't play the K if RHO plays the A from AQ when K is my choice.

I realize that playing the 9 under the T or Q is a looney tunes play I was just trying to point
out that the 9 can never be considered a winning choice when rho has QT but your individual cases
indicate the 9 as a "correct" play in order to achieve the 9 being "right" 4 times. In fact, I was
so focused on that one perceived discrepancy I failed to notice the same "problem" when rho holds the
AQT and you also have a 9 there which can never be right. The actual totals are J=4 k=3 9=2. keep
up the good work

[Cyberyeti]

gszes, on 2014-March-06, 07:57, said:

I realize that playing the 9 under the T or Q is a looney tunes play I was just trying to point
out that the 9 can never be considered a winning choice when rho has QT but your individual cases
indicate the 9 as a "correct" play in order to achieve the 9 being "right" 4 times. In fact, I was
so focused on that one perceived discrepancy I failed to notice the same "problem" when rho holds the
AQT and you also have a 9 there which can never be right. The actual totals are J=4 k=3 9=2. keep
up the good work

I simplified slightly in my terminology, the logic is right, there are three strategies (rather than specific cards) which I symbolised by 9, J and K.

Beat RHO's card as cheaply as possible, play the 9 if RHO plays the A
Beat RHO's 10/Q as cheaply as possible, play the J on the 8, play the 9 on the A
Play the K unless RHO plays the A in which case you play the 9