Author Topic: The shortest-chain-of-the-World ( problem )  (Read 6098 times)

X1

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The shortest-chain-of-the-World ( problem )
« 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.)

Dan_Lehman

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Re: The shortest-chain-of-the-World ( problem )
« Reply #1 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*
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X1

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Re: The shortest-chain-of-the-World ( problem )
« Reply #2 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...   
« Last Edit: January 12, 2013, 11:21:29 PM by X1 »

X1

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Re: The shortest-chain-of-the-World ( problem )
« Reply #3 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.
« Last Edit: January 13, 2013, 12:34:42 AM by X1 »

Dan_Lehman

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Re: The shortest-chain-of-the-World ( problem )
« Reply #4 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*
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X1

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Re: The shortest-chain-of-the-World ( problem )
« Reply #5 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.
« Last Edit: January 13, 2013, 01:31:29 AM by X1 »

X1

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Re: The shortest-chain-of-the-World ( problem )
« Reply #6 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 ?  :)
« Last Edit: January 13, 2013, 01:42:01 AM by X1 »

X1

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Re: The shortest-chain-of-the-World ( problem )
« Reply #7 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."
« Last Edit: January 13, 2013, 02:58:02 AM by X1 »

Dan_Lehman

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Re: The shortest-chain-of-the-World ( problem )
« Reply #8 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*
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X1

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Re: The shortest-chain-of-the-World ( problem )
« Reply #9 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.

« Last Edit: January 13, 2013, 03:56:19 AM by X1 »

Dan_Lehman

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Re: The shortest-chain-of-the-World ( problem )
« Reply #10 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*
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Knotman

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Re: The shortest-chain-of-the-World ( problem )
« Reply #11 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

X1

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Re: The shortest-chain-of-the-World ( problem )
« Reply #12 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...  :)
« Last Edit: January 13, 2013, 04:40:28 PM by X1 »

X1

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Re: The shortest-chain-of-the-World ( problem )
« Reply #13 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 !  :) )
« Last Edit: January 13, 2013, 04:43:07 PM by X1 »

Dan_Lehman

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Re: The shortest-chain-of-the-World ( problem )
« Reply #14 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*
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