bonanova

And to you, cricket!

Let me take the moment to wish you all a very Happy New Year. I'm going out for a walk in some freshly fallen snow. My planned visit to NYC for celebration there is off, so I'm making the best of it. Talk to you all next year.

Acute angles would limit the triangle you start with. The process works with say a 30 - 70 - 110 triangle. Google morley's theorem proof for more on this. Here's a beautiful demo.

An argument given in my previous post shows that you can't be sure p[A=6] = 1/2. Depending on how magician chooses his number, it could be 1/2 [random choice]. But, It could be 6/11 [pick higher number or pick 6 if it's there]. It could be 1/36 [pick lower number].

The OP admits to either answer, 2/11 or 1/6, or even 0, in some cases. The algorithm used by magician [on which OP is silent] determines whether the eleven outcomes actually have equal probability. My previous post was too long to read, being written late at night, so I'll excerpt here how [i see] what the magician was doing could change things. Consider my simulation problem where magician sees 5-6. He has the liberty to say "one of your dice shows 5" or "one of your dice shows 6" My next line of code was going to be a random choice function. But it could have been pick the first value, or pick the higher value, or pick a 6 is one is present, else pick a .... whatever! That's when the light came on. What you logically can infer from hearing a chosen number depends on how the number was chosen. So, leave the magician now, and consider the Subject. By premise, he hears "one of your dice shows 6". What is he to think? If the subject reasons: if the magician sees that either of my dice is a 6, then he'll tell me about it, then all eleven cases consistent with A=6 or B=6 have equal footing. Because in each of the eleven cases, necessary and sufficient conditions are present for the magician to make his "six" statement -- including the 6-6 case, where magician's algorithm doesn't matter: he must say "six". Thus p[7] = 2/11 If the subject reasons: the magician will tell me about the higher/highest of the dice, then, again, all eleven cases have equal footing. Because if either die is 6 it will be the higher/highest [highest applies to 6-6 where neither technically is higher]. Again p[7] = 2/11 If the subject reasons: the magician, having not told me how he chooses the number to call out, can be assumed to have made a random selection, then the eleven cases no longer are on equal footing. The subject considers the 6-6 case. If magician saw that, he must call out the six. But if magician saw any of the other cases, say A=5 B=6 or A=6 B=5 then half of the time he will call out the 5. Similarly with 4, 3, 2 and 1. So it is only with 50% confidence that I can place the other 10 possibilities into my outcome table. That fact changes p[7] into [.5 + .5]/[.5 + .5 + .5 + .5 +.5 +.5 +.5 +.5 +.5 +.5 +1] = 1/6. If the subject reasons: the magician will tell me about the lower/lowest of the dice, then, only the 6-6 case will be mentioned, and since it totals 12, p[7] = 0! Because the OP is silent on the magician's algorithm, the subject cannot be certain of how to calculate p[7], and neither can we. Now, I admit did not believe the assertions that the 11 cases are not always on equal footing, given by others. To me they were conclusions with words wrapped around them, rather than clear expositions of the problem from start to finish. Once I came to the line of code in the simulation that addresses it, I had to put it into my own words, and of now course it is perfectly clear.

Yeah, that's it. Morley's theorem. Pretty cool.

Is it related to this?

Hmmmmm .... restrictions. Well, I can start you off on the solution, but restrictions are just use the pencil as a pencil, the straight edge to make sure a line you want to be straight is straight, and the angle trisector to give you the two lines that trisect any angle you may have drawn. Now the method ... if this helps: Draw at least one straight line. Make some angles. Trisect some or all of the angle(s) Make an equilateral triangle. That adds some restrictions,I guess. Like, don't fold the paper. Don't try to draw arcs or circles. Don't use the [assumed square] corners of the paper. Don't give up. Does that help?

Bingo.

Yup. Good ones, tho.

rossbeemer: We may assume the magician was not lying. Scraff: With your last quote, I think all the cards are now on the table. Now for a longish post that I think will be my last, here. [spoiler=Formalization of 7's argument, and its final consequence Now, probability is changed by knowledge. We start with 6x6 = 36 EL [equally likely] outcomes. Each of Magician's statements eliminates 25 of those cases, leaving 11. Both sides agree with that statement. I'll say more about their comparative likelihood later. We can construct 7's outcome table by adding together the probabilities of the 11 outcomes for each of the six, multiplied by 1/6. The manner in which the 66 = 6x11 outcomes distribute themselves on 7's outcome table is telling: one major diagonal term 1-1, 2-2, etc from each, and two off-diagonal terms 3-5, 1-2, 5-6, etc collectively from the six. The 36 outcomes for 7 occur equally, except for the "doubles" outcomes. That is, the off-diagonal outcomes happen twice, while the "doubles" cases happen only once. They sum to 66 [6x10 + 6x1]. I think this touches on Scraff's objection to 7's reasoning - as if there were an assumption they are all on equal footing. They are not. But that's OK, because we've counted [accounted for] them all. But now look at the NE to SW diagonal - the outcomes where 7 is the total. There are 2's in each of those boxes. That is, 12 of the 66 outcomes have 7 as a total. So if all 66 outcomes are equally likely [recall so far we've only counted them as said there are 66] then p[7] = 12/66 or 2/11 for Subject 7. Notice the if. It's the same if that leads to p[7] = 2/11 for the first 6 subjects. OK that's a long [rigorous] path to saying p[7] for Subject 7 is the average of the p[7]s for the first six. Note that whatever the probability distribution of the 11 outcomes for each of the first 6 subjects [the distributions are all the same by symmetry] the the same probability distribution will come out in the average. That is, all p[7] values will be the same. Subject 7's argument is thus valid: If p[7] = 2/11 for the first six, p[7] = 2/11 for Subject 7; if p[7] = 1/6, 0, 0.0001, ... whatever, for the first six, 7's is the same. I said that each of the magician's statements eliminates 25 outcomes. That's not in dispute. Differences arose in assertions of the remaining 11 outcomes being equally likely. More to the point: if not, why not. I assert now that they are, maybe, or they're not, maybe, equal. That is, the OP does not tell us enough to be sure. In the "one of your dice is 4" case, if the magician looked for a 4 and called it out if it was there [see how luck enters - what if the dice were 3-5?] then all 11 outcomes [14, 24, 34, 44, 54, 64, 41, 42, 43, 45, 46] have the same likelihood, and p[7] = 2/11. If the magician glanced at the dice and chose between the dice at random, then the special case 44 is twice as likely as the others: he's compelled to say 4, where in the other 10 cases he could have equally likely have called 1, 2, 3, 5 or 6. And then p[7] = 1/6. Thus, if Magician looked for a 6, 5, 4, 3, 2 and 1 in sequence as he talked to the first six subjects AND FOUND HIS NUMBER IN EACH CASE [more and more luck] then p[7]=2/11. if Magician chose at random from the two dice AND IT JUST HAPPENED THAT WAS ABLE TO AND DID CHOOSE 6, 5, 4, 3, 2 and 1 in sequence [how much luck is involved there???] then p[7] = 1/6 in each case. But we don't know for sure, because the OP does not tell us the Magician's algorithm. Now the final point in formalizing 7's argument: what do we do with case 1? We've said that 7's p[7] must be the average of the first six. Yet we can't live with an unconstrained p[7] different from 1/6. The answer is that in case 1 the argument is invalid, while in case 2 it's valid. How convenient, you say - when the answer is wrong, you say it doesn't apply, but if you like the outcome you say it's OK. Well, maybe I've done that once, in the course of my life. But not here. There's a valid objection in case 1. Here it is. It's the apples and oranges thing. To justify taking the average of 6 cases, they must be of the same type. In our case this means the SAME ASSUMPTIONS must hold in each case. They don't: first we looked for 6's then we looked for 5's and so on. That removes the justification for averaging. In case 2, we're doing precisely the same thing each time, and averaging is permitted. The fact that luck is necessary does not introduce bias: we can re-roll the dice until luck shines on us. Certainly each roll of fair dice has the same p[7]. Finally, note that you can compute the average p[7] for both cases. It's just not valid to do so in case 1. Bottom line is no valid logic compels us to deduce p[7] = 2/11 for Subject 7, even though there is a valid set of circumstances for which p[7] = 2/11 for the first six. That is the conditions of case 1. That, in essence is the solution to the puzzle: why and how we can escape unconstrained p[7] different from 1/6. Now it's usually implied in logic puzzles that when a process is not specified, we assume the absence of mechanisms that produce bias. That is, if there is a mechanism that produces bias, we expect it to be identified in the puzzle. Thus, the preferred algorithm for magician is that he choose at random from the two dice when he talks with each subject. Now we come to p[7] becoming 1/6 for the first six subjects in case 1. In both cases, there are are absolutely 11 outcomes. A=n or B=n absolutely removes 25 outcomes. But let's consider the likelihood, for the n=6 case, of the outcomes 5-6 and 6-6. In the 5-6 case, under the conditions of case 1, 50% of the time Magician will say "one of your dice is 5". But in the 6-6 case, 100% of the time Magician will say "one of your dice is 6". Notice, this is not the same as saying A=5, B=6 is the SAME CASE as A=6, B=5. Arguing that way says there are ONLY 6 OUTCOMES, with the A and B values separately not mattering. That statement is false. There are two distinguishable cases, just as they were distinguishable in the unconstrained 36-outcome case. What LEADS TO THE SAME RESULT but, as desired IS ALSO TRUE is this: one ascribes, rightly, twice the likelihood that "one of your dice is 6" suggests a 6-6 configuration than you, rightly, do to it suggesting a 5-6 configuration. Thus, the 5-6 and 6-5 cases carry the same "logical weight" TOGETHER as the 6-6 case does BY ITSELF. Several solvers [most likely] saw this, and invented ways to make it happen: 5-6 and 6-5 are the SAME CASE, said some, but just saying so is (a) wrong and (b) not surprisingly, not justified. 6-6 should be COUNTED TWICE, said others, but with no justification given, other than p[7]=1/6 if you do so. Approaches like these might get you partial credit on a test. I know they did for me, on occasion. Both assertions give the desired answer; but neither stands up to scrutiny. Finally, What opened my eyes was an attempt to sketch a simulation program and "remove doubt." I love simulations, because you have to program them [i.e. specify your premises and argument precisely everywhere] and because they do distinguish between sufficiently different outcomes. Easy. Repeat this a million times: Roll the dice. Say to your subject, one of your dice shows a ...........long pause for thought........... Ouch! the dice came up 3-5. What do I say - 3, or 5? Eyes begin to squint from the bright light that all of a sudden was shining. End of simulation algorithm. See why I love simulations? Finally, finally: OP is incomplete - it doesn't give guidance on which die to identify to our subject. If we go one way it creates p[7] = 2/11. If we go another way it creates p[7] = 1/6. What do we do? Well, we could just say the puzzle is flawed, and walk away disgusted. Better, take the usual and most reasonable choice: Assume that for areas where OP is silent, no bias is introduced. Under the assumption of random choice between A and B, p[7] = 1/6 in all cases. p.s. [those initials stand for mea culpa]: My contribution to the heat-but-little-light aspect of the discussion was to insist the 11 remaining cases HAD TO HAVE equal probability. They started out that way, and "Nothing the magician said changed that." Is what I think I said. It still has sort of a compelling ring. But I would rather think of it this way. Start with "One of your dice is a 6" and put allowable cases into the outcome table. Clearly I could not put any of the forbidden 25 in, but as I put the others in, I would have the opportunity of asking what probability they have for being there. The 6-6 case certainly must be there. Whatever the magician was up to, he'd have to name a 6. But the 5-6 case is there only if magician happened to choose the 6 to call out - 50% of the time if magician was random in his choice. Thinking this way, it's clear that 5-6 [AND 6-5] BOTH belong in the table, but their likelihoods of producing the magician's statement are half of the 6-6 case. So again, in case 1, the 11 cases do not have equal probability: p[7] = 1/6; in case 2 they do: p[7] = 2/11.

I have no idea what an angle trisector looks like. The OP says use it [only] to trisect angles. But, given a line, put the trisector on it, trisect it for a 60o line, move down the 60o line and trisect it. One of those 60o lines will be parallel to the first line. We're pretty close to the first method, disclosed in one of the early posts. But we need to find the second method. Or, given there might be a third method, any method that's fundamentally different from method 1.

Can this now be laid to rest? <_< Sure. We can stop whenever. But did we debunk his argument? I'm not just trolling for more posts: I find it persuasive. I don't have a tightly-worded rebuttal. I paraphrase your answer as [he was given no clue] therefore [1/6 cannot change.] That ignores his argument. It says, since my argument is right and we differ, you must be wrong. If [he had heard nothing] then clearly [he was given no clue.] But [he heard the first six conversations]. I am trying to prove, to my satisfaction, that [hearing them] does [not lead to a clue]. I haven't succeeded yet. It's interesting, probably pivotal, that had he [not heard anything] he could nevertheless reason exactly the same!

Sounds like a proof to me.

This is a proof, not an opinion. It is based on correct logic and statistical methods. Please tell me you all can accept the truth when it's presented, regardless of how many people before you have believed a falacy. So, Bonanova is correct, there is no problem to solve, other than to answer why so many people get sucked into this blatant misuse of statistics. You ask why I don't say whether 2/11 is correct for subjects 1-6. I did make a case; but it's in a spoiler, since it's part of the puzzle. The puzzle is solved either of two ways: [1] Prove that p[7] = 1/6 for all seven subjects. [2] Either (a) Give up the notion that Subject 7's p[7] has to be 1/6, or (b) Debunk S7's argument. You took approach [1], giving a proof [one that satisfies you] that p[7] = 1/6 for all seven subjects. You've finished the puzzle. If I was perceived to put you [or anyone else] down, that's not ever my intent; I apologize. I believe it's within the bounds of respect for a person to point out what might be a weak point in an argument, but certainly I will not impugn anyone's intelligence. I reserve for myself the right to make a mistake; and, frankly, I exercise that right fairly regularly. So I this give that same latitude to others.

Comments in red in following spoiler. You may have proved it and I don't understand, so I ask a question.

Can't believe I did that. Wait. *checks calendar* Yeah, I can believe it. "at least" does not belong there. It doesn't let the bonus row of 4 trees make much sense. Wow - sorry for the headache that caused. Now that several of you have over-achieved the stated OP, shall I show my 18 rows of exactly 3 trees - with a bonus row of 4? Or shall the headaches continue?

Any clue that is imparted by a statement is also imparted by a statement that is its logical equivalent. At least one of your dice is 6 is logically equivalent to A=6 or B=6: either or both are 6; i.e., not neither are 6. Also equivalent is both of your dice are not distinct from 6; more precisely: it is not the case that both are distinct from 6. All these statements do nothing other than to remove the 25 cases where neither of the dice show a 6. Their common outcome table is the truth table that proves their logical equivalence. For any value of n, not just for 6. Subject 7 says that p[7] = 2/11 once it is known A=n or B=n for some [any] value of n. He argues that with certainty his dice permit the statement for some value of n. Then since, for any n, 25 cases are removed, and among the remaining cases 2 are favorable, his p[7] must be 2/11. Do you agree that "at least one of your dice is 6" and "both dice are not distinct from 6" [in your words] are logically equivalent?

Yeah, I got that on a re-read. Apologies. That a "6" shows with probability of 1/6, sounded like p[7]=1/6 because of two counts from double 6 on first read. I should have edited my post to that effect. Let this admission accomplish that. I compiled the summary in probably too short a time; in haste, to direct things back to the question.

P.S. I REALLY like this one. Anything that brings me that close to doing math deserves some accolades. Nice formalism. Now, can it be used to construct a proof?

How do you read I may be present in one place, but not in others I took it to mean that if "you" are present in one place, then "you" are not present elsewhere. But it could mean just that "you" might not [need not] be present elsewhere.