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.
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".
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 :
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”.
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 :
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 !
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.
