## Search found 602 matches

- Thu Sep 10, 2015 10:32 am UTC
- Forum: Logic Puzzles
- Topic: Anti-Gambler's Fallacy
- Replies:
**162** - Views:
**27948**

### Re: Anti-Gambler's Fallacy

People have said that we should be bayesian about this, and I agree. There is no earthly way that you could convince me to keep playing after 8000 consecutive losses, because nothing could grant me a prior certainty strong enough to fight that data. People have mentioned seeing the coin flipped Grah...

- Tue Sep 08, 2015 5:51 am UTC
- Forum: Logic Puzzles
- Topic: Anti-Gambler's Fallacy
- Replies:
**162** - Views:
**27948**

### Re: Anti-Gambler's Fallacy

If you quit this game then I presume you believe I have supernatural powers or something (I knew the future and knew which numbers you'd guess, so I pre-arranged the cards in a way that would ensure you'd lose), because, remember, you're not gambling your whole 200000 at once, just 100 from it. It ...

- Tue Sep 08, 2015 12:21 am UTC
- Forum: Logic Puzzles
- Topic: Anti-Gambler's Fallacy
- Replies:
**162** - Views:
**27948**

### Re: Anti-Gambler's Fallacy

Since I have entered a spooky realm where probabilities no longer mean anything, yeah, I'd quit. I'd quit by 100. I'd be weirded out by 10, incredulous by 20, and questioning the veracity of the rules by 30.

- Mon Aug 24, 2015 11:46 pm UTC
- Forum: Logic Puzzles
- Topic: Who is What?
- Replies:
**21** - Views:
**4293**

### Re: Who is What?

If someone refers to "the engineer", and then it turns out that there were two engineers, that's not an implicit false assumption. You have actively lied to me through grammar in that case.

- Mon Aug 24, 2015 11:43 pm UTC
- Forum: Logic Puzzles
- Topic: Rethinking Blue Eyes - A Logic Puzzle
- Replies:
**97** - Views:
**15642**

### Re: Rethinking Blue Eyes - A Logic Puzzle

You made a claim that A should know that B knows that C can see at least two people with blue eyes, due to the symmetry. Consider the following case: A has brown eyes, and B, C, and D each have blue eyes. 1) How many pairs of blue eyes can C see? 2) How many pairs of blue eyes does B know that C can...

- Mon Aug 24, 2015 10:05 am UTC
- Forum: Logic Puzzles
- Topic: Who is What?
- Replies:
**21** - Views:
**4293**

### Re: Who is What?

Use of the terms "the programmer" and "the engineer" later in the puzzle strongly (conclusively?) indicate that exactly one brother is a programmer, and exactly one brother is an engineer.

- Mon Aug 24, 2015 10:02 am UTC
- Forum: Logic Puzzles
- Topic: Rethinking Blue Eyes - A Logic Puzzle
- Replies:
**97** - Views:
**15642**

### Re: Rethinking Blue Eyes - A Logic Puzzle

Let's go with your four-person example. Alice, Bob, Charlie, and Diane are on the island. No one knows the color of their own eyes, but each sees that the other three have blue eyes. There's no guru to make any sort of announcement. Each one is thinking about what the others know or don't know. Alic...

- Thu Aug 20, 2015 4:05 am UTC
- Forum: Logic Puzzles
- Topic: Three Cards Trick
- Replies:
**15** - Views:
**3214**

### Re: Three Cards Trick

Where do you draw the line, then? Since the cards given are random, the Assistant has two arbitrary cards to work with and must be able to encode any of the other 50 cards somehow. As Cradarc said, there are 3! = 6 orderings of the cards, so somehow more information has to be encoded than in just th...

- Thu Aug 20, 2015 2:24 am UTC
- Forum: Logic Puzzles
- Topic: Three Cards Trick
- Replies:
**15** - Views:
**3214**

### Re: Three Cards Trick

It sounds like your solution doesn't even really rely on there being three cards. The assistant could just as easily put only the face-down card on the table, or report to the magician "I put a card face down on the table".

- Thu Aug 20, 2015 2:18 am UTC
- Forum: Logic Puzzles
- Topic: Who is What?
- Replies:
**21** - Views:
**4293**

### Re: Who is What?

I looked up Smullyan's solution, and it's... lacking.

It makes me not want to read his book, honestly.

It makes me not want to read his book, honestly.

- Thu Aug 20, 2015 1:26 am UTC
- Forum: Logic Puzzles
- Topic: Three Cards Trick
- Replies:
**15** - Views:
**3214**

### Re: Three Cards Trick

Can you give an example of the communication the Assistant gives to the Magician over the phone in your phone example? I feel like this is the crux of the matter, maybe.

- Sun Aug 16, 2015 8:00 pm UTC
- Forum: Mathematics
- Topic: Hard combinatorics
- Replies:
**24** - Views:
**3594**

### Re: Hard combinatorics

The algorithm seems simple to write. Turn the problem into one on graphs: the vertices are the quintuplets, and two are linked by an edge if they have exactly two elements in common. Then, you're hunting for the largest clique (https://en.wikipedia.org/wiki/Clique_%28graph_theory%29) in the graph, a...

- Tue Aug 11, 2015 10:50 pm UTC
- Forum: Mathematics
- Topic: Hard combinatorics
- Replies:
**24** - Views:
**3594**

### Re: Hard combinatorics

The original post said nothing whatsoever about probability, which is why I was confused. It talked about the distribution of m, which, when you're talking about random events, almost always refers to a probability distribution of some sort. Edit: The problem of finding the distribution given a ran...

- Tue Aug 11, 2015 3:40 am UTC
- Forum: Mathematics
- Topic: Hard combinatorics
- Replies:
**24** - Views:
**3594**

### Re: Hard combinatorics

I think the question is something along the lines of: Take k quintuplets at random (5000 <= k <= 10000), and extract from them a maximal subset of the desired form; call the size of this subset m_k. Potentially, m_k depends on which quintuplets were chosen; what is the probability distribution of m_k?

- Wed Aug 05, 2015 4:32 am UTC
- Forum: Mathematics
- Topic: Conjecture "2^n"
- Replies:
**10** - Views:
**2723**

### Re: Conjecture "2^n"

That's slick, though I'm rather partial to mine as a constructive, straightforward proof. I`m talking about Hagen Von Eitzen proof not Almahed one or Alaqqad. Very elegant and simple. Yeah, I know. His is non-constructive, that is, it asserts that there is some position at which the divisibility ha...

- Tue Aug 04, 2015 6:42 am UTC
- Forum: Mathematics
- Topic: Conjecture "2^n"
- Replies:
**10** - Views:
**2723**

### Re: Conjecture "2^n"

That's slick, though I'm rather partial to mine as a constructive, straightforward proof.

- Fri Jul 31, 2015 7:39 pm UTC
- Forum: Mathematics
- Topic: Conjecture "2^n"
- Replies:
**10** - Views:
**2723**

### Re: Conjecture "2^n"

This is true. S_n(k) = 2^k/3 (2^{n-k+1} - (-1)^{n-k+1}), since the sum performed in calculating S_n(k) is of a geometric series. Write n = 2^w*3^h*d, where 2^w is the largest power of 2 dividing n, 3^h is the largest power of 3 dividing n, and d is odd and not a multiple of 3. Letting phi(d) be the ...

- Tue Jul 21, 2015 10:59 pm UTC
- Forum: Logic Puzzles
- Topic: Rethinking Blue Eyes - A Logic Puzzle
- Replies:
**97** - Views:
**15642**

### Re: Rethinking Blue Eyes - A Logic Puzzle

Quintopia, while the islanders leave early in your scenario, and indeed correctly claim they have blue eyes, they do not know they have blue eyes, since knowledge is not justified true belief. As such, the islanders have violated the terms of the island (even though they don't know it). I assume th...

- Mon Jul 20, 2015 6:10 pm UTC
- Forum: Logic Puzzles
- Topic: Rethinking Blue Eyes - A Logic Puzzle
- Replies:
**97** - Views:
**15642**

### Re: Rethinking Blue Eyes - A Logic Puzzle

Quintopia, while the islanders leave early in your scenario, and indeed correctly claim they have blue eyes, they do not know they have blue eyes, since knowledge is not justified true belief. As such, the islanders have violated the terms of the island (even though they don't know it). I assume thi...

- Fri Jul 17, 2015 11:51 am UTC
- Forum: Logic Puzzles
- Topic: Rethinking the Prisoner's Dilemma
- Replies:
**31** - Views:
**5926**

### Re: Rethinking the Prisoner's Dilemma

Offtopic: Superrational agents can't leave early, because leaving isn't volitional. You leave exactly when you determine the color of your eyes, and not a moment before or after. Since everything is deterministic, agents can't change their behaviors, so there's no way to coordinate a...

- Mon Jul 13, 2015 10:07 am UTC
- Forum: Logic Puzzles
- Topic: Rethinking the Prisoner's Dilemma
- Replies:
**31** - Views:
**5926**

### Re: Rethinking the Prisoner's Dilemma

Superrationality is a specific type of irrationality that may occasionally lead to better outcomes than similar games with all rational players, in the case that every player is superrational. Basically, superrationality is deductively begging the question. The argument for why cooperate is the logi...

- Sun Jul 12, 2015 8:19 pm UTC
- Forum: Mathematics
- Topic: Help with a lemma about rationals and coprime integers
- Replies:
**4** - Views:
**2023**

### Re: Help with a lemma about rationals and coprime integers

Multiplying by -1 doesn't actually produce another example, because they're not coprime - they all share the divisor -1! Of course, maybe you exclude the negative unit from the definition as well. Typically, one excludes all units. The positive examples also have the divisor -1, since 5 = (-5)(-1),...

- Thu Jul 09, 2015 5:20 am UTC
- Forum: Mathematics
- Topic: Help with a lemma about rationals and coprime integers
- Replies:
**4** - Views:
**2023**

### Re: Help with a lemma about rationals and coprime integers

Well, you can multiply your one example by -1 to get a second example, but that'll be it. I'll show that there's only among the positive rational multiples, by showing that two such rational multiples are necessary equal. Let (a,b,c,d) and (e,f,g,h) be two rational multiples of (w,x,y,z) that have a...

- Wed Jul 08, 2015 7:55 pm UTC
- Forum: Mathematics
- Topic: Question about calculation of the limit Indeterminate
- Replies:
**3** - Views:
**1748**

### Re: Question about calculation of the limit Indeterminate

He's actually doing x^sin(x), but none of his binary operators rendered correctly in wherever he copied this from, I'm guessing.

- Mon Jun 22, 2015 10:53 pm UTC
- Forum: Mathematics
- Topic: Leading numbers
- Replies:
**8** - Views:
**2571**

### Re: Leading numbers

You're exactly using the inductive algorithm for proving CRT (the one that goes "first solve one congruence, then solve that one plus another simultaneously, then those two plus another simultaneously, etc.") when you construct your sequences. It's not so much that you're invoking CRT as t...

- Sun Jun 21, 2015 10:51 pm UTC
- Forum: Mathematics
- Topic: Leading numbers
- Replies:
**8** - Views:
**2571**

### Re: Leading numbers

...But that's the Chinese Remainder Theorem.

- Sat Jun 20, 2015 9:01 am UTC
- Forum: Logic Puzzles
- Topic: Hat puzzle - 6 (or n) randomly coloured hats in a circle.
- Replies:
**14** - Views:
**9681**

### Re: Hat puzzle - 6 (or n) randomly coloured hats in a circle

First, the standard solution in a nutshell, so that everyone is on the same page as me for my modified solution. The players number the colors from 0 to n-1 and also number the players from 0 to n-1. When each player is up to guess their hat color, they guess as though the sum of all the hats is con...

- Thu Jun 18, 2015 4:47 pm UTC
- Forum: Mathematics
- Topic: Why is 0! = 1?
- Replies:
**6** - Views:
**2574**

### Re: Why is 0! = 1?

In the same way that an empty sum (that is, a sum of no things) should be the additive identity, 0, an empty product (that is, a product of no things) should be the multiplicative identity, 1.

- Wed Jun 10, 2015 4:30 pm UTC
- Forum: Logic Puzzles
- Topic: What does this say?
- Replies:
**4** - Views:
**2480**

### Re: What does this say?

Initial analysis: It almost certainly has something to do with the positions of keys on a keyboard, but my first few guesses as to the nature of the relationship haven't yielded anything. Actually, I'm certain it's not a substitution cipher, so maybe my intuition about the keyboard is m...

- Thu Jun 04, 2015 4:59 pm UTC
- Forum: Logic Puzzles
- Topic: My write-up of the "Blue Eyes" solution (SPOILER A
- Replies:
**1368** - Views:
**424461**

### Re: My write-up of the "Blue Eyes" solution (SPOILER A

In saying it can be safely discarded, you're presuming knowledge that someone, somewhere doesn't have. As a blue-eyed person on the island, I know there are more than 96 blue-eyed people, sure. Bob and Carol and David and Erica know it too, though I can't immediately figure that out without modeling...

- Mon May 11, 2015 7:21 pm UTC
- Forum: Mathematics
- Topic: Can anyone solve this probability problem?
- Replies:
**15** - Views:
**2970**

### Re: Can anyone solve this probability problem?

Yakk wrote:Note that the largest die size possible is 6^(6*5/2) = 470184984576. Which is 300 times larger than the average. So at least the average is plausible.

What? The largest possible die is 6^(2^5). On a best roll you square the remaining dice, and this could happen 5 times at best.

- Thu May 07, 2015 4:41 am UTC
- Forum: Mathematics
- Topic: Greatest Mathematician of all time?
- Replies:
**65** - Views:
**20281**

### Re: Greatest Mathematician of all time?

I'm always a fan of Cauchy. (obviously)

Alec Baldwin is the single greatest actor of all time. Source: Team America: World Police.

Alec Baldwin is the single greatest actor of all time. Source: Team America: World Police.

- Mon May 04, 2015 2:11 am UTC
- Forum: Mathematics
- Topic: Can anyone solve this probability problem?
- Replies:
**15** - Views:
**2970**

### Re: Can anyone solve this probability problem?

If we roll a single die with a_1 sides, our expected output is E_1(a_1) = (1+a_1)/2. If we roll two dice, each (initially) with a_2 sides, then our first roll is equally likely to be any number from 1 to a_2. This modifies the second die, to be either a_2, 2 a_2, ..., or a_2^2. The expected values o...

- Mon May 04, 2015 2:10 am UTC
- Forum: Mathematics
- Topic: Equations Like f(2x)=4f(x)
- Replies:
**8** - Views:
**5618**

### Re: Equations Like f(2x)=4f(x)

Putnam 2010 B-5 asks the question: Is there a strictly increasing function f : R -> R such that f ' (x) = f(f(x)) for all x?

- Mon Apr 06, 2015 5:54 am UTC
- Forum: Mathematics
- Topic: Prove by induction
- Replies:
**4** - Views:
**1624**

### Re: Prove by induction

Like with many induction proofs, this problem can become simpler if you attempt to show a stronger statement. Specifically,

**Spoiler:**

- Tue Mar 31, 2015 6:09 pm UTC
- Forum: Individual XKCD Comic Threads
- Topic: 1505: "Ontological Argument"
- Replies:
**278** - Views:
**44471**

### Re: 1505: "Ontological Argument"

Even if we allow "exists" as a property, the argument still runs into problems, because it posits the existence of God in order to prove God exists. It's very subtle in how it does this, so I can imagine that people who come up with the argument really believe it. In short, the argument gi...

- Sun Mar 29, 2015 10:55 am UTC
- Forum: Mathematics
- Topic: Paths between two points
- Replies:
**18** - Views:
**3502**

### Re: Paths between two points

A PDF is a probability distribution function. What is the probability distribution that's being modeled? "Every path is equally likely" sounds great when you don't think too hard about it, but it's a surprisingly hard thing to formalize. For instance, have you ever thought about trying to ...

- Fri Feb 27, 2015 8:26 am UTC
- Forum: Individual XKCD Comic Threads
- Topic: 1492: "Dress color"
- Replies:
**362** - Views:
**62396**

### Re: 1492: "Dress color"

The proper plural form [is] math. Whoa there. "Math" isn't a plural noun. You don't have one mathematic, two mathematics. "Mathematics" is a mass noun, and so is "math". "Maths" is some weird contrivance that shouldn't exist. On topic, when I looked at the &q...

- Sun Feb 15, 2015 10:28 pm UTC
- Forum: Logic Puzzles
- Topic: Three princesses
- Replies:
**446** - Views:
**206966**

### Re: Three princesses

Oh, I'm well aware of the difficulties (I even stated one of them in my post) and I already read solutions for the strong random assumption, it just bugs me that it doesn't fit the way the puzzle is worded. What? The accepted solution works no matter how the middle princess behaves . The fact that ...

- Thu Feb 12, 2015 10:39 pm UTC
- Forum: Mathematics
- Topic: Can anyone understand why e shows up here?
- Replies:
**9** - Views:
**2552**

### Re: Can anyone understand why e shows up here?

If you use the Principle of Inclusion-Exclusion to count how many arrangements of n objects fail the n conditions "The first object is in the first position", "The second object is in the second position", etc., and then factor out an n!, you'll find that the number is exactly n!...