[CSGibson]

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Interested in hearing situational analysis - if you play the ace from hand in a trump suit, missing the KQJ5, and the 5-K pops up, what do you think the chances are that you are losing 2 tricks in the suit? How about if the 5-Q pops? 5-J? This is assuming that someone would play randomly from equals, but lowest otherwise...

[jdonn]

Yes restricted choice applies, which is another way of restating your assumption that someone plays randomly from equals. I'll use the simplifying assumptions that 3-1 either way = 50% and 2-2 = 40%, which are very close to the actual odds and make the math a lot easier.

QJ5 - K = 50% (3-1) * 50% (rho singleton) * 25% (singleton is king) = 6.25%
J5 - KQ = 40 % (2-2) * 50% (king on right) * 33.3% (queen also on right) * 50% (king played) = 3.33%
5 - KQJ = 50% (3-1) * 50% (lho singleton) * 25% (singleton is 5) * 33.3% (king played) = 2.08%

6.25% + 3.33% + 2.08% = 11.67%

QJ5 - K = 6.25%/11.67% = 53.57% (15/28)
J5 - KQ = 3.33%/11.67% = 28.57% (8/28)
5 - KQJ = 2.08%/11.67% = 17.86% (5/28)

You calculate similarly if RHO plays the jack (consider KQ5 - J, K5 - QJ, Q5 - KJ, 5 - KQJ) or the queen (KJ5 - Q, K5 - QJ, J5 - KQ, 5 - KQJ)

[jdonn]

Btw, it's not necessarily optimal for RHO to play randomly from equals here, do you see why? Consider the the following.

Odds of:
J5 - KQ = 6.67%
Q5 - KJ = 6.67%
K5 - QJ = 6.67%

Assume RHO doesn't know the KJ are equals (if he knows it then he can play randomly from all these holdings and the following doesn't apply.) If he plays randomly from KQ and QJ but always J from KJ then RHO is playing K 3.33% of the time, Q 6.67% of the time, and J 10% of the time. So perhaps RHO should always play J from KJ, Q from QJ, and K from KQ so that he plays each an equal amount of the time and you can't exploit his plays to gauge the odds of a 2-2 break.

Or perhaps he should just make a (sometimes fake?) suit preference signal since no matter what he plays a 2-2 break is still against the odds relative to 3-1 breaks so he might assume you will play for 3-1 no matter what.

And how can you tell if he knows KJ are equals or not?

Or even if he knows KJ are equals maybe his partner will have Qxx, assume a K that was dropped was singleton, and go up queen on the second round to leave himself a small card to exit, so he can't necessarily risk K from KJ.

Or maybe he knows you know J means 2-2 could be more likely so he also plays the J more often from KQJ to compensate.

Bridge can be tricky.

[CSGibson]

Actually, I did see that playing "randomly" was not necessarily best, which is why I posed the question. I was wondering if there was an appropriate mixed strategy, or whether it was always right to drop one honor or another from equals in this situation, keeping in mind that the person with the doubleton honor is not sure that this situation applies, as opposed to the one where declarer is only missing two honors.

[kenberg]

Probability problems are tricky and I have more than once had to eat my words, but here is how I would analyze the K problem.

Cutting to the chase, I claim the probability of the suit breaking 3-1 when the play is king on the right, 5 on the left, is close to 8/11.

The conditions of the problem amount to the assumption that when the king is played, and the 5 follows, the K will be from one of
K
KQ
KQJ

In all three possible holdings the location of all four outstanding cards is stipulated. That is, the first holding is K on the right, QJ5 on the left. These holdings. with the location of all cards stipulated, are almost equally likely a priori.

The conditions stipulate random play in all these holdings (where there is a choice).

Imagine 18 deals where the king might be played;
K alone on 6 deals, always played.
KQ tight on 6 deals, K played 3 times.
KQJ on 6 deals, K played 2 times.

Conclusion: In the 18 deals where the king could be played, the king will be played 6+3+2=11 times. On 3 of these, the suit is 2-2. The probability of one loser is 3/11, the probability of two losers is 8/11, the odds on losing two tricks are 8 to 3.

Of course it is not the case that every 18 deals will be distributed as stated, this is a device for simplifying thought. But in 18,000 deals the distribution of holdings will be, in percentage terms, pretty close to 6,000 of each type.

Added: The three holdings were described as being almost equally likely. I fact the KQ is slightly more like than either of teh other two. The remaining 22 cards from the two hands must be placed, and there are more ways to do this with 11 empty spaces in each hand than with 12 in one hand, 10 in the other. So, if the rest of what I say is right, the probability of losing only one trick is slightly better than 3/11. I imagine that the small effect is is overwhelmed by the fact that in some arrangements of the remaining cards the opponents would be in the auction.

[nige1]

CSGibson, on Oct 12 2010, 02:05, said:

Interested in hearing situational analysis - if you play the ace from hand in a trump suit, missing the KQJ5, and the 5-K pops up, what do you think the chances are that you are losing 2 tricks in the suit?  How about if the 5-Q pops?  5-J?  This is assuming that someone would play randomly from equals, but lowest otherwise... [/ONESUIT]

Presumably, an opponent will always play the knave from KJ doubleton but may play randomly from KQJ, KQ, or QJ. As far as I remember, opponents can play true cards most of the time. False-carding occasionally is sufficient to keep declarer guessing.