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Questions With Answers (cf. the "Universe Threads")

The difficulty here seems to be finding the proper language with which to analyze the problem. Ie, what questions do we ask, and how do we answer them.

With Monty Hall, it's pretty easy. The question is "Should we switch doors?", and the answer is obtained by looking at how switchers vs non switchers fare over repeated trials.

In this Sleeping Beauty problem, I can't seem to find the right question. My best guess so far is, "Should Sleeping Beauty guess heads?".

To answer this, I bought a huge sack of roofies, gathered up several homeless women, stole a coin from one of them, and did some experimenting. 32 of these women were instructed to say "Heads" upon awakening, and the remaining 29 were instructed to guess tails. Since both groups faired roughly equally well (yes, the women who correctly answered tails were correct twice , unlike women who correctly guessed heads, but...w/e, this doesn't seem relavant regarding the bold question), my inclination is to guess A.

So, to paraphrase this, if this was one of those scenerios where Sleeping Beauty would be executed if she guessed incorrectly, and if I was one of those conventional logic puzzle "wise men", then I would see no reason to tell her, going into this, to favor heads over tails, or tails over heads. ATM, to me at least, they seem equally likely, thus making A my answer.


However...when I first saw the Monty Hall problem, my instinct was 50/50 as well. It wasn't until I saw a worked out solution that I realized what was going on.



So, I'm probably wrong here...which makes my NEW answer B.
 
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^ I believe the question to ask here is how much subjective capacity does Beauty have?

Beauty will not have a memory of herself and the environment (drug is taken) when heads lands, but she will have a memory of her environment and herself when tails lands (no drug taken). As weeks go by, the value of the coin being heads will approach 1/3 because she will be able to induce it by her own memory.

I have chosen 1/3 by deduction from the answers given. If the answer where to be 1/2, then this would mean Beauty lives in a vacuum.

The actual calculation is not as obvious to me at the moment...


EDIT: So my answer is also B
 
I'm sure 1/3 is the answer the puzzle maker was looking for. After all, this is more or less the same thing as Monty Hall.


But still, it don't quite sit right with me. Something's missing.
 
w01f: It seems to be the case that you're not completely understanding the hypothetical framework here. In this hypothetical, all physical aids are to be removed, and as memory is a physical aid, it is to be removed as well. It is posible that there are always shortcomings in problems that allow for little tricks like that, but here we're looking past that to get to the actual problem of the "paradoxical answers."

BA: Nobody is satisfied with Monty Hall problem until Bayes' Theorem is used. So I think the bolded question should be "Does there exist an analgoous theorem or sequence of theorems which can show via posterior probabilities that 1/3 is correct?" Is it the same type of problem as Monty Hall, or is it one where the first instinct IS correct!?!? Hmmm....

There is a competing consideration, which speaks in favor of the view of (B). Suppose that whenever she is woken Beauty is offered even odds on the coin having fallen Heads. Then, if immediately on being woken Beauty has a policy of betting on Heads, she is more likely to lose than to win.

Consider a similar case where the chance of multiple wakings is much higher. Suppose that Sleeping Ugly is treated like Sleeping Beauty except that if the coin lands Tails then she is woken on 999 rather than just two successive days. Then, arguing as (B), on waking on Monday not knowing which of the 999 days it is, her probability for Heads should be 1/1000. Can this be right????
 
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The answers given as (A) and (B) are elaborated as "initial reactions," and even though one is correct and one is wrong, the correct answer can (and should) instead be explained through rigorous proof. Meaning that yes, the initial reaction is "correct," but a valid argument exists to back it up.

Below is just which answer is correct. Choose to look or not, but all discussion of that should also exist in NSFW tags in future posts, indefinitely.

NSFW:
The answer is (A), or 1/2.

That's what so great about this problem. It comes off of the tongue so much like Monty Hall, but instead you prove that it IS your intuition that is correct. Understanding how these two are different in structure allows one to understand why they are different in answer. And makes you a heck of a lot better at math once you truly understand it all.


There. Now I want a rigorous proof as to why.
 
Gut instincts on 1/2 involve analogous situations like this:

Pile A is a fully shuffled stack of 1 million cards, numbered from 1 to 1 million.

Pile B is a single card, numbered zero.


Some goofball is going to pick a card from one of the two Piles, and if you guess the card's number, the goofball gonna give you a million bucks (and the goofball picks the Pile based on
the toss of a coin).


Does any zero-like number pop out at you?

_____________________________________________________________

So, the question is exactly how analagous is this?

Suppose Pile A had two cards, 1 & 2, and suppose Pile B was just the zero card. Is the situation identical to the situation faced by the sleeping woman?
 
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Can you get 88 using only the integers 1, 2, and 3, using each of them only once, and using the standard arithmetic operations, including factorials and repeating decimals? To be more precise, use only 1, 2, and 3, each used once, and use any of the following functions in whatever order and however many times you wish:

plus a+b
times a*b
subtract a-b
negate -a
divide a/b
factorial a!
exponentiation a^b
roots a^(1/b)

and the following three operations, which work on raw digits only, not on more complicated expressions:

concatenation of digits, as in 23 from 2 and 3
the decimal point (as in .2)
the repeating decimal point (as in .2^. which means 0.2222222...)

and of course as many parentheses as you wish.


To get you started thinking about how things can be used here is the example to get to 87:

87 = (.1^.)^(-2)+3!
 
^Not sure if this is allowed by the rules, but it's the best I've got so far:


NSFW:
answer: (.1222...) x [(3!)!]




Edit: Oh snap! I just got the joke. Your puzzler was post #88 ITT.


Well played, my good sir!
 
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Binge Artist said:
^Not sure if this is allowed by the rules, but it's the best I've got so far:


NSFW:
answer: (.1222...) x [(3!)!]




Edit: Oh snap! I just got the joke. Your puzzler was post #88 ITT.


Well played, my good sir!
Click to expand...

!!!????

!!!

Wow. That is a complete coincidence! I got the puzzle over E-mail from one of my actuary friends months ago. I'll have to tell her, I think she'll get a kick out of that.
 
Binge Artist said:
^Not sure if this is allowed by the rules, but it's the best I've got so far:


NSFW:
answer: (.1222...) x [(3!)!]




Edit: Oh snap! I just got the joke. Your puzzler was post #88 ITT.


Well played, my good sir!
Click to expand...

Very interesting answer, but not allowed within the rules. You cannot use a compound integer and a repeater in the same unit. So, 12 is allowed, as is 23 or 13. And .1 repeating is allowed, as is .2 or .3 repeating. But not .12 repeating.
 
Dead_Flowers said:
Pick two points at random on a line segment. What is the probability that the three segments created by the points will form a triangle?
Click to expand...



NSFW:
My best guess ATM is 1/4.

Let the points be A & B, and let the line segment be of length 1. We distinguish 2 cases: A < 1/2, and A > 1/2.

If A < 1/2, we must have 1/2 < B < 1/2 + A.

If A > 1/2, we must have A - 1/2 < B < 1/2.

Graphing these inequalities on a 1 x 1 plane yields a region of area 1/4. Hence, my guess.
 
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Binge Artist said:
NSFW:
My best guess ATM is 1/4.

Let the points be A & B, and let the line segment be of length 1. We distinguish 2 cases: A < 1/2, and A > 1/2.

If A < 1/2, we must have 1/2 < B < 1/2 + A.

If A > 1/2, we must have A - 1/2 < B < 1/2.

Graphing these inequalities on a 1 x 1 plane yields a region of area 1/4. Hence, my guess.
Click to expand...

Precisely.

NSFW:
There is no restriction on the first point A, the second point B must be on the opposite half of the segment (P=1/2), and l B-A l < 1/2 (P=1/2). Multiply the probabilties of each condition and we get 1/4!
 
Binge Artist said:
Ah yes, probabilities of 4. You must be an actuary yourself!


BTW, modifying my last answer to your 88 puzzle yields this:

NSFW:
(.222... - .1) x [(3!)!] = 88
Click to expand...

Impressive. I swear I tried everything but I gave up after a while and not coming up with it.
 
Anyway, new Puzzler.


21,200 is a five digit number with the property that the first digit states how many zeros are in the number, the second digit states how many ones are in the number, the third digit states how many twos are in the number, the fourth digit states how many threes are in the number, and the fifth digit states how many fours are in the number.

Does a similar ten digit number exist? (the tenth digit is how many nines are in the number).
 
binge artist said:
anyway, new puzzler.


21,200 is a five digit number with the property that the first digit states how many zeros are in the number, the second digit states how many ones are in the number, the third digit states how many twos are in the number, the fourth digit states how many threes are in the number, and the fifth digit states how many fours are in the number.

Does a similar ten digit number exist? (the tenth digit is how many nines are in the number).
Click to expand...

NSFW:
6,210,001,000
 
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