Suit combination

[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]

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]

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]

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]

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