[manudude03]

You have the suit:

xx

vs

AT9xx

to play for 1 loser to make the contract (this is not the trump suit).

You play the A and a small one West playing 2 low and East playing J-K.
You cross back into the long hand and now have to guess the suit. You know that West had 4 trumps compared to East's 0, but don't know any more about their shape.

You can either play for 3-3 or take the ruffing finesse.

What's your best chance of making it?

[gwnn]

(very possibly a little pedantry:) What do you mean by '1 loser' ? Don't you mean 3 winners?

[manudude03]

Yes, probably a better word for it.

[jdonn]

Someone else can tell you the exact odds. But it's far more likely that east has KJ doubleton than that east has KQJ tripleton AND has played the J and then the K. Restricted choice and all.

[mikeh]

Josh is technically correct, but most bridge players develop habits in restricted choice scenarios. We all 'know' we are supposed to mix up our sequence of plays but humans are very bad random number generators.

Thus against all but the best and the worst of players, I'd expect J then K from KQJ more often than J then Q. Of course, we also have to weigh KQJ, QJK, and so on, so it's not a 2 horse race in terms of how he'd play with KQJ.

Add to that the information that east had 4 trumps and either 3 or 4 in this suit, compared to East having 0 trump and 2 or 3 cards, and I think that against most people, this is as close to a 'pick 'em' as is possible to get.

I still go with the hook since it is technically sound, is definitely right against strong opps and weak opps (on restricted choice grounds) and means you don't have to spend 10 minutes trying to estimate the personal proclivities/abilities of RHO with KQJ....and then go wrong approximately half the time anyway. Life is too short.

[benlessard]

use automaton

http://www.automaton.gr/tt/en/odds.htm

considering that with KQJ you can play JK,JQ,and QK (order is irrelevant) so you basically have to divide by 3, less if you think your opps dont randomize enough (not able to play Q+K from KQJ) you also have to take into account that KQJ in a side suit make a pretty nice lead.

[barmar]

Doesn't the 4-0 trump split change the odds in favor of East having the Q, because of vacant spaces?

Also, I think that the order that East plays his cards from KQJ probably depends on whether you start with the A or lead a low card towards the AT. If you start from the top, I think he might try starting with the K.

[MFA]

Applying the principle of restricted choice to vacant spaces problems is quite easy, since we just have to multiply the relevant opponent's # of empty spaces with the number of ways he could potentially have randomized. That number is 2 in simple restricted choice situations and 3 in double restricted choice situations like this (with 3 equal honours where any of these could be saved for the third round).

So vacant spaces makes this 11:6 in favour of the drop, since we can place seven cards with west and two cards with east so far.

We multiply 6 by 3 because of the double restricted choice, and then it becomes 18:11 in favour of the finesse.

If we like we could then proceed to speculate if the defender is randomizing well enough or if he is more likely to have KQJ for some specific sequence of the honours, etc. And we could evaluate if it is consistent with the bidding to hold KJ and 11 cards in two suits. But the starting point is a solid 18:11 for the finesse.

EDIT: Small miscount of vacant spaces corrected (early in the morning ).

[benlessard]

if East has 13 vacant to west 9.

KJ---qxxx is 2.3 %

KQJ---xxx is 1.6 %

In theory we have to divide 1.6 by 3 because he can play QK,JK,JQ to get .533

.533/ 2.833 = 18.8% chance that is dealt KQJ and play JK.

2.3/2.833 = 81.2 % that is dealt KJ.

[MFA]

benlessard, on Sep 25 2010, 12:31 PM, said:

if East has 13 vacant to west 9.

KJ---qxxx is 2.3 %

KQJ---xxx is 1.6 %

Something is wrong. East is the one with KJ or KQJ in the suit, and since he is known to be short in trumps, KQJ as a specific holding is more likely than KJ as a specific holding.

[Fluffy]

benlessard, on Sep 24 2010, 04:18 AM, said:

considering that with KQJ you can play JK,JQ,and QK (order is irrelevant)

order is relevant, King then jack won't be palyed except from KQJ.

[benlessard]

Quote

Something is wrong. East is the one with KJ or KQJ in the suit, and since he is known to be short in trumps, KQJ as a specific holding is more likely than KJ as a specific holding.

Your right i inverted east and west.

Qxxx--KJ = 0.88 = 62.3 %

1.6% divided by 3 is still good = 37.7 %

Quote

order is relevant, King then jack won't be palyed except from KQJ.

If you can find opponents friendly enough for this I agree, of course on the otherside you can find opps that can mastermind with K from KJ. So since i expect less than 1% to play KJ from KQJ or KJ it doesnt change a thing.

[MFA]

62.3% to 37.7% is about the same as the 18:11 I suggested. I can recommend learning the method I am using, since the calculations are so easy that they can be performed at the table for instance within seconds.

[manudude03]

Scoring: MP

---

Ace of clubs led followed by a second club. Cashed the ♠A and discovered the 4-0 break, then cashed ♦A and exited a small diamond from hand (East playing J-K). A heart came through and I decided to try for the overtrick, had a feeling the K wasn't onside so rose up with the ace and ran the ♦T. -1 when East had KQJ tight.

[Bbradley62]

But your analysis of the diamond suit should not be done in a vacuum. You can both increase your chance of making 11 tricks and guarantee your contract:

Trick 7) Ruff the third diamond. If diamonds split 3-3 you can cash 11 tricks. Otherwise:

T8) cash ♣Q, pitching ♥x. If West ruffs, you cash 10 tricks.
T9) ruff ♥T back to dummy. If West over-ruffs, you cash 10 tricks.

T10) ruff fourth diamond;
T11) ruff ♥Q (per force high, with West having to under-ruff);
T12) ♠K;
T13) ♠J

This yields 11 tricks if West is 4243 and makes 10 if he is 4342 or 4144. Of course, if you have ♠9 in your hand, you can ruff high on both T9 and T11 and lead toward J9 on T12. This increases your chance of making 11 tricks from 62% to about 68-69% (depending on the ♠9) while limiting the downside to making 10 tricks.