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:
... 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!
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.pdfMonty 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 ...
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 knotwhich 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*
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