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.pdf**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 ...

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*

====