Restricted Choice

[Zelandakh]

Spisu, restricted choice is the bridge name for something called Bayes' Theorem. You can read about BT here. Once you have read that and digested it, please feel free to come back and explain to us why BT is mathematically incorrect. I have no doubt that you will be hailed as one of the greats of the field if you succeed.

 

As you can see from the theorem itself, it is required for there to be 2 events. This is why it is pointless to discuss RC/BT on the first round of the suit. It is fundamental to understand this before you move on to trying to disprove it. Unfortunately, I am also fairly confident that Bill is correct but would love to be proved wrong.

[Spisu]

Seriously?!?! You don't see the logical dissonance between this post and your opening post?

 

If the answer to the "2 kids/ boy girl puzzle" is 2:1 for the second child to be the opposite sex, why is the answer to "2 missing honors/ left right puzzle" not 2:1 for the missing honor in the opposite hand?

 

I am surprised at the comments here. This is not that hard. The odds are obviously 2:1 for the second of a double finesse after losing the first finesse, and I have never said a word to the contrary. Why not give a thought that your example was perhaps an actual reason for those odds, contrary to the claims of RC which asserts the odds are based on play from equals.

[Spisu]

The principle of restricted choice is not applied, has never been applied and never will be applied on the first play of the suit. The entire principle is a statement of the impact of an earlier play on the table of probabilities that apply to the possibilities in a later play. That in turn requires that there must have been an earlier play of the suit, and it is only to the later play that the principle applies. So your point eludes me.

Sorry if I was impatient. But if you read a sentence or two into restricted choice in most places, it professes that plays of one of two equals reduces odds the other equal is present by half. But the fact is and the point I presented is that 50% (+-) of all 2 honors divided between hands (normally with side cards of course) are combined honors, and nothing can reduce the frequency of an honor played from a combined honors' hand below its frequency/odds of 50% (they are the only two cards, combined, and in half of all such hands)....nor can a single (lone) honor increase the frequency of divided hands above 50%. There are of course consequences of this which should be apparent.

 

Nothing I have said or written suggests impacts the odds for the second finesse which are easily seen a priori, independent of either equal played. But RC does not set those odds.

[Stephen Tu]

Spisu you are just completely off base. RC is just application of Bayes theorem as described above. It is analysis that absolutely *does* depend on play from equals, the probability of a particular holding being held by opponent after the first round play is the *product* of both the a priori odds and the odds that they chose a particular honor. With holding only 1 honor the second term of the product is 100%, with 2 honors it is usually calculated as 50%, though in actuality an opponent might have quite a bit of bias towards one or the other which doesn't affect your best choice of play until it gets really extreme.

 

If you were playing against an opponent who was known to never randomize, and say always played the K from KQ tight, then on AJt9x vs. xxxx one could do slightly better by playing for the drop every time the first finesse lost to the K rather than taking a second finesse, finessing twice only if the first hook lost to the Q. You would pick up Kxx onside and KQ tight offside, which is better against such an opponent than finessing twice regardless and picking up Kxx and Qxx onside. RC is merely an observation that opponents can and should vary their play from equal cards, and that if you are comparing a priori odds you need to multiply by a factor taking this to account. If your first finesse lost to the K, you compare the odds of stiff K vs. *half* of the odds of KQ tight, or alternately you pretend you had bad vision and can't distinguish between equals and compare odds of KQ tight vs. *both* stiff K and stiff Q. There is nothing fallacious about RC. It does depend on opponent having choices from equals, and not having choices when having only one honor to win with.

[Spisu]

There are several similar puzzles. I like Bertrand's Box:

 

There are three boxes. One box contains 2 gold coins; one box contains 2 silver coins; and one box contains 1 gold coin and 1 silver coin.

 

Choose a box at random. Then choose a coin at random from that box. The coin you have chosen is gold. What is the probability that the other coin in your box is also gold?

 

This is a good example but probably not in the way you intended. It illustrates the fallacy regarding the RC claim equal cards are distinguishable (well, to declarer but not defender, a notable contradiction).

 

The equals quality of statistical significance for the boxes is that 2 boxes are the same (each Au or AG) and 1 is different (AU/AG), and nothing to do with Au/Ag itself. It's 2:1 when you accept the obvious "sameness" that the other coin will be the same as you see. The equivalent in RC (had they been more astute) would have been to seize on the fact that a K and Q are equals like the "sameness" of 2 Au and 2 Ag over 1Au/Ag, then not going astray at seeing a gold (He played a KING!), and flying off at a tangent over what the specific significance of that gold (KING!) might be...and getting "creative" in the process.

[Spisu]

Spisu you are just completely off base. RC is just application of Bayes theorem as described above. It is analysis that absolutely *does* depend on play from equals, the probability of a particular holding being held by opponent after the first round play is the *product* of both the a priori odds and the odds that they chose a particular honor. With holding only 1 honor the second term of the product is 100%, with 2 honors it is usually calculated as 50%, though in actuality an opponent might have quite a bit of bias towards one or the other which doesn't affect your best choice of play until it gets really extreme.

 

If you were playing against an opponent who was known to never randomize, and say always played the K from KQ tight, then on AJt9x vs. xxxx one could do slightly better by playing for the drop every time the first finesse lost to the K rather than taking a second finesse, finessing twice only if the first hook lost to the Q. You would pick up Kxx onside and KQ tight offside, which is better against such an opponent than finessing twice regardless and picking up Kxx and Qxx onside. RC is merely an observation that opponents can and should vary their play from equal cards, and that if you are comparing a priori odds you need to multiply by a factor taking this to account. If your first finesse lost to the K, you compare the odds of stiff K vs. *half* of the odds of KQ tight, or alternately you pretend you had bad vision and can't distinguish between equals and compare odds of KQ tight vs. *both* stiff K and stiff Q. There is nothing fallacious about RC. It does depend on opponent having choices from equals, and not having choices when having only one honor to win with.

 

I am familiar with Bayes and the restricted choice folks'claim it is based on the Bayes postulate known also by some as the "Equidistribution of Ignorance". The only thing missing I've noted is any confirming data to support a valid basis for that other than the claim. Note that assuming an opponent plays randomly is not mathematically valid. So if you have evidence of a Bayesian basis other than presumptions or guesses, please share. I'd love to see it.

 

BTW, the freq dist for lone vs double honors shows KQ, K/Q, Q/K, QK each 25%...so the frequency of a specified honor such as the K in East is 1/2 of 25% or 12.5%, while a KQ is 25%, twice that of a specified lone honor. The math of RC depends on your not noticing the frequencies are different.

 

For 400 hands, East holds 400 total honors 200 are divided( 100 Ks and 100 Qs) and 200 KQ combined honors. Simple math for first finesse shows 100 lone Ks played will match with 100 Ks from KQ (1:1), 100 Qs match with 100 Qs from KQ, leaving 100 KQs to nail you in the 33% of losing second finesses. There is no "half" relationship of KQ.

[kuhchung]

perhaps it's time to take nige1's lead and ask Spisu to put his money where his mouth is

[Stephen Tu]

So you agree the 2nd finesse works about 2/3 of the time, I take it, and agree with restricted choice conclusions to take 2 finesses if ajtxx opposite xxxx, and to finesse 2nd round on akt9x opposite xxxx if first round drops an honor offside, and that the second finesse is better by almost 2:1? And are only arguing about the calculation method?

 

Look there are two ways to look at the issue. One is that you cannot visually distinguish between equal honors, you look at only the a priori odds, and calculate the best line overall, regardless of what honor shows up first round, because you cannot see which one it is. The other is that you *do* distinguish between honors, and you compare only the combinations that can actually exist given the prior play in the suit, but correct for the odds that an opponent actually chose to show you the honor you identified.

 

One method counts Kxx - q and Qxx - K vs. xx-KQ. Which you seem to favor.

The second method visually sees the opponent won the Q first round, counts *only * Kxx-Q vs. (xx-KQ AND opponent choose the Q), the *times half* you are complaining about.

Each method leads to the exact same conclusion, exact same answer, exact same ratios! They are both equally valid, neither is fallacious. In math often there are multiple ways to calculate a correct answer, more than one way can be valid. It's like you are comparing 4:2 and the other guy is comparing 2:1 and even though you are both getting the same 2:1 answer you are saying his way of calculating is wrong.

 

I don't understand what you mean by your declaration that assuming an opponent plays randomly from equal honors "isn't mathematically valid". What would you consider to be mathematically valid?

[Phil]

Note that assuming an opponent plays randomly is not mathematically valid. So if you have evidence of a Bayesian basis other than presumptions or guesses, please share. I'd love to see it.

 

I'd suggest the burden is on the RC heretics to demonstrate that not only people play non-randomly with equals but more importantly *which* card they choose. Otherwise a claim of non-randomness is invalid.

 

One of my partners has been known to mutter at the table, they ALWAYS falsecard with the higher honor. This may be true on low level club games- the same group that tries to drop singleton kings offside with 3 out- but in any real game there is no discernible pattern.

[Spisu]

So you agree the 2nd finesse works about 2/3 of the time, I take it, and agree with restricted choice conclusions to take 2 finesses if ajtxx opposite xxxx, and to finesse 2nd round on akt9x opposite xxxx if first round drops an honor offside, and that the second finesse is better by almost 2:1? And are only arguing about the calculation method?

 

Look there are two ways to look at the issue. One is that you cannot visually distinguish between equal honors, you look at only the a priori odds, and calculate the best line overall, regardless of what honor shows up first round, because you cannot see which one it is. The other is that you *do* distinguish between honors, and you compare only the combinations that can actually exist given the prior play in the suit, but correct for the odds that an opponent actually chose to show you the honor you identified.

 

One method counts Kxx - q and Qxx - K vs. xx-KQ. Which you seem to favor.

The second method visually sees the opponent won the Q first round, counts *only * Kxx-Q vs. (xx-KQ AND opponent choose the Q), the *times half* you are complaining about.

Each method leads to the exact same conclusion, exact same answer, exact same ratios! They are both equally valid, neither is fallacious. In math often there are multiple ways to calculate a correct answer, more than one way can be valid. It's like you are comparing 4:2 and the other guy is comparing 2:1 and even though you are both getting the same 2:1 answer you are saying his way of calculating is wrong.

 

I don't understand what you mean by your declaration that assuming an opponent plays randomly from equal honors "isn't mathematically valid". What would you consider to be mathematically valid?

 

Random choice of play is an oxymoron. And mathematical validity of the sort of weather prediction or a batting average predicting a specific at-bat hit can exist. These can be calculated/updated with the valid use of Bayes, but the RC chose to assume randomness with no basis. But assumptions require verification...You might assume as a starting point that one player chose 50-50 between equal honors, then keep records of his plays over many hands to see, and update that 50% to whatever (THAT is a Bayesian update)...And then you would know one person's proclivities..Then you could do the same on all the bridge players you ever play with...or, 1. you could do what RC did and presume, then claim it is mathematical but isn't, or, 2. just accept the fact that equals are divided 50% and the targeted hand has both honors only 25% when the targeted hand is known to have any honor (all a priori 2:1).

[1eyedjack]

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.

[Spisu]

Spisu:

Say you have akt98 opposite xxxx in dummy. You bang down the ace and an honor falls on the left. Do you play for the drop or the finesse? This is one of the basic restricted choice combinations.

 

1. Do you agree that out of all possible original holdings, LHO will be dealt QJ tight 6.78% of the time, the stiff Q 6.22% of the time, and the stiff J 6.22% of the time?

 

2. If you agree with 1, does it not make sense that playing the drop works 6.78% of the time, compared to 12.44% of the time for the finesse? Or basically 64.7% advantage to finesse, close to but not quite 2:1, counting only deals where an honor falls on the left?

 

If you do not agree with these statements, please explain why.

 

3. Do you agree that holding the stiff J, LHO will play the J 100% of the time from that holding?

 

4. Do you agree that holding the QJ tight, most LHO will play the J much less than 100% of the time from that holding?

 

Non-restricted choice (fallacious) reasoning:

LHO dropped the J. He either had QJ or J tight. The chances of these are about the same. Actually QJ slightly more common. So play the drop for a slight edge.

 

Restricted choice reasoning:

Wait a minute, from the stiff J he always has to play the J. From the QJ he might have played the Q half the time. So it's actually more likely that he had the stiff J and was forced to play it, than that he BOTH was dealt the QJ AND randomly picked the J to play. Finesse for a big edge.

Or simply go by the original deals, stiff Q + stiff J (finesse line) is almost twice as common as QJ tight (drop line).

 

"Had to play it" suggests random discovery in this case, unlike other most other so called RC situations. So in 2 of 3 cases you've found a singleton. Notably the originators changed their tune in this case and abandoned their focus on specific cards played and came up with "Quack". The odds are 2:1 for the randomly discovered play of a singleton.

[Spisu]

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.

[Spisu]

I'd suggest the burden is on the RC heretics to demonstrate that not only people play non-randomly with equals but more importantly *which* card they choose. Otherwise a claim of non-randomness is invalid.

 

One of my partners has been known to mutter at the table, they ALWAYS falsecard with the higher honor. This may be true on low level club games- the same group that tries to drop singleton kings offside with 3 out- but in any real game there is no discernible pattern.

 

It is patently absurd to come up with a stipulation and then claim it must be true unless someone can prove it's false.

[Stephen Tu]

"Had to play it" suggests random discovery in this case, unlike other most other so called RC situations. So in 2 of 3 cases you've found a singleton. Notably the originators were befuddled by this case and abandoned their focus on specific and came up with "Quack". The odds are 2:1 for the random play of a singleton. But if you knew how the player would play that specific doubleton you could even have a certainty, but you probably don't.

 

You don't need certainty. The 50% randomization that one method of calculation assumes is just a baseline assumption. If one wants to be totally comprehensive then one can specify a range of behavior of the opponent, say "play for the finesse 2nd round is always best as long as person plays that card between x% and y% of the time". If the opp biases his plays one way or the other, you'll just win somewhat more on the 2nd round when one honor appears, and somewhat less when the other honor appears, which will completely cancel out. The overall success rate will remain the same when summed over both honors if you are taking the line suggested by restricted choice principles, whether calculated by the way you appear to think is invalid or the alternate method.

 

It's just a meaningless philosophical question whether you just want to calculate all equals as quacks, or identify the specific honor and assume that opp is playing randomly. In the end your conclusion for the correct play is the same either way, your overall success rate is the same either way. Only if you can identify and encounter an extreme opponent who plays the same card from doubeton honor nearly always will your conclusion about the best tactic change. "If this player plays the Q from QJ > 91.7% of the time, play for the drop if the Q appeared".

 

Doesn't mean restricted choice is bunk or using 50% as the assumed randomization rate which will also give you the appropriate combined success rate for either honor appearing is an invalid way to calculate. If you know an opponent plays 75% one honor and 25% the other, then all that does is give you different rates of success depending on which honor appeared. The average overall success rate will still be equal to the success rate using 50% as the assumption.

[Spisu]

An intelligent player at Reading Bridge Club claimed that the rule of restricted choice was bunkum. We offered to set up a typical matrix, then shuffle the remaining cards and back our respective judgements at £10 per deal. After some homework, our friend declined our challenge :)

 

Why would he play differently if RC is "bunkum"? The odds are 2:1 without RC. The big question for some here is why aren't the odds 4:1 if RC works? Louis Watson recommended it strongly in 1934 describing the first finesse as a preparation as I recall.

[Spisu]

Seriously?!?! You don't see the logical dissonance between this post and your opening post?

 

If the answer to the "2 kids/ boy girl puzzle" is 2:1 for the second child to be the opposite sex, why is the answer to "2 missing honors/ left right puzzle" not 2:1 for the missing honor in the opposite hand?

 

You're confusing the issue of the odds for a second finesse and the validity of RC. The odds are 2:1 for the second finesse whenever a W does not have both honors and an E wins the first trick. That has nothing to do with pointing out the fact that 50% of all plays of 1 of 2 equals in open play come from combined honors undermines RC.

[Stephen Tu]

It's not 4:1 because apparently what your idea of how one method of RC is calculating isn't actually what RC is saying, you are totally confused about it. Let's take AJT9x opposite xxxx, finesse twice vs. finesse follow by drop, and specify that first finesse lost.

 

You are saying, it was either Kxx- Q Qxx- K or xx-KQ. So 2-1 for two finesses. (Not exactly 2-1 because 2-2 breaks are slightly more frequent than 3-1 because of the effect of other cards, needing 13 cds in hand. But 2-1 is a close enough for an estimate, and for the rest of this post I will just use this as an approximation). That's certainly reasonable, we aren't disagreeing with you at all about this, it's totally valid way to come to correct conclusion about strategy.

 

The other method of calculating is saying:

First finesse lost to the Q. Therefore it was not Qxx-K because duh. So it's either Kxx-Q or (xx-KQ AND opp chose the Q, which as a baseline guess we'll take as 50%)

So 1 case to half of 1 case, or 2-1, same as your conclusion

Or:

First finesse lost to the K. Therefore it was not Kxx-Q because duh. So it's either qxx-K or (xx-KQ AND opp chose the K, which as a baseline guess we'll take as 50%)

So 1 case to half of 1 case, or 2-1, same as your conclusion

 

Or: First finesse lost to the Q. Therefore it was not Qxx-K. Let's not assume opp plays randomly perfectly even from KQ. Let's say they play p% the Q, and 100-p% the K.

So if first finesse lost to the Q, it's 1 case (Kxx - Q) vs. p/100.

But first finesse lost to the K, it's 1 case (qxx - K) vs. (100-p)/100.

So overall, it's going to be 2 cases vs. ((p/100) + (100-p)/100). Well ((p/100) + (100-p)/100) = 1. So again, 2-1, for any value of p, same as your conclusion.

 

It's 2:1 any way you calculate, so I don't see your beef with the alternate methods of doing the calculation.

 

The only time one starts to get a different conclusion about strategy is when you start considering that specific 2-2 break is slightly more likely than a specific 3-1 break, combined with an opponent who is not randomizing hardly at all and playing say 95% the higher honor, or 95% the lower honor.

[Spisu]

Spisu, restricted choice is the bridge name for something called Bayes' Theorem. You can read about BT here. Once you have read that and digested it, please feel free to come back and explain to us why BT is mathematically incorrect. I have no doubt that you will be hailed as one of the greats of the field if you succeed.

 

As you can see from the theorem itself, it is required for there to be 2 events. This is why it is pointless to discuss RC/BT on the first round of the suit. It is fundamental to understand this before you move on to trying to disprove it. Unfortunately, I am also fairly confident that Bill is correct but would love to be proved wrong.

 

Well, I merely showed that the first honor play from equals (the underlying basis of RC) is a fallacy based on the reality that such plays from combined equals are 50% of all plays involving 2 equals. Knowing that at or before trick one could be useful to some. But restricted choice is certainly NOT Bayes Theorem as you say, though they do claim it is based on a Bayes Postulate. I have never hinted that "BT" is incorrect and I would think that properly used it is completely valid.

[Spisu]

It's not 4:1 because apparently what your idea of how one method of RC is calculating isn't actually what RC is saying, you are totally confused about it. Let's take AJT9x opposite xxxx, finesse twice vs. finesse follow by drop, and specify that first finesse lost.

 

You are saying, it was either Kxx- Q Qxx- K or xx-KQ. So 2-1 for two finesses. (Not exactly 2-1 because 2-2 breaks are slightly more frequent than 3-1 because of the effect of other cards, needing 13 cds in hand. But 2-1 is a close enough for an estimate, and for the rest of this post I will just use this as an approximation). That's certainly reasonable, we aren't disagreeing with you at all about this, it's totally valid way to come to correct conclusion about strategy.

 

The other method of calculating is saying:

First finesse lost to the Q. Therefore it was not Qxx-K because duh. So it's either Kxx-Q or (xx-KQ AND opp chose the Q, which as a baseline guess we'll take as 50%)

So 1 case to half of 1 case, or 2-1, same as your conclusion

Or:

First finesse lost to the K. Therefore it was not Kxx-Q because duh. So it's either qxx-K or (xx-KQ AND opp chose the K, which as a baseline guess we'll take as 50%)

So 1 case to half of 1 case, or 2-1, same as your conclusion

 

Or: First finesse lost to the Q. Therefore it was not Qxx-K. Let's not assume opp plays randomly perfectly even from KQ. Let's say they play p% the Q, and 100-p% the K.

So if first finesse lost to the Q, it's 1 case (Kxx - Q) vs. p/100.

But first finesse lost to the K, it's 1 case (qxx - K) vs. (100-p)/100.

So overall, it's going to be 2 cases vs. ((p/100) + (100-p)/100). Well ((p/100) + (100-p)/100) = 1. So again, 2-1, for any value of p, same as your conclusion.

 

It's 2:1 any way you calculate, so I don't see your beef with the alternate methods of doing the calculation.

 

The only time one starts to get a different conclusion about strategy is when you start considering that specific 2-2 break is slightly more likely than a specific 3-1 break, combined with an opponent who is not randomizing hardly at all and playing say 95% the higher honor, or 95% the lower honor.

 

Personally I am interested in the basic principles without complicating distributional factors. We need to get the principles right, and the outlying hands with added complexities can come next. So the hand you show is not on my radar at the moment. But it's not just "a different way to get the numbers". People are drawing false conclusions in general play when they see an equal that is as likely to be a loner(divided honors) or from combined honors. Double finesses target a hand and force a play from one hand which is perforce half as likely to hold both honors as the 2 opponent hands' combined. Principles need to be correct.

[Phil]

It is patently absurd to come up with a stipulation and then claim it must be true unless someone can prove it's false.

 

Except we have math on our side.

[Stephen Tu]

People are drawing false conclusions in general play when they see an equal that is as likely to be a loner(divided honors) or from combined honors. Double finesses target a hand and force a play from one hand which is perforce half as likely to hold both honors as the 2 opponent hands' combined. Principles need to be correct.

 

No, we aren't drawing false conclusions. We are drawing correct conclusions, in fact the same conclusion. You are apparently blind to the fact that both arguments are drawing the *same* conclusion, not different conclusions, and that both are correct.

 

If you lose a finesse to an equal. it is *not* "as likely to be a loner or from combined honors". It's twice as likely that the honor was a loner. That is the whole gist behind restricted choice. A loner will be played 100% of the time. Combined, that particular honor would be chosen to play approximately half the time against an opponent that randomizes. It's only "as likely to be a loner or from combined honors" if an opponent nearly always chooses a particular honor from the combined honors and he plays that honor he favors. Then a second finesse is basically 100% against the disfavored honor, but only 50% against the other. Combined odds will still be 2:1 to succeed.

 

I don't know who you think is saying that an equal is as likely to be a loner as combined. That implies second finesse is only 1:1, not 2:1. RC is not claiming that. It is claiming 2:1, exactly the same as you. The principle is correct.

 

The second finesse is 2:1 to succeed using both arguments. This is a correct conclusion. Neither is false. If RC gives the exact same answer as your 2:1 answer, how is it a false conclusion? If your argument is right, and RC agrees with same answer, then either both are right or both are wrong! Logically RC cannot be wrong while your argument being correct if it is claiming the same result!

[1eyedjack]

Spisu, by now I expect that you have had a chance to digest my post in another thread, here

 

http://tinyurl.com/hjps5mv

 

That post contrasts two situations which have in common that you hold king to 5 opposite ace to 4,

 

They differ in that in case 1 the only significant missing card is the Queen, but in case 2 you are missing both Queen and Jack.

 

In both cases you cash the ace and then lead low toward the king.

 

In both cases 2nd hand fallows low to both tricks. 4th hand follows to the first round with a low card in case 1, and with Jack in case 2.

 

I attempt to demonstrate in that thread, that in case 1 the drop is about as good as the finesse on the second round (slightly better in fact but I gloss over that), while the finesse is heavy favourite in case 2. The only difference in the underling conditions is that in case 2, the fourth hand's play of Q on first round is potentially from equal cards (QJ) where in case 1 the first round play of x is not. I cannot escape the conclusion that the difference in initial underlying conditions is the cause of the difference in the resulting probabilities, which is all that PRC states. DO YOU DISAGREE?

 

Incidentally, was there any point to your post #38 in this thread? If it was simply to preserve my comment for posterity lest I have some inclination to remove it, I can assure you that I have no such intention.

[benlessard]

I think you need to do it step by step.

 

1.......void....KQxx....4.783...1.....4.783

2.......x.......KQx.....6.217...2.....12.435

3.......xx......KQ......6.783...1.....6.783

4.......Q.......Kxx.....6.217...1.....6.217

5.......Qx......Kx......6.783...2.....13.565

6.......Qxx.....K.......6.217...1.....6.217

7.......K.......Qxx.....6.217...1.....6.217

8.......Kx......Qx......6.783...2.....13.565

9.......Kxx.....Q.......6.217...1.....6.217

10......KQ......xx......6.783...1.....6.783

11......KQx.....x.......6.217...2.....12.435

12......KQxx....void....4.783...1.....4.783

 

Spisu do you agree that this chart look like the a priori distribution of all the split possible when we are missing KQxx. Note that the 4th column is the permutations.

 

3--KQ5 is slightly less likely than KQ--53

 

However note taht on line 2 x--KQx is multiplied by 2 to represent both 3--KQ5 & 5--KQ3 added together.

[Zelandakh]

I have never hinted that "BT" is incorrect and I would think that properly used it is completely valid.

Could you provide an example of a case where the correct application of RC provides the wrong answer. That you do not think RC is BT tells me that you simply do not understand it mathematically so I think we will find your error fairly easily from such an example.