CrystalMath
CrystalMath
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January 14th, 2014 at 8:34:31 PM permalink
Quote: kubikulann



A group of size N has been shown by MathExtremist and the Wizard to have a probability of (1 - 1/2N-1) of not failing. The probability of global win is the product of all (infinite) group probabilities. (No communication means no correlations needed to be computed.)
This product must be higher than 90%, or its logarithm higher than ln 0.90 = -0.1053.
ln(1-x) can be approximated by -x; it is necessarily higher, anyway. So we are looking for NK's such that the SUM of ln(1-1/2NK-1) >
SUM(-1/2NK-1) > -0.1053.
It is known that the geometric series yields a + a² + a³ + ... = a/(1-a) if a²<1. With a<=0.0953178 you get a/(1-a) <= 0.1053, so you define 1/2N1-1 <= 0.0953178, or N1-1 >= 3.39111. The first group must be at least of size N1=4.39111 and MathExtremist was right to choose 5. The next groups need not be NK=N1K . It is sufficient to define NK=(N1-1) K + 1 . Or even NK>=3.39111 K + 1.


I was thinking along these lines, too, but doesn't this fail the requirements, because you will have too high a chance to not make any guesses. The problem stated that you must have a 90% chance of having an infinite number of correct guesses, but given a starting group of 5, the chance of not making a guess is near 62.5%. At best, this gives you a 37.5% chance of having an infinite number of correct guesses.


Quote: kubikulann


Note: not every francophone is French, exactly like Canadians are not United-Staters. Is Blanchard not a French name?


Sounds French, but if he had more French in him, he would have said "deja vu" instead of "vous" here.
I heart Crystal Math.
kubikulann
kubikulann
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January 15th, 2014 at 6:41:19 PM permalink
You're right, there is a problem.
The number of correct guesses is finite if there exists a M above which all groups abstain. The probability of this must be low.

Probability of abstention in a group of size N = p(N) = 1 - (1+N)/2N-1
Probability of abstention in ALL groups of size greater or equal to M = q(M) = PRODUCT{N=M to inf} p(N)
Consequence: as q(M+1) >= q(M) and it must be low, we will be interested in the largest M, i.e. M tending to infinity.

- ln q(M) = - SUM ln p(N) = (approximately) SUM ((1+N)/2N-1)
In our proposed succession of NK = Kn* :
=[1/2N-1]M { (1+Mn*)(1/2N-1) + (1+(M+1)n*)(1/2N-1)² + (1+(M+2)n*)(1/2N-1)³ + (1+(M+3)n*)(1/2N-1)4 + ...}

=[1/2N-1]M { (1+Mn*)(1/2N-1) - (1+Mn*-n*)(1/2N-1)² } / { (1 - 1/2N-1)² }

=[1/2N-1]M { 1+Mn* - (1+Mn*-n*) 1/2N-1 }(1/2N-1) / { (1 - 1/2N-1)² }

This tends to zero as M tends to infinity.
So, q(M) tends to 1. There is a quasi-certainty that the number of correct guesses will be finite.


Next step: find another group size progression?
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24Bingo
24Bingo
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January 16th, 2014 at 12:26:32 PM permalink
Your more fundamental problem is you've got the probability of failing, not that of not failing. The probability of a group of wizards not failing is the probability that every single individual guesses right, i.e., 1/2^n. For any nontrivial group this will be less than 90%, so you're finished before you start.

No matter which wizards answer, an infinite number of wizards will have an infinitesimal probability of all being right, unless they have some information about their own hat. I'm not sure what information we're supposed to think they have, if not by communication. Even if they could see every single wizard, the landscape seen by a white-hat wizard and a black-hat wizard would be indistinguishable, since removing a finite number from an infinite population won't change the relative populations. I'm wondering if this is the same kind of voodoo pseudo-math by which allegedly a child becomes 16% less likely to be a boy based on not knowing whether they're older or younger than their brother (i.e., each wizard looks at the surrounding wizards, and gambler's-fallacies away).

(No spoiler markup because if I'm wrong, it doesn't matter, and if I'm right, this problem doesn't merit it.)
The trick to poker is learning not to beat yourself up for your mistakes too much, and certainly not too little, but just the right amount.
kubikulann
kubikulann
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January 16th, 2014 at 2:49:11 PM permalink
Quote: 24Bingo

same kind of voodoo pseudo-math by which allegedly a child becomes 16% less likely to be a boy based on not knowing whether they're older or younger than their brother

Pfff! I'm tired of people who, not understanding a result and being explained several times, continue to brag about how their little personal intuition still must be correct against all reasoning and it must henceforth be science that is wrong! No offence intended,but please take a 1O1 course in probabilities before talking about "voodoo pseudo-math". When all scientists say A and you think not-A, isn't it time to accept you are making the mistake?
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kubikulann
kubikulann
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January 16th, 2014 at 2:55:51 PM permalink
Next time they'll talk about "voodoo pseudo-physics" that says time is relative and "voodoo pseudo-biology" that says humans are evolved from primates.

Ignorance is bliss, they say. Well, for the ignorant maybe, but not for the informed to be endured from the ignorants.
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Twirdman
Twirdman
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January 16th, 2014 at 2:57:22 PM permalink
Quote: kubikulann

Pfff! I'm tired of people who, not understanding a result and being explained several times, continue to brag about how their little personal intuition still must be correct against all reasoning and it must henceforth be science that is wrong! No offence intended,but please take a 1O1 course in probabilities before talking about "voodoo pseudo-math". When all scientists say A and you think not-A, isn't it time to accept you are making the mistake?



I have to agree with you there. It is one thing to be ignorant, which is already bad enough, it is quite another to be proud of your ignorance and say well I must be right because thats how I think it should work. Math can be incredibly difficult and even some really simple examples of elementary probability are incredibly counter intuitive that does not mean they are wrong. Intuition is nothing against the power of mathematical proof.
kubikulann
kubikulann
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January 16th, 2014 at 2:58:38 PM permalink
Quote: 24Bingo

Your more fundamental problem is you've got the probability of failing, not that of not failing. The probability of a group of wizards not failing is the probability that every single individual guesses right, i.e., 1/2^n.

Nope.
What is required is just that one individual per group is right (and the others abstain), provided there are infinitely many such groups.
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Twirdman
Twirdman
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January 16th, 2014 at 3:04:16 PM permalink
Quote: kubikulann

Nope.
What is required is just that one individual per group is right (and the others abstain), provided there are infinitely many such groups.



Technically don't even need to go that far. Don't you simply need each group to have a non zero probability of having one answer. Its perfectly acceptable in project to have an infinite number of the groups fail to answer as long as an infinite number of groups do answer.
kubikulann
kubikulann
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January 16th, 2014 at 3:08:20 PM permalink
Quote: Twirdman

I have to agree with you there. It is one thing to be ignorant, which is already bad enough, it is quite another to be proud of your ignorance .

I wonder how many are truly proud of their ignorance, and how many are just playing up a countenance because they hate to recognize their ignorance.

I have colleagues (historians, language & literature, philosophy) who, sadly, boast about knowing nothing of math. I am terrified when this attitude is present in so-called educated people (University professors, please!).
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kubikulann
kubikulann
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January 16th, 2014 at 3:12:42 PM permalink
Quote: Twirdman

Technically don't even need to go that far. Don't you simply need each group to have a non zero probability of having one answer. Its perfectly acceptable in project to have an infinite number of the groups fail to answer as long as an infinite number of groups do answer.

You are right. That's what I meant when writing "infinitely many such groups". I didn't ask "all groups". But you are right in pointing that the groups where there is not one right-guessing member must be groups where everyone abstains.
Of course, the requirements of the problem allow for more than one right-guesser per group, but I stick to the "Pascal triangle" idea of Wizard and MathEx.
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