Let's discuss Mathematics

[dutchfire]

Yes, but the 100 tosses don't necessarily reveal what E(X) is. If the coin was fair, it would be irrelevant, what the other one chooses.

I think PS's method is fair, but here's another one, which I hoped you'd come up with:

They do double tosses, HT means ParadigmShifter wins, TH means DutchFire wins, and HH and TT means they'll toss again. If the coin lands H with the probability p, the probabilities of HT and TH are the same, namely p(1-p).

I suppose ParadigmShifter's method is quicker, but what's remarkable with this method is that you can make any random thing a perfect coin! Let's say you have a wheel divided into three sections, A, B and C. They look like equally big, but they don't even have to be! Now you choose that Ax means win for other party and xA for the other. Here x is B or C, and they do double spins, xx means they'll spin again.

Notice, that this assumes only that there is a constant p, which is the probability for one of the options to turn out, so this could -at least in theory- be used to find out whether a coin really has a constant probability to end up H or T up. Or to be more precise that how probable it is that there is such probability. I have read that there have been empirical investigations into the coins and dice, and the result has been that the used ones were unlikely fair. I'd be happy to know whether this thing has been investigated empirically.

EDIT: Shoot off does the same trick too, though.

And 0.999.....=1-thing, there has been debates about it here, and the last one of them was so depressing that I didn't visit this place for months.

I used this method (it's called von Neumann normalization, or something), in a project of mine, building a physical random number generator. I believe I posted about it here...

 

[dutchfire]

Hmm, can't find it.

Another interesting one:
100 prisoners are locked in prison. In this prison, there are 100 identical rooms. In every room, there are 100 numbered boxes, each box containing the name of one prisoner in such a way that every prisoners name is in one box* in every room (the rooms are identical, so each name is in the same box in every room).
The prison director proposes the following scheme: "Each prisoner will be led into a separate room and will be allowed to open 50 boxes. If every prisoner opens the box containing his own name, all prisoners will be released."
Initially, the prisoners don't really like this scheme, each individual prisoners has a 50% chance of opening the correct box, so in total they should have a 1/2^100 chance of success. However, luckily, there is a mathematician in the room. He devises a clever scheme that greatly increases the chance of them winning (to more than 30%!) and tells all the prisoners.

Note (Big hint!): His scheme would also work if all other prisoners were robots/computers that could follow a simple algorithm. However, the scheme can't work if the clever mathematician is only allowed to give each prisoner a list of boxnumbers he should open.


*All prisoners have different names

 

[LulThyme]

I didn't come up by myself with the method I proposed, it was an exercise in a book to prove that it's fair and to calculate what's the expected number of tosses before the conclusion is arrived at, so I knew the answer before the question. It would have been too easy to ask you that way.

When you asked this question, I thought you already knew this, so I didn't mention it, to avoid spoiling things, but this trick actually has a name:
Neumann's coin trick
in reference to the first known reference (by von Neumann).

 

[ParadigmShifter]

Hey! I am as great as von Neumann then!

This device will make me famous!


Link to video.

Dutchfire's problem sounds similar to one I heard before about lightbulbs. I'll have a drunken think about it after listening to the Big Black song.

 

[ParadigmShifter]

Dutchfire, do the prisoners get to see the results of each box opening before opening another?

 

[IdiotsOpposite]

Idiotsopposite: What transcendetal numbers do you know? Can you reduce this question to knowledge of them? (Assuming that it is allowed, which I do assume, since otherwise this is quite hard).

EDIT: Oops, didn't notice the page had changed.

e and pi are the only transcendental numbers most people know about.

Is that a hint, do you know how to show it Atticus? Or just speculative?

EDIT: This looks like the correct theorem, which is an extension of one of the ones I posted links to earlier

http://en.wikipedia.org/wiki/Baker's_theorem

Well, I know that ln(2) and ln(3) are transcendental, and I believe that a integer multiple of a transcendental number is also transcendental. This means that if the natural log of all primes is transcendental, then the natural log of all composite numbers is transcendental, at least so I believe. (Because, for example, ln(4)=ln(2^2)=2ln(2) ) Although... if a and b are transcendental, is a+b also transcendental? I think that's rather necessary.

And hey, I believe that any good mathematician can prove anything he wants if he has the will (and it's correct.)

 

[Mise]

Dutchfire, do the prisoners get to see the results of each box opening before opening another?

I don't even see how that would help?!

 

[ParadigmShifter]

Me neither, but it helps to get all the info beforehand Especially when a bit tipsy.

 

[ParadigmShifter]

Well, I know that ln(2) and ln(3) are transcendental, and I believe that a integer multiple of a transcendental number is also transcendental. This means that if the natural log of all primes is transcendental, then the natural log of all composite numbers is transcendental, at least so I believe. (Because, for example, ln(4)=ln(2^2)=2ln(2) ) Although... if a and b are transcendental, is a+b also transcendental? I think that's rather necessary.

And hey, I believe that any good mathematician can prove anything he wants if he has the will (and it's correct.)

Yeah, x=1 is the only rational number such that log(x) is algebraic, but if you know that the proof is trivial

if a and b are transcendental, a+b might not be. b = -a, b = -a + k, where k is algebraic, etc.

EDIT: You are correct that any algebraic rational multiple of a transcendental number is also transcendental, however. That is easy to show.

EDIT2: Might be true for algebraic multiples too, but that would need a proof.

 

[dutchfire]

Dutchfire, do the prisoners get to see the results of each box opening before opening another?

Yes .

 

[IdiotsOpposite]

Damn. I was hoping that I could prove numbers like log(6) were transcendental by manner of them being log(2)+log(3).

 

[dutchfire]

And hey, I believe that any good mathematician can prove anything he wants if he has the will (and it's correct.)

Gödel disagrees

 

[Mise]

I'm assuming the solution would have them all open each box with an equal probability, rather than simply opening at random?

 

[ParadigmShifter]

Damn. I was hoping that I could prove numbers like log(6) were transcendental by manner of them being log(2)+log(3).

Might be true if the numbers aren't rational multiples of each other. Needs a proof though...

EDIT: But the proof is probably covered by one of the 3 theorems I posted earlier. I think one of those proves the result you are after though.

EDIT2: When I say log I mean log to the base e of course.

 

[IdiotsOpposite]

Might be true if the numbers aren't rational multiples of each other. Needs a proof though...

EDIT: But the proof is probably covered by one of the 3 theorems I posted earlier. I think one of those proves the result you are after though.

I think that for two transcendental numbers a and b, if b-a is not algebraic (or 0) then a+b is transcendental. But that would also need a proof.

 

[dutchfire]

I'm assuming the solution would have them all open each box with an equal probability, rather than simply opening at random?

We can number the prisoners and the boxes have already been numbered, so a configuration in this case is just a permutation of the numbers 1-100. What you should have is that the final chance doesn't depend on the numbering of the prisoners (or of the boxes).

 

[ParadigmShifter]

Let's simplify it for tipsy PS.

Minimum number of prisoners is 4 if looking at the contents of a box is to have any effect (and opening an integer number of boxes).

 

[dutchfire]

That's right.

 

[Mise]

The prisoners can't communicate with each other I guess?

 

[ParadigmShifter]

OK, so wlog (without loss of generality), prisoner 1 opens box 1.

If he doesn't see his name... hmm.

I suspect first move is for prisoner x to open box x, and for subsequent openings depend on something I can't imagine at the moment.

Good so far?