9 card fit missing KQ

[leswt]

I watched a hand yesterday where there was a 9 card fit missing KQ, declarer finessed first round, losing to the K, and then finessed again for the Q (loosing it was a 2-2 split with KQ offside)

 

Many of the kibitzers thought this was wrong, "nine never" and all.

 

Isn't this a rule of restricted choice situation?

 

There was no useful information from the auction about the distribution of the suit

 

Thanks

[1eyedjack]

There is seldom no useful information from the auction and play to date. But that said, from the information in this post, I would say that declarer probably played with the odds.

[gwnn]

three plays:

1. finesse then finesse

2. finesse then ace

3. ace

 

Cases that are possible to pick up but aren't picked up by all three: (LHO holds)

 

KQxx, KQx: picked up by 1 and 2, but not 3

xx: picked up by 2 and 3 but not 1*

Kxx, Qxx: picked up by 1 and 3 but not 2*

 

So 1>3>2, which is a well known bridge fact. However, people tend not to think of other possible holdings and just berate any declarer who plays red cards with minus numbers instead of green cards with equal signs on them.

 

Note that it is often better to play ace first because KQxx might be troublesome anyway, and because it gets out the trumps faster (you don't want to allow a defensive ruff when they have Hx-Hx).

[gwnn]

Oh I see that I asterisked xx and Kxx, Qxx, but I didn't write anything about them.

 

I wanted to say that when, say, the finesse loses to the K, then Kxx onside ceases to be a possibility, but then restricted choice makes xx half as likely than Qxx.

 

The analysis above refers to a time before you touched the suit and are comparing various strategies.

[inquiry]

Gwnn got it right.

 

Here is a similar summay. Let's say Ace is in dummy. There are some situations where you have to lose two tricks. RHO (hand behind dummy) holds KQx or KQxx... there is no way to guess correctly. These percentages are 12.4 and 4.8. Other than these two holdings, you can always lose just one trick assuming you guess correctly. The hands were you CAN guess correctly are (west holding). There are other hands you can not misguess. Hands like LHO holding singleton honor or KQ doubleton, or Kx or Qx.

 

So less ignore the hands where you can not misguess or can not guess right. That leaves us with just the following few combinations.

 

LHO holds

xx

Hxx

HHx

HHxx

 

CASE ONE: LHO holds two small. The winning play is either ACE first, or finesse first then Ace.

 

CASE TWO: LHO holds three small, the winning play is either ACE first (dropping stiff honor offside) or finesse both times

 

CASE THREE: The winning play is to finesse first, then of course win ACE.

 

CASE FOUR: The only winning play is to finesse both times.

 

 

Each case is not equally likely, of course. West holding xx is only 6.8% chance, West holding Hxx is 12.4%, holding HHx is also 12.4%, and HHxx is 4.8%. I don't like working with percent numbers like this, I prefer using combination.

 

There is only one way WEST can have HHxx, or xx. There are two ways he can have HHx or Hxx. Thus there are only six possible combinations we have to deal with. The likelyhood of the xx combination, however, is greater than the likelyhood of the HHxx combination because the later requires all four cards in the suit be dealt to one person. However, case two and three are equally likely, Without figuring out the likelyhood of the different hand patterns (the precentage things showed earlier), we see that:

 

Play ace or double finesse works in case two, so let's eliminate that one.

 

That leaves us with case one (playing ace wins), and case three and four, double finesse wins. Case one covers one specific hand pattern. Case three and four cover three hand patterns. It is true that case four is less likely than case one, but case 3 (with two possible holdings) is more frequent.

 

The advantage of the double finesse of play ace (we will ignore finesse and then ACE as it is basically same as play the ace first), is quite clear. If we take the total combination in play (4) it wins on three of them, while Ace wins on one. If we take the percentages of "successful play" when it matters, we find play the ACE wins 6.8%(out of 24% of hands where it makes a difference) compared to 17.2% (out of 24% where it makes a difference).

 

 

This makes it look like a huge difference, but it really is not. Here is a break down table:

 

WEST HOLDS, those in RED play ACE first wins, those in BOLD UNDERLINE, finesse twice wins

-

x

xx (6.8)

H (6.2%)

Hx (6.8%)

Hxx (12.4%

HH (6.8%)

HHx (12.4%)

HHxx (4,8%)

 

Adding up the percentages, you get double finesse is 76%, play Ace and another 65.6%, a net difference of 10.4%. The 24% we had in play earlier was 6.8 vs 17.2 or the same 10.4%. But you should be able to figure out finesse is correct line (not necessarily, of course, the winning line) without so much math...

[leswt]

Thanks to everyone for these great replies

[JLOGIC]

9 never refers to having 9 missing the queen, it is percentage to play for the drop. In the situation you describe, you are absolutely right, restricted choice makes it right to double finesse.

[kenrexford]

Obviously, however, with the odds being only a smidge different, any information suggesting that the person behind the Ace must have or likely has at least two in the suit or must have or likely has at least one honor may change the correct line.