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General => Chit Chat => Topic started by: X1 on January 11, 2013, 02:16:38 AM

Title: The shortest-chain-of-the-World ( problem )
Post by: X1 on January 11, 2013, 02:16:38 AM
The shortest-chain-in-the-World problem.

   We are about to make the shortest chain of the world, made by two links, and two links only !

   We have a tall pile of separate links, but we do not know how strong they are. All that we know is that one third of them can withstand a pull of only 1 kilogram, one third of them can withstand a pull of only 2 kilograms, and the last third of them can withstand a pull of only 3 kilograms.
   So, the average weight one link can withstand is 2 kilograms, right ? [ "Yes"... ].

   We now choose two links at random, and we connect them to form the shortest-of-the-World chain.
  So, this chain be able to withstand a pull of 2 kilograms at average, right ? [ "No" ! ]

   If you answer "Yes" to the last question, read the problem once more ...
   If you answer "No", proceed to the next question :
   
   What is the average weight this two-link chain can pull ?
(
   We can replace the "chain" by "knotted line", where "links" will be replaced by "bends"... but the essense of the problem will remain the same.)
Title: Re: The shortest-chain-of-the-World ( problem )
Post by: Dan_Lehman on January 12, 2013, 09:37:05 PM
[ Third time's a charm?  --Net/computer problems have
defied me on two prior attempts ... . ]

I see that this problem was raised & answered in another
thread, but as the OP in that concerns knots and not
probabilities, this is the better one to continue such
discussions as have already swamped the other.

I'm happy to see that (I think) I got the right answer;
in any case, I concur in what X1 got.  My reasoning
ran as follows, assuming that the number of links
is so great as to make negligible the change to probability
of choosing a particular type a 2nd time after removing
one such link on first choice.


I see nine logical possible choices,
which I'll set in length indicating strengths:

1+1
1+2
1+3
2+1
2++2
2++3
3+1
3++2
3+++3

There are five double-links with a 1-link (strength =1);
only one double-link of 3-links (strength =3);
so the remaining three have strength =2).
(5x1) + 3 + (3x2) = 14 / 9 = 1.55... .

If one were to double the number of 2-links --say, to give
the distribution more of a Bell curve character,
I see that circumstance --w/same question-- as:

1+1
1+2
1+2
1+3
2+1
2+1
2++2
2++2
2++2
2++2
2++3
2++3
3+1
3++2
3++2
3+++3

Here, one needs sixteen cases in order to give the
increased quantity of 2-links their higher probability.
There are now seven double-links with a 1-link (strength =1);
still only one with both 3-links (strength =3);
and so the other half of eight has strength =2.
(7x1) + 3 + (8x2) = 26 / 16 = 1.625
--less of an increase than one might suppose!?

I think that this is an accurate reasoning, though
clumsy for dealing with probabilities --percentages--,
but I've yet to learn better.


  - - - - - - -

Now, against some of the arguments that such reasoning
as shown here for the (perhaps surprising) effect of low
values of "double links" that were advanced in the other
thread must come the realization of actual probabilities
and confidence intervals.  Here, the 1-link is fully a third,
then a quarter of the possible choices; in practice, the
low values should occur far less often --and perhaps be
far less different--, which will then be less significant.

 - - - - - - -

Also presented in the other thread was the good ol'
Monty  Hall guess-the-door problem, which has been
a longstanding debatable issue.  It really is a subtle
and, well, debatable issue; many highly educated
people have argued either side.  The subtle point
hinges on a maybe not-clearly-described condition
--which itself can seem irrelevant-- : that what is
revealed in offering a 2nd choosing is deliberately
a non-winning option.  That the revealer's knowledge
should influence things can seem, well, irrelevant.

I.e., suppose the circumstance were that it would
be the case that it would be revealed what was in
the "door next to" the 1st choice --let us say that
2 is next to 1 is next to 3 is next to 2.  2/3 of the
time this revelation would match the traditional
puzzle and be a losing option ; the odds of the
winner being in the other unchosen door, however,
are but 50/50, which is what is the *wrong* answer
to the traditional problem,
where the revelation is assured of being a loser
--and so is whichever of the unchosen doors fits!
These cases can look much alike.
And I think that were the situation revised to be that
there were three, graduated-in-value results (so, some
middle one that at least was better than *nothing*),
the problem pretty much dissolves --EITHER non-chosen
door could be revealed, there would at least be this
not-so-bad option if not the grand-prize one,
and no way to reason odds, really.

 
--dl*
====
Title: Re: The shortest-chain-of-the-World ( problem )
Post by: X1 on January 12, 2013, 11:20:03 PM
assuming that the number of links is so great as to make negligible the change to probability
of choosing a particular type a 2nd time after removing one such link on first choice.

  Yes, we always have to assume this in such simple problems - otherwise they will be really hard - as everybody that has attempted to calculate probabilities in poker knows !  :)

I see nine logical possible choices

   The term "logical" in not the right one here... You should better say : nine possible choices, each having the same probability with any other. End of the story...
   As the strength of the chain is equal to the strength of its weakest link ( as everybody do know - and that is why I was forced to rephrase the 2 "knot" problem to the 2 "link" problem, to offer a hint...), there are 5 chains of ( a weakest link that is able to support ) 1 kilogram only, 3 pairs of 2, and 1 pair of 3. So, 5x1 + 3x2 + 1x3 = 5 + 6 + 3 = 14. So, the average strength of a randomly chosen pair of links is 14/9 - lower than the "expected" "median" 2...   
Title: Re: The shortest-chain-of-the-World ( problem )
Post by: X1 on January 13, 2013, 12:28:13 AM
   The Monty Hall "guess-the-door" problem has ONLY one solution - no question about it. One can write a computer program, and actually play this game many times - if he will not follow THE correct solution against a player who will, he will lose very quickly !  :)
   
   Let me present two lines of thought that lead to this solution.
   
   The first is a very general one, which was the one I followed. It sounds rather simple to my ears, but that is most probably due to the fact that, after a certain age, everything one has NOT thought, sounds too complex !  :)
   At the time he made his  first choice, the player used ALL the information he was offered at that stage. The ONLY / BEST thing he could do was to select one out of the three doors at random. At the second stage, he was offered another, new piece of information, relevant to the problem, which was not available to him before. This new information was neither included in nor deductible from the one he already had.
   Here comes the solution : Just because he had used every piece of information he had while he made his first choice, and now he has an additional, new one, the ONLY way to actually incorporate it into his given new sum of information, is to change his first choice. If he does not, it would mean one of three things :
   He has ignored the new information - which is obviously not the best thing to do, in whatever problem !
   He thinks that he had already included this information when he made his first choice. However, this would mean that the new information was not new - which is wnot the case.
   He believes that the new information will lead to the same choice, because it is compatible to the previous one. After all, it is but an information about the same situation, anf the situation had not changed between his first and his second choice. Yes, it is compatible, and it is about the same situation - but it was not actually known when the first choice was made. If it were offered at the first place, the first choice would possibly have been different.
   Unperformed experiments have no results, and we can not compare/add/subtract un-made, hypothetical choices, to actual, made ones.
   The only way one can "add" to his already-made first choice something, whatever this might be, so he will incorporate the whatever new piece of "added" information offers, is to actually change his first choice. We need not know what exactly this new piece of information was able to tell us. We need only to "add" something to our first choice - and there is nothing else to do to actually achieve this, than to alter our first choice.

   The reader who will think that the father is talking nonsense ( as it happens too often in this forum ... :)), should also listen to the solution of the son (my 17-years-old son).
   When we have chosen one out of the three doors in the first place, what really had we expected to happen, if the door was to open right afterwards ? Not what we had hoped to happen, what we expected to happen, what we would bet our money/pride on, that it would probably happen !   What would be the expected outcome of our choice ? Not the less probable outcome, of course... :) That means that, although we hoped we will make the correct choice, if we were rational beings, we should have expected that we will make the wrong choice. So, when we are informed that one other choice is also another wrong choice, to choose the correct one, we should better choose the only one left.
Title: Re: The shortest-chain-of-the-World ( problem )
Post by: Dan_Lehman on January 13, 2013, 12:56:07 AM
   The Monty Hall "guess-the-door" problem has ONLY one solution - no question about it.
. . .

The devil's in the subtlety of wording such that "the"
becomes a suggestion against what in fact might be
plural, not singular.

Let me re-state the problem and see if it helps, or muddles?

Suppose I offer you this choose-a-door option, telling you
that I will open the "next" door,
and then allow you to reconsider --unless that door has
the prize, then you lose this contest.  How's that?

So, you make your choice --say, Door_A--, and I open
Door-B and we see a goat (non-prize), and now I say
to you "Would you like to change your selection?"
What's your answer?


--dl*
====
Title: Re: The shortest-chain-of-the-World ( problem )
Post by: X1 on January 13, 2013, 01:23:05 AM
So, you make your choice --say, Door_A--, and I open Door-B and we see a goat (non-prize), and now I say to you "Would you like to change your selection?"
What's your answer?

   I can not see your trap - although I suspect you have (carefully) placed a concealed one, under some word(s)... :)
   Why is this a re-phrase of the original problem ?
   I will search the net and see if I can find any references to it. Where are you, Structor ?

  Of course, because I get a new piece of information, on top of what I already had, I will do the ONLY thing I possibly can, to incorporate it ( it : as I have said, I do not need to know its meaning or details, I am forced to act according to the facts that, a) it is relevant to the problem, and, b) it was not incorporated into the original information I had ) into my new, actual choice - so I will change/alter my first choice . I will now choose door C.

P.S. "next" means any one of the two left, the two I had not chosen, I suppose...It has nothing to do with the location in space of the (in-line ?) doors.
Title: Re: The shortest-chain-of-the-World ( problem )
Post by: X1 on January 13, 2013, 01:40:30 AM
   So, let me now offer again the other line of thought.
   When I had chosen the door A, I knew that the most probable outcome would be that it contains a goat. That is, although I hoped it contained the price, I knew that I should expect it contained a goat - because that was the most probable outcome. So, I hope it contains the prize, but I expect it contains a goat, and my future actions would/should be based on rational expectations, not hopes... 
    When I actually see that door B contains a goat, because I expect that, most probably, A would also contain a goat, I chose C - I want to run away from goats, remember ?  :)
Title: Re: The shortest-chain-of-the-World ( problem )
Post by: X1 on January 13, 2013, 02:42:03 AM
http://en.wikipedia.org/wiki/Monty_Hall_problem

   "Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice?
 
( the host is constrained always to open an unchosen door revealing a goat and always to make the offer to switch. Very often it is also assumed that the car is initially hidden completely at random, and that, if the host has a choice of door to open (which happens if the player initially picked the car), then this choice is completely random, too. Some authors, either instead of or together with these probability assumptions, assume that the player's initial choice is completely random.)

==================================================================

Another interesting line of thought that I was not aware of :

 Increasing the number of doors

 " That switching has a probability of 2/3 of winning the car runs counter to many people's intuition. If there are two doors left, then why is each door not 1/2? It may be easier to appreciate the solution by considering the same problem with 1,000,000 doors instead of just three . In this case there are 999,999 doors with goats behind them and one door with a prize. The player picks a door. His initial probability of winning is 1 out of 1,000,000. The game host goes down the line of doors, opening each one to show 999,998 goats in total, skipping over only the player's door and one other door. The host then offers the player the chance to switch to the only other unopened door. On average, in 999,999 out of 1,000,000 times the other door will contain the prize, as 999,999 out of 1,000,000 times the player first picked a door with a goat?the chance that the player's door is correct has not changed. A rational player should switch.
 
Intuitively speaking, the player should ask how likely is it, that given a million doors, he or she managed to pick the right one. It's as if the host gives you the chance to keep your one door, or open all 999,999 of the other doors, of which he kindly opens 999,998 for you, leaving, deliberately, the one with the prize. Clearly, one would choose to open the other 999,999 doors rather than keep the one."
Title: Re: The shortest-chain-of-the-World ( problem )
Post by: Dan_Lehman on January 13, 2013, 03:17:55 AM
So, you make your choice --say, Door_A--, and I open Door-B and we see a goat (non-prize), and now I say to you "Would you like to change your selection?"
What's your answer?

   I can not see your trap - although I suspect you have (carefully) placed a concealed one, under some word(s)... :)
   Why is this a re-phrase of the original problem ?
   I will search the net and see if I can find any references to it. Where are you, Structor ?
::)
Wow, there surely is a lot of meta-problem solving going
on above (why would one suspect a "trap"??!),
rather than doing the given one!

Quote
P.S. "next" means any one of the two left, the two I had not chosen, I suppose
...It has nothing to do with the location in space of the (in-line ?) doors.

"Next" is as defined earlier : Next (door#) = (door# modulo 3) +1
(or for other than 3 doors, "<#_of_doors>" replaces "3").
I.e., 1=>2, 2=>3, 3=>1, closing the circle.


--dl*
====
Title: Re: The shortest-chain-of-the-World ( problem )
Post by: X1 on January 13, 2013, 03:54:33 AM
why would one suspect a "trap"??!,rather than doing the given one !

Timeo Danaos et dona ferentes.

Φοβου τους Δαναους και δωρα φεροντες.
( for members that prefer to read anything I say as if it is said in my "native tone"... :))

http://en.wikipedia.org/wiki/Timeo_Danaos_et_dona_ferentes

   You should know in advance where the ABoK rare copy is, otherwise you can not always open a door ( the B ) with a knot tyer - as I am sure you wish...  :) !
   However, even if you do - that is, even if you choose to open the one of the two other-than-the-one-I-had- already-chosen doors ( the B or the C, as I had chosen the A ), and you do this in random ( your choice was decided by flipping a coin, between the option door B and the option door C ), and it happened that this door was the door B, that contains the knot tyer and not the rare ABoK copy ), even then, it would be better to me to alter my initial selection, and select C.

   Read the simpler explanations offered in the Wikipedia article - and then read again my most simple one - that you have to alter your decisions/selections, every time you are offered a new piece of information you have not already included in your original choice.

Title: Re: The shortest-chain-of-the-World ( problem )
Post by: Dan_Lehman on January 13, 2013, 07:38:51 AM
... even if you choose to open the one of the two
other-than-the-one-I-had- already-chosen doors
( the B or the C, as I had chosen the A ),
and you do this in random ( your choice was decided by flipping a coin,
between the option door B and the option door C ),
and it happened that this door was the door B,
that contains the knot tyer and not the rare ABoK copy ),

even then, it would be better to me to alter my initial selection,
and select C.

Well, you sank into it, trap or not.  Because I know that
you can run out the *random* possibilities here, and see
that there are equal chances for all --for B being void,
and the prize being A or C.  On only 1/3 of the cases
is C the winner (as for the other two doors' probabilities);
that you got a peek at B, by random chance, doesn't
change this --or enhance A's chances--, but it does
leave these two as candidates, of now equal standing (50/50).
(You'll be no worse off by changing.)


Quote
that you have to alter your decisions/selections,
every time you are offered a new piece of information
you have not already included in your original choice
.

"Information" is a term with shadings of meaning, but
the "information" that you get must be helpful.
In my example, it wasn't a coin flip after the fact,
but I think it amounted to a similarly unbiased/uninformed
choice --the stipulation to open a door determined by
your choice ("next"), something done w/o being informed
by door contents.  (The computer program that runs through
all the possibilities to show that Monty Hall's puzzle wins
on changing will show no such win here, as the prize
takes it turn at A, B, C doors.)

And this should show how subtle these statistical issues
can be --head-hurtingly subtle.


--dl*
====
Title: Re: The shortest-chain-of-the-World ( problem )
Post by: Knotman on January 13, 2013, 10:58:47 AM
....On only 1/3 of the cases is C the winner (as for the other two doors' probabilities);....

This is correct, 1/3 of the cases is C and 2/3 of the cases the winner is either A or B

....that you got a peek at B, by random chance, doesn't change this --or enhance A's chances--.....

Correct.  Since you now know that the winner isn't B the 2/3 probability is now apportioned to A.  Knowing some extra information along the way can't change the 1/3 chance you picked the correct door in the first place. 

Darren
Title: Re: The shortest-chain-of-the-World ( problem )
Post by: X1 on January 13, 2013, 03:42:31 PM
Well, you sank into it, trap or not. '
 Because I know that you can run out the *random* possibilities here, and sethat there are equal chances for all --for B being void, and the prize being A or C.  On only 1/3 of the cases is C the winner (as for the other two doors' probabilities); that you got a peek at B, by random chance, doesn't change this --or enhance A's chances--, but it does leave these two as candidates, of now equal standing (50/50).
(You'll be no worse off by changing.)


Oh, my KnotGod !
Do me ( and yourself ) a favour, DL...PLEASE
Stick to the knots, and leave numbers to others. In other words, do not change your daily job to become a mathematician, or even a tax lawyer... :) ( THAT is the new information you should include into your rationale ( I did...), given that it is the n - th time I had explained that you were wrong right from the beginning, and you are still wrong - but you have not understood it neither then, nor now.)

In my example, it wasn't a coin flip after the fact, but I think it amounted to a similarly unbiased/uninformed
choice --the stipulation to open a door determined by your choice ("next"), something done w/o being informed by door contents.

You hoped it would have happened, you may even loved to see it happening, you may even have prayed to God to change the odds, for me to have fallen into your chlldish trap...but sorry, DL, sorry, dear God, I had not.  :)


   You should know in advance where the ABoK rare copy is, otherwise you can not always open a door ( the B ) with a [goat]
   However, even if you do - that is, even if you choose to open the one of the two other-than-the-one-I-had- already-chosen doors ( the B or the C, as I had chosen the A ), and you do this in random ( your choice was decided by flipping a coin, between the option door B and the option door C ), and it happened that this door was the door B, that contains the [goat] and not the [car] ), even then, it would be better to me to alter my initial selection, and select C. 

   IT DOES NOT MATTER HOW YOU had managed to provide ME a useful information. The information is usefull, it is new, so I have to use it.
   Where is the trap you have fallen ( again ?)
   In order to achieve this, to be able to choose the "wrong" door by chance, you have to pray to God - AND the God has to be willing to help you, because he is usualy occupied by tying the new knots I provide him... :) OR, you are talking only about half of the events, where you had been lucky and succeeded to choose the door with a goat, and not with a car . You ignore the other half cases, where your "example" would not stand, would not be as you have described it, because you would have chosen the door with the car.
   So, in a series of many trials ( because we are talking about many trials here - the "probabilities" of single, miraculous events - the monopoly/speciality of God) - do not bother us...), where you examine all the equally possibly events,  there are half of the trials that do not belong to the situation you have described for your example. You count the probabilities in only half the equally possible and probable cases ! You count a fraction of reality ! Wake up ! ( Or should I say : Come down ( of the back of this animal) - after all, "you will not get worse off " by doing by this... :))
   
   P.S. I would be glad if you examine at least half of the new knots I had presented in this Forum...but you stick to your point, you are even proud of this attitude, and you do not. Just another lost possicility of a probably useful event, that you have chosen to ignore - neither God himself nor can not alter THIS sequence of past events...  :)
Title: Re: The shortest-chain-of-the-World ( problem )
Post by: X1 on January 13, 2013, 04:23:53 PM
....that you got a peek at B, by random chance, doesn't change this --or enhance A's chances--.....

   Would you able to do this in all cases ? How ? Tell me, please, because this line of communication of me with the future ( or with the God ) has been cut off... :) If you will not, and you had only succeeded at this, single case, once, by pure chance, you are not talking about something it could have repeated many times, and happens every time.  You are not talking about the conditions of a problem it always remains the same. So, how on Earth you are talking about probabilities , if you are talking about a single event, if you suppose that something had only occurred by chance or by devine intervention  ? Probabilities can only have a meaning when the events are not MIRACLES.
   The "example" DL got out of his sheaves was specific about it. When the host ( the host, NOT another player !) will open the door, this door will (always) containg a goat. It is not a single, miraculous event. If the door contained a car, I suppose the player would have won already and would have left with it to California, without been forced to choose anything !   :) 
   You do not know in advance how and why you have offered another piece of information. Timeo Danaos et dona ferentes. Dl has his reasons...That should not interfere with your decisions. All you should do, all you have to do, is to take the new information into account - if it is not a re-phrased variation of the original information, of course. Just because you had used all the information you had in the first place exhaustively ( you used every piece of it... and the only thing left for you was to choose at random, because every door has the same probability to contain a car ), and because now you have another, new piece of information, you have to "add" it somehow to the one you already had. If you add something to something ( both being not-zero ), you get something different ! So, with the different information you now have, you should act differently - unless you stick to your point, and keep riding ! The ONLY way you can act differently, is to change your first choice.

   The same problem can be posed with four doors, and with you been offered the option to choose every time after a door that you had not chosen opens, and reveals that it contains a goat. You should change your previous choice every time, just because every time you acquire a new piece of information, and the only way you can do something about it, the only way you can actually change your actions because the information that is related to them has changed, is to change your previous choice.

   ( In a way, that situation explains, in a way, why I have zero response about the new knots I publish... after all, they are more difficult to understand than elementary probabilities !  :) )
Title: Re: The shortest-chain-of-the-World ( problem )
Post by: Dan_Lehman on January 13, 2013, 07:26:06 PM
I am mistaken in my surmise the X1 could "run out the
probabilities ...", and make a sensible reply, conceding
the point.  No, not this time; beyond that, he inflates
his response with gratuitous insults.

For others, who might care about the issue,
let me explain.

In the problem setting where one chooses one of three
doors, and then a coin toss decides which of the other
two doors is opened,
the chance of the winner being among this pair is 2/3;
of being in the door randomly revealed is 1/3.
So, upon seeing a loser revealed, the contestant
knows only that there is now a choice between doors
each of which has 1/3 probability of hiding the prize,
so now an even chance between these remaining doors.
This follows from the random revelation.

In contrast, if the host (Monty Hall) reveals a loser door,
the chance of that is 100% --there was no chance;
he will for sure show a loser--, and so the 2/3 probability
remains with the other of those two doors.  You see,
in the random situation, there is a 1/3 probability that
the coin toss will reveal the winner; this 1/3 is removed
by Monty in choosing the *other* door in such a case,
never revealing a winner.  (In my "Next(choice)" construction,
this amounts to Monty breaking with that function,
to reveal Next+1 instead.)

One can construct a table of the possible locations of
the prize, and plot the choices; the revelation of a loser
removes those cases where that door should have a
winner, leaving equal chances otherwise; it is the
point that the elimination is done by chance and not
knowing choice that changes this case from the
traditional puzzle.


--dl*
====
Title: Re: The shortest-chain-of-the-World ( problem )
Post by: X1 on January 13, 2013, 09:06:50 PM
   
he inflates his response with gratuitous insults.

 
(edited, when my blood pressure came down to normal levels again... :))

  ... It is a tricky problem, indeed, and the people who did not get it right at the first place always feel somehow insulted / betrayed - and then they try to mud the issue, by inventing various / numerous "loopholes" and "variations". It has been done many times in the past, as one can read in the article in Wikipedia, and it will be done again in the future - human narure does not change onernight !  In fact, that problem is widely known exactly because of this childish/wild behaviour : Riding-whatever mathematicians and/or would-like-to-be-omniscient people, tried to formulate their own versions, to trap back the people they feel they had trapped them...
    However, I had no intention to trap anybody. I was talking about the problem of the two bends, placed the one after the other in the same line, and I was arguing that the "average" of the strenghts is meaningless, because it will be pushed lower, towards the weakest "link". I just thought that the two problems had something in common : the idea that "adding" averages and calculating probabilities of not-performed experiments (1), or not-taken choices, can be misleading...
   It was proved to be a wrong idea !   :)

1. A. Peres, 1978
    Unperformed experiments have no results
   
Title: Re: The shortest-chain-of-the-World ( problem )
Post by: X1 on January 13, 2013, 10:27:57 PM
   When I thought it would be better to compare the two-links-with-three-value-strengths problem,  with the infamous choose-one-door three-doors-two-outcomes "Monty Hall" problem, I was not aware of the problems this problem generates to the people who had not solved it at the first place...  :) and then try hopelessly to discover some "loophole" of the correct argument, or some "variation" of the simple initial problem, that could lead to another solution. I underestimated the fact that I had spent my New Year evening trying to explain it to a friend of mine, that has a PhD in engineering, in vein ! So, I should perhaps expect a similar behaviour from some members here - but I have to confess that I had hoped for something else.
   Now I am reading the Wikipedia article ( I was not even aware of the name of the problem till now... ), I see that the behaviour of Dan Lehman is rather typical ... so I should not blame his "expected" behaviour, as I did. People all over the world try to escape by teeth and nails - because, somehow, this problem offends their ego, their self-respect, in a deeper way one would have thought. There are even many studies of psychologists who try to explain this wild behaviour - so we are talking about something rather common.
   Reading this article, I was amazed by the number of "traps" people had invented, to "revenge" the ones they feel they had trapped them in the first place. If I knew this, I would never have thrown more fuel to the fire we have already in this Forum. My original "two-in-line-knots", or "two-links" problem was more than enough !  :)
   Somebody publishes a series of experiments, where one set of knots ( the eyeknots, the loop knots ) is shown to be stronger, on average, than another ( the eye-to-eye knots / the bends ). However, those results do not tell that if we put in line TWO eyeknots, two loops, the one linked to the other by the linkage of their bights, to form a compound bend, they will still be stronger than ONE eye-to-eye knot, one bend. I pointed it right away (1), and I have repeated it in other words (2),(3)... but Dan Lehman had not understood the issue (4)...and have chosen to insist defending his mistaken view ever since (5). So, I should have not be surprized he has chosen a similar escape in the more tricky Monty Hall problem - where, at least, he seems that he is in good company. The interested reader is advised to read the Wikipedia article, to learn how we all can persuade ourselves that the other guy is always "not big"... but WE are !  :)

   1.  http://igkt.net/sm/index.php?topic=4150.msg25822#msg25822
   2.  http://igkt.net/sm/index.php?topic=4150.msg25830#msg25830
   3.  http://igkt.net/sm/index.php?topic=4150.msg25879#msg25879   
   4.  http://igkt.net/sm/index.php?topic=4150.msg25828#msg25828
   5.  http://igkt.net/sm/index.php?topic=4150.msg25864#msg25864
Title: Re: The shortest-chain-of-the-World ( problem )
Post by: Dan_Lehman on January 14, 2013, 08:51:00 PM
   
he inflates his response with gratuitous insults.

  ... It is a tricky problem, indeed, and the people who did not get it right at the first place
always feel somehow insulted / betrayed - and then they try to mud the issue,
by inventing various / numerous "loopholes" and "variations".
...
In fact, that problem is widely known exactly because of this childish/wild behaviour :
... to trap back the people they feel they had trapped them...
However, I had no intention to trap anybody.
...

A heap of deprecations serving no good purpose,
and which suggest refusing to follow reason and
concede an argument, fearing a loss of pride.

There was no trapping, nor on my part any misunderstanding;
the "Monty Hall" problem has been understood by me for some
time.  But the problem statement is key to distinguishing it
from seeming equal problems that however differ in result:
e.g., to say that the host "opens a door" is ambiguous as to
a key ingredient --randomness or not.

And so I asked the similar problem wherein the door opening
was done by chance --something that X1 denied making any
difference, when in fact it is a critical point : there is no benefit
to switching (nor penalty from doing so), in this case.  One
can read about this via Wikipedia, if not "running out the
probabilities" oneself, on some paper, say.

--to wit:
Quote
... even if you choose to open the one of the two
other-than-the-one-I-had- already-chosen doors
( the B or the C, as I had chosen the A ),
and you do this in random ( your choice was decided by flipping a coin,
between the option door B and the option door C ),
and it happened that this door was the door B,
that contains the knot tyer and not the rare ABoK copy ),

even then, it would be better to me to alter my initial selection,
and select C.

Well, you sank into it, trap or not.  Because I know that
you can run out the *random* possibilities here, and see
that there are equal chances for all --for B being void,
and the prize being A or C.  On only 1/3 of the cases
is C the winner (as for the other two doors' probabilities);
that you got a peek at B, by random chance, doesn't
change this --or enhance A's chances--, but it does
leave these two as candidates, of now equal standing (50/50).
(You'll be no worse off by changing.)

Or, better for your pride, but not your chances of winning
the prize copy of ABOK --but maybe you'll meet some
"big" knot tyer, and learn something!   ;D

But of course you won't heed the repeated argument, from me;
you can read about it elsewhere, though, via URLinks from the
Wikipedia page you now know about.  E.g.,
http://probability.ca/jeff/writing/montyfall.pdf (http://probability.ca/jeff/writing/montyfall.pdf)
Quote
Monty Fall Problem: In this variant, once you have selected one of the three doors,
the host slips on a banana peel and accidentally pushes open another door, which just
happens not to contain the car.
Now what are the probabilities that you will win
the car if you stick with your original selection, versus if you switch to the remaining
door?
In this case, it is still true that originally there was just a 1/3 chance that your original
selection was correct.  And yet, ...
the probabilities of winning if  you stick or switch are both 1/2, not 1/3 and 2/3.


All of this --and indeed the considerable, continued discussion of the
Monty Hall issues-- should illustrate how some seemingly subtle,
minor ingredients in a problem statement can be critical to the
solution (or, put another way, to actually describing the intended problem)!


.:.  QED , at last?!   ::)




Now, how did we get into all this statistical debate, at all?
--from this [nb: I've corrected X1's "eye-to-eye" to be "end-2-end",
by which is meant other's "bend", i.e. a single, ends-joining knot]:

Somebody publishes a series of experiments, where one set of knots
( the eyeknots, the loop knots ) is shown to be stronger, on average,
than another ( the [end-2-end] knots / the bends ).
However, those results do not tell that if we put in line TWO eyeknots, two loops,
the one linked to the other by the linkage of their bights, to form a compound bend,
they will still be stronger than ONE [end-2-end] knot, one bend.

I pointed it right away (1), and I have repeated it in other words (2),(3)...
but Dan Lehman had not understood the issue (4)...
and have chosen to insist defending his mistaken view ever since (5).
So, I should have not be surprized he has chosen a similar escape in the more tricky Monty Hall problem
--where, at least, he seems that he is in good company.


And this gratuitous misstatement of my thinking and repeated
denial of reason again confronts us with an apparent need
to (further) explain --although the explanations are ignored.
Here is what was said, in part.

Given these results, it would seem that the strongest means of bending two lines together
should be with two end-of-line loops.

  No ! When you will use two loops, the reduction in the strength of the compound bend
will be larger than the average you have reported. From any pair of loops / links,
you will have to take into account only the weaker, and then to calculate the average
of the weakest links, and the weakest links only.

Yes, not "no"!  Look at the values : except for the fig.8
end-2-end knot
, the other such knots are way below
the strengths of the eyeknots --weaker values or not!


To emphasize, "the other such knots are way below the strengths
of the eyeknots"
!!


Which points to our having more than mere average values
from the testing, but sD values, which give a predicted shaping
to the distribution of expected results.  And given the large
differences that were observed, even taking into consideration
the value-diminishing effect of 2-link sampling, the eye-in-eye
(or also, we should believe, twin eye --CLDay's term) knots
come out on average stronger than single-knot end-2-end joints,
in these observations, by statistical reasoning.


That reasoning uses an expected bell curve,
as referred to ...

Quote
My view is that you should have ignored that single test,
just follow the expected bell curve, and so keep the standard deviation low.

Good, let's do that, indeed --and make the reasoning afforded
by such statistical analysis via the standarDeviation (sD).


Of a normal bell curve, one finds 95% of the results within
2 sD of the mean; one sD contains 34%, the 2nd 14%, and
at a 3rd and beyond are 2% = 50%, reflected on the other
side of the mean.  As in our situation we're concerned about
only the divide between a lower range and the rest, we can
put our divide at 2 sD below the mean and have 98% vs 2%
--quite some diminution of the lower values!


But look as where 2 sD below the observed AVG values puts
us : at eyeknot strengths of the overhand & LazyDog of about
80% & 75%, respectively.  In comparison, the end-2-end knots
corresponding to these have means (not 2 sD below!) of about
70%.  .:.  To my thinking, this supports the proposition that
the stronger means to joining lines is by interlinked (one way
or another) eyeknots, in some cases, at least.  Notably NOT
among those cases by this current data is the fig.8 knot
which tests high in both forms (which surprises me somewhat!?).




And let's take this more refined sD/Guassian-distribution thinking
to the 2-Links problem in the OP.  Note that it assumes values
that are quite different (1 is half of 2 and a third of 3), and and
equal distribution of these values.  I showed, in my answer
to this problem, what happens if one simply doubles the middle
value (2) in population.  But let's now try to model the implications
of a bell curve and it sD implications.


To model a 98% distribution above the low value,
we can use 39 of 40 links, with 38 at value 2, one each at 3 & 1.
To assess the effects of this on expected average strength of
a 2-link joint, we have:


1+1 =1
1+2 =1   x38
1+3 =1
 -------------------> sums to 40x1 =40kg
((
2+1 =1
2++2 =2  x38
2++3 =2
 -------------------> sums to (1 + (38x2) + 2) =79kg
)) x38 !

 -------------------> sums to 38 x 79 =3002kg



3+1 =1
3+2 =2   x38
3+3 =3

 -------------------> sums to (1 + (38x2) + 3) =80kg


total kg = (40 + 3002 + 80) = 3122


There are (40 + 1520 + 40)=1600 cases
dividing the total kg sum of 3,122
. . .
for an average of 1.95kg.


So, yes, taking into consideration the effect of 2-links
alters the expected average, but only slightly, once
the distribution of such values is figured in.




--dl*
====

Title: Re: The shortest-chain-of-the-World ( problem )
Post by: X1 on January 14, 2013, 09:46:10 PM
fearing a loss of pride.

Yes, FEAR is the correct word, I agree.  :)
Here comes a unicornman !  :)

Or the unicorn man fears to stand on his own feet ? That is the question...
1. If you have chosen somebody that would fear something, anything - except time -, you have chosen poorly. Look elsewhere, or into the mirror.
2. If you think that all people are feeling like you, remember it the next time you will insult somebody.
3. If you are fearing your loss of pride, relax : your pride is so great, that any, however great that might be, loss of a portion of it, will leave the net sum unchanged.

the "Monty Hall" problem has been understood by me for some
time. 
   
   I do not doubt this...However, we do not know how much time you were trying to understand it, do we ?  :)
   Apparently, you have not understood my solution, and that is why, wisely perhaps, you have not said one word about it - but many words about/against me, as always...
   Speak about what I do, about my knots, not about me... because, apparently, you choose people poorly.

maybe you'll meet some  "big" knot tyer, and learn something!   ;D

  Was this your great final one-liner ?  :) Because if it was, either you underestimate yourself, or you underestimate Ashley. I leave to the reader to guess, judging from your behaviour, which is the most probable, the "expected" outcome !  :)
   To be anle to teach something about anything, you should be a teacher . Frankly, you are the worse teacher I had met in my life !  :)
   However, this will not disturb your loneliness - it will secure it.

   The childish trap with the two players, where the one is choosing after the other, was easily detected - but I did not believe you were so naive to place this one, and not any one of the many much better you could think of - or, as you knew the problem "for some time ", could have copied from the Wikipedia article !
   So, I had excluded it , and I was very strict about it :
 
   The "example" DL got out of his sheaves was specific about it. When the host ( the host, NOT another player !) will open the door, this door will (always) contain a goat.

   That was my hope, that you had not placed that childish trap... although, judging from your behaviour since then, It seems that I should have "expected" this...
   Timeo Danaos et dona ferentes.
   
   Stand on your own feet, Dan Lehman, do not fear the expected outcome - because unperformed experiments have no results.

   Q.E.D.

   P.S. I would hope that you could spent a small percentege of this verbiage about any of the knots I publish in this Forum...but , frankly, I do not expect it any more. And we should never add or subtract hopes with expectations, as the Monty Hall problem has taught us.
Title: Re: The shortest-chain-of-the-World ( problem )
Post by: X1 on January 16, 2013, 07:55:10 AM
The childish trap with the two players,

   My statement that the variation of the original game DL pulled out of his sheaves ( to "trap back" whom he had imagined had tried to trap him at the first place), was a "two players" variation, was not accurate. In fact, it is a "three players" variation - the first player is not allowed to alter his choice, the second is allowed to alter his choice, but he always decides to not actualize this option ( because he decides it will not matter ), and the third is also allowed to alter his choice, and he does so, every time. 
   One need not "run out the probabilities" :), as DL did, in those long pages, to solve this problem !
   The first player chooses right from the beginning, and he retains his choice till the end, because he is not allowed to do anything else. There are 2 goats and one car behind those closed doors, and even a child knows that his chances are 1 in 3. The second player does exactly what the first did - although for a different reason ( he decides to do it, while the first player was obliged to do it ). However, the reason he does whatever he does, can not alter his chances !  :). So, his chances are the same as those of the first player, 1 in 3.
   So, the first player has 1 in 3 chances (=1/3), and the second has 1 in 3 chances (=1/3). What are the chances of the third player ?  :) One need not be a rocket scientist, riding a spacecraft ( this time... :) ), to conclude that the remaining player will have the remaining chances, and 1 - 1/3 - 1.3 = 1/3 ! So, in this childish variation of the original game, every player has the same chances, in all possible three cases, ( just as it should have been expected ) : 1) choosing and not be allowed to alter one s choice, 2) choosing, allowed to alter, but not deciding to alter one s choice, and, 3) choosing, allowed to alter, and deciding to alter one s choice.
   Why should we "run the probabilities', as DL, did, to see this ? I suppose that, if one does not understand something, he finds it "safer" to rely to some numbers, even if they are not needed. However, this use of "statistics" can be misleading, and hide under the rug the lack of deeper understanding. I have seen people in panic to count with their fingers, just like children do, just to be sure they will not be mistaken...
Title: Re: The shortest-chain-of-the-World ( problem )
Post by: Dan_Lehman on January 17, 2013, 05:39:18 PM
X, there is no connection of all your words about some
"two", now "three players" situation to anything I wrote:
I simply answered to your false assertion that in the case
where whichever of the two unchosen doors was revealed
was selected by chance --a coin toss, say; or as that cited
article puts it, a slip on a banana peel ("Monty Fall")--
there was still favor in switching : no, odds then are equal.


Then my presented running-out-numbers addressed
your continued warnings of dire consequences of using
two eyeknots in tandem to join ropes vs. a single knot,
that one must consider the weaker strength only.

That I did, using the assumptions coming with a bell
curve and standard deviations, enlarging the sample
count to a unit of 40 which allows 39/40 to model
the 98% of the population within 2sD of the Mean
plus the top end (strongest).  There, I showed that
with your 2-links situation, given this appropriately
proportioned population of 1kg/2kg/3kg links, one
would see the Mean only diminished from 2kg to
1.95kg --which is not so bad, and, esp. in light of
the large differences seen in the test data between
the eyeknots & end-2-end knots, supports J.P.'s
proposition.


--dl*
====
Title: Re: The shortest-chain-of-the-World ( problem )
Post by: X1 on January 18, 2013, 12:18:49 AM
 
there is no connection of all your words about some "two", now "three players" situation to anything I wrote:

   There is, but you just do not see it ( I can not be sure about why is this happening...).  However, apparently you bite hard into this problem, for whatever reason :)..., so I suggest you read the variation of the problem as I had described it in my last post, and see that it is the same variation you had presented by yourself, at your "contre-attaque"  :). I do not doubt that one can describe this variation in fewer words, in an ever more condensed form - but the essence will remain the same. Your variation is about two players ( two people that, in principle, can not know which door hides which object ), not one host and one player. I had transformed it in the "three players variation", because I believe that helps one understand it better/deeper, without the use of "counting" - by fingers or otherwise  :). It helps one understand that each player has the same chances than anybody else, so that those chances are 1 in 3, and, lastly, answer your specific question, about the chances of the second and the third player ( the player who does not change his choice, and the player who does ). And it does it without even having to "run the probabilities".

I simply answered to your false assertion that in the case where whichever of the two unchosen doors was revealed was selected by chance ... there was still favour in switching


   The false assertion was that I will fall into your trap - and it was proven false, because I had not, and I said it so, word by word :  Read my lips :

   The "example" DL got out of his sheaves was specific about it. When the host ( the host, NOT another player !) will open the door, this door will (always) contain a goat.

   THAT was what I had read out of your text - and that was what I expected you to do, because I expected your trap to lie elsewhere, where it would be more concealed ( and clever/cunning): in the use of the vague word "next".  I thought that you had tried to use a variation of the problem which would contain conditional probabilities, that would relate the initial choice of the player(s) with the conditions the host will place after he is informed about them. Conditional probabilities are always harder to evaluate. However, it turned out you had changed the original problem in a less complex / more childish way - that I had not expected. Perhaps you thought such a variation would be better for you, because a more obvious trap would be ignored... and you were right on this.

   I repeat, for the last time, and I hope that you will get it this time !  :)
   If there is one host, who knows what is behind which door, and one only player, who does not, the player should alter his original choice after the host opens a door that contains a goat. He (the player) can not know how this had happened, and he should not suppose that the host oppened this door by chance, of course - however, he should think that, even if that is what had happened, indeed, he would be no worse if he alters his choice, and he would be better if the host had opened one of the two doors he knew it contains a goat - so he should alter his initial choice. So, ONE player - a door which is opened by choice or chance and reveals a goat - this player should always alter his first choice.
   If there is one host and two players, and those two players have different/opposite strategies ( the first would choose at random, and would open the door he has chosen, and the other will also choose at random, but he has offered, and exercised, the option to alter his original choice after the first player has opened a door, and this door happens to contain a goat ), at the long run, the two players will win about the same number of times : 1 in 3.
   If there is one host, who does not even know which door hides a car and which hides the car, and three players , and the first player choses at random, and opens the door, then the second chooses at random, but he does not open the door he has chosen, and then the third player chooses at random, and he always alters his first choice if the first player has chosen a goat - at this variation of the original ptoblem, all players will have the same probability to win the car, 1 in 3.
   I believe my re-formulating of the variation you had posted into this three-player form, is more "balanced", and explains your two-player variation without the need to "run the probabilities", i.e., without the need of "counting".

[re. the ] 2-links situation, given this appropriately proportioned population of 1kg/2kg/3kg links

  Here we come again !  ( I should have expected that ! ) Your beloved method of twisting the given data as if they were ropes  :), is in action here, again. Would you, ever, be able to address the problem which is posed to you, without trying to figure out "loopholes" and "variations" ?  You did it in the Monty Hall problem, and you do it again here... , because, apparently, that is always your favourite escape route.
   WHO ON EARTH had spoken about this "appropriately proportioned population"  :) of the 1kg/2/kg/3kg/ links ? Is it something like this "abbreviated topological correspondence" of yours ? Read my lips :

 
we know ... that one third of them can withstand a pull of only 1 kilogram, one third of them can withstand a pull of only 2 kilograms, and the last third of them can withstand a pull of only 3 kilograms.
   So, the average weight one link can withstand is 2 kilograms, right ? [ "Yes"... ].

   Of course, another distribution with the same "average" strength of the links, would lead to another result. We do not know which is the expected or the actual distribution of a large sample of similar tests. I had supposed it would be a Gaussian ( normal, bell shaped ) distribution, but this was only a wild guess - so we can not depend on it ! It may well be a more "square" or a more" flattened" one, we just do not know. However, the whatever different result will not always be closer to the 18/9 than the 14/9 is... More weaker-than-average links will tend to "push" the value even lower - and the stronger-than-average links will not help in this : it is the weaker links that count and determine the strength of each compound chain/knot. That is what I had repeated over and over again - the "average" does not mean much - and we should not add and subtract "averages of strengths", like they were weights !  :)
   
 
To my thinking, this supports the proposition that the stronger means to joining lines is by interlinked (one way or another) eyeknots, in some cases, at least. 

   Reading again the looong previous posts, just to check again if I had missed something, I had noticed that "loophole", which was left as an insurance premium !  :) :"in some cases, at least." (sic). How convenient, this "some" and this "at least"... ( For the "one way or another", that hides the escape route of the "shared eye legs" "variation", I had already responded elsewhere (1)) .I could easily say the exact opposite, that "the stronger means to joining lines is by eye-to-eye knots, in some cases, at least" - but I will not, because :
 
 I fear no jamming knot
 or any loss of pride
 or spill by the unicorn
 of which I do not ride.


   Music from the movie " All that Jazz" ( 1979)
   Any other inguistic improvements and rhythm s ideas are welcomed... :)

   An advice : do not be so "sure" about J.P.s conjecture - or, if you have chosen to show persistence on your (wrong) first idea, prepare to figure out a plausible "loophole" or "variation", like you did in my 1-2-3 strength link problem, or in the Monty Hall problem. With your experience on knots, I hope you will find a much better escape route than the excuses you managed to pronounce on the simple, well stated problems of this thread.

 1)  http://igkt.net/sm/index.php?topic=4150.msg25955#msg25955   
Title: Knot tyer s song.
Post by: X1 on January 18, 2013, 06:01:32 PM
I fear no jamming knot
or lose the long-lost pride
or spill from the unicorn
on which I do not ride.