Register | Sign In


Understanding through Discussion


EvC Forum active members: 65 (9162 total)
2 online now:
Newest Member: popoi
Post Volume: Total: 915,819 Year: 3,076/9,624 Month: 921/1,588 Week: 104/223 Day: 2/13 Hour: 0/1


Thread  Details

Email This Thread
Newer Topic | Older Topic
  
Author Topic:   A Deep Thought, Monty Hall, and Trisecting Angles
Otto Tellick
Member (Idle past 2331 days)
Posts: 288
From: PA, USA
Joined: 02-17-2008


Message 16 of 18 (600985)
01-18-2011 1:52 AM
Reply to: Message 3 by Modulous
12-21-2006 10:46 AM


Re: A better way to understand Monty...
Modulous writes:
I find considering it in terms of ten boxes/cups/doors/whatever You pick one and Monty says you either get to open that door, or the other nine. Which do you choose?
This (finally) gives me a better sense of the statistical reality underlying the Monty Hall problem. But still, when stated so as to retain the relevant properties of the original scenario, it leaves one with a sense of unease:
A more complete statement of Modulous's variation: There are (say) 10 doors, you pick one, Monty (who knows where the prize is) opens one of the other 9 (no prize) and says you can stick with your one, or instead switch to picking all 8 remaining doors. That's a no brainer.
A scenario more like what Monty actually does: You pick one door out of 10, Monty opens 8 other doors (no prize) and says you can now stick with your one, or switch to the last remaining door. That still feels, intuitively, like a simple 50/50 coin-toss type of choice, because when faced with the dichotomy, one easily forgets the relatively low probability of the initial choice being "right", and treats this new choice as "independent" of the first one (when in fact it isn't).
I recently saw a nice YouTube video by QualiaSoup called Flawed Thinking by Numbers, which also explained the problem reasonably well, and even cited a study of responses to the problem, showing that 88% of people faced with the 3-door scenario stuck with their initial choice (did not change after one of the other choices was eliminated).
Curiously, in citing the study, QualiaSoup did not mention how many times the initial choice was actually wrong. Perhaps the particular quantity of wrong first choices in that study should be considered irrelevant, but it would have been nice to at least see something like the expected discrepancy between people's misguided decisions and the actual outcomes...
(Plus, it's still a matter of probability, and knowing you had better odds when switching is hardly a comfort for those cases when the initial choice happened to be right.)

autotelic adj. (of an entity or event) having within itself the purpose of its existence or happening.

This message is a reply to:
 Message 3 by Modulous, posted 12-21-2006 10:46 AM Modulous has not replied

  
Noetherian Atheist
Junior Member (Idle past 4553 days)
Posts: 7
From: London
Joined: 08-19-2010


Message 17 of 18 (601597)
01-21-2011 8:24 PM
Reply to: Message 7 by Chiroptera
12-21-2006 4:26 PM


Hi Chiroptera,
The polynomial is of degree 3, which is not a power of 2.
Uh-huh, with you so far.
If an irrational number is not a root of a irreducible polynomial of degree a power of 2...
Very interesting, but what's that got to do with the case at hand? All you know is it's a root of an irreducible polynomial of degree 3. A bit missing here I think.
I got the impression that Percy's incredulity at the impossibility of trisecting an angle (albeit he misunderstaood the proof) stemmed from it somehow being proven that there couldn't be "some other way of doing it" (whether taking a cube root or trisecting an angle). How could you prove there's no other way?
Obviously this has been proven, as you show. However, step 10 is one hell of a leap! So I'm not sure it will help Percy much unless we could show how constructible lengths relate to polynomials (of any kind). I'm not saying I can do this - it's been > 20 years.
Lastly, I think you may have over complicated things. The point is that a number which is the root of a polynomial of degree 2 or less is constructible. No need to mention irrationals or irreducible polynomials (well, not at step 10 at least).
If I knew how to do them, there'd be smiley faces above - not trying (but possibly succeeding) to be a pain in the arse with criticisms, simply hoped to clarify.
It's been too long since I've had algebra.
Such a shame, you're missing out on all the fun. What do they make you do these days?

This message is a reply to:
 Message 7 by Chiroptera, posted 12-21-2006 4:26 PM Chiroptera has not replied

Replies to this message:
 Message 18 by nwr, posted 01-21-2011 9:03 PM Noetherian Atheist has not replied

  
nwr
Member
Posts: 6408
From: Geneva, Illinois
Joined: 08-08-2005
Member Rating: 5.1


Message 18 of 18 (601600)
01-21-2011 9:03 PM
Reply to: Message 17 by Noetherian Atheist
01-21-2011 8:24 PM


Noetherian Atheist writes:
All you know is it's a root of an irreducible polynomial of degree 3.
My Galois theory is a bit rusty. As I recall, given an algebraic number, there is a minimal polynomial (a polynomial of least degree) that it satisfies (that part is trivial), and any polynomial that it satisfies is a multiple of the minimal polynomial. So it cannot be a root of an irreducible polynomial of degree a power of 2 (else that polynomial would not be irreducible).
Hmm, wouldn't a Noetherian (atheist or not) know that?

Jesus was a liberal hippie

This message is a reply to:
 Message 17 by Noetherian Atheist, posted 01-21-2011 8:24 PM Noetherian Atheist has not replied

  
Newer Topic | Older Topic
Jump to:


Copyright 2001-2023 by EvC Forum, All Rights Reserved

™ Version 4.2
Innovative software from Qwixotic © 2024