Restricted Choice

[Zelandakh]

For anyone that is at all confused by Spiso's attempts to confuse and is still bothering to read this thread, here is some real maths, hopefully without error!

 

Bayes' Theorem is p(B|A) = p(A|B)p(B)/p(A).

 

For the problem set by Stephen, one way is to use:

A = LHO has xx, RHO has at least one honour

B = LHO has xx, RHO has KQ

 

So p(A) = 0.1922, p(B) = 0.0678, p(A|B) = 1 and p(B|A) ~= 0.35

 

We could of course also calculate the odds of the second honour being on the left instead by using:

C = LHO has Hxx, RHO has H

 

That gives p(C ) = 0.1244, p(A|C) = 1 and p(C|A) ~= 0.65

 

All simple. The issue that RC works on is looking at the case of the specific honour:

 

D = LHO has Qxx, RHO has K

 

Now if we were to be naive, we might think p(D) = 0.0622, p(A|D) = 1 and p(D|A) ~= 0.32. RC reminds us that this calculation is wrong and that the figure needs to be doubled. That is all it does, it forces non-mathematicians to use the correct interpretation of Bayes' Theorem. There are of course multiple correct ways of arriving at the same answer, as evidenced by the previous cases. But bridge players are usually not so much interested in obtaining the precise probabilities as in knowing the best line of play. And the beauty of RC is in providing that answer without having to know any maths at all. So no, there is no fallacy here. The underlying maths is sound.

[1eyedjack]

For the problem set by Stephen, one way is to use:

A = LHO has xx, RHO has at least one honour

B = LHO has xx, RHO has KQ

 

Are you sure that there is no typo there? If definition of LHO is identical in each of cases A and B, are not RHO hands likewise identical between the cases, by definition?

[Stephen Tu]

I think Zelandakh means lho has xx and either an honor in addition or not for case A.

[1eyedjack]

I think Zelandakh means lho has xx and either an honor in addition or not for case A.

Ah yes, obvious when thinking what we are trying to achieve.

[Zelandakh]

I think Zelandakh means lho has xx and either an honor in addition or not for case A.

Exactly, A is meant to be the situation we find ourselves in at the point of decision - xx-H + H in an unknown hand - and this is necessary to have the correct conditional probability at the end of it.

[jogs]

I did some sums a few years ago to work out how far an opponent had to depart from random, in terms of the frequency of preferring one card over another equal card, in order to suggest that applying restricted choice only provided a break-even success rate. It would depend on other external factors, no doubt, but in the example I chose to consider (missing QJxx) it worked out that a player had to play one card about 12 times more frequently than the other (when presented with a choice) in order to bring this about. I never had my sums peer reviewed, so perhaps now would be a good time, if I can remember how.

 

Say North has ♠KT987 and 8 non-spades opposite South ♠A654 and 9 non-spades, and no other information. South cashes ♠A to which all follow, East with ♠Q, West low. West also follows low the next Spade. Crunch time.

 

Now suppose that East is known to play Q from QJ doubleton with probability p where 0<=p<=1 (and therefore J from that holding with probability 1-p)

 

The question to resolve is what is the value of p such that when the Q appears on the first round, the success rate in playing for the drop on the second round equals that of the finesse.

 

West has 11 vacant spaces in which to hold the J. East has 12.

 

Under PRC, East's vacant spaces are effectively reduced by a factor p. If p=0, the finesse is certain to succeed. If p=1, the drop is slight favourite by a factor 12:11. If p=0.5 (ie random case), then the finesse is favourite by the ratio 11:6, as predicted by classic PRC.

 

The break even point arises when the ratio 11:12p = 1:1

 

ie when p = 11/12.

 

I conclude therefore that even substantial deviations from random can be afforded before it upsets the recommended finesse suggested by PRC.

Question: You thought you detected a B-I-T by East. East plays an honor(queen or jack). Do you now play East for the other honor?

[sfi]

Question: You thought you detected a B-I-T by East. East plays an honor(queen or jack). Do you now play East for the other honor?

 

Surely anyone above a rank beginner has thought about which one to play many times, and doesn't need to hesitate at the table. Unless it's an obvious enough hesitation that everyone notices it (and possibly comments on it), I'm going to play the hitch to mean 'I need to think about playing my singleton in tempo so he doesn't work out the honours are split' or for them to be playing games.

[Jinksy]

Come on mods put a stop to this shite and ban the (not very good) troll

 

At the very least if you're determined to give him the benefit of the doubt, move it to another forum - I suggest Novice and Beginner.