AxiomOfChoice
Joined: Sep 12, 2012
• Posts: 5761
January 16th, 2014 at 3:24:56 PM permalink
It's not quite enough for each group to have a non-zero probability of answering. You also need the sum of the probabilities to diverge (it's possible for infinitely many non-zero probabilities to sum to a finite number -- which would not be enough).

Eg, if you number the groups 1, 2, 3, ... , and group n has a probability of 1/2^n of answering, that's not sufficient.
Twirdman
Joined: Jun 5, 2013
• Posts: 1004
January 16th, 2014 at 3:28:07 PM permalink
Quote: AxiomOfChoice

It's not quite enough for each group to have a non-zero probability of answering. You also need the sum of the probabilities to diverge (it's possible for infinitely many non-zero probabilities to sum to a finite number -- which would not be enough).

Eg, if you number the groups 1, 2, 3, ... , and group n has a probability of 1/2^n of answering, that's not sufficient.

Yeah your right I was thinking about having the sequence of probabilities have a positive limit. Though think your right and you just need the sum to diverge.
kubikulann
Joined: Jun 28, 2011
• Posts: 905
January 16th, 2014 at 3:38:00 PM permalink
Reperiet qui quaesiverit
24Bingo
Joined: Jul 4, 2012
• Posts: 1348
January 17th, 2014 at 1:11:28 AM 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?

Two things:
1. You're talking from the wrong end.
2. You're right.

1. My frustration with this thread goes back mainly to that other "riddle" thread, where you and others bang on ad infinitum about how "I flip two coins until one comes up tails and show you that one" is the same problem as "I flip two coins once and you see that one came up tails," and act like I'm the idiot for not getting it. They aren't the same. They won't be. And I don't think I'll be seeing a "Project Steve" on that any time soon.

2. In this thread, though, I was so on edge by the fact that practically everyone had demonstrated a bald-faced inability to recognize the gambler's fallacy dancing naked in front of you just because the events weren't arranged chronologically that I assumed this was that, not taking into account how the wizards' probabilities of guessing could be manipulated; I hadn't even opened the Wizard's solution, so I wasn't sure what you were talking about. In fact...

What needs to happen is twofold: the probability of no group guessing wrong needs to converge, as you had it, and that of a group guessing right needs to diverge. So we need a function such that Π‎(1-21-ak) converges to something above 0.9 and Π‎(1-ak*21-ak) diverges to 0. In other words, we need an f(x) such that, ∫(log(1-21-f(x)) dx) converges but ∫(log(1 - f(x)*21-f(x)) dx) blows up in the negative direction as x goes from 1 to infinity. (Once we've an appropriate function it can be scaled.) Just not sure what that function would be. I can't help but think of "Gabriel's horn," but not sure how to apply it here.

But that kid could still be as easily a boy as a girl.
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
Joined: Jun 28, 2011
• Posts: 905
January 17th, 2014 at 6:01:23 AM permalink
The content of your "Spoiler" is perfectly summing up the problem. Great.

Quote: 24Bingo

1. My frustration with this thread goes back mainly to that other "riddle" thread, where you and others bang on ad infinitum about how "I flip two coins until one comes up tails and show you that one" is the same problem as "I flip two coins once and you see that one came up tails," and act like I'm the idiot for not getting it. They aren't the same. They won't be. And I don't think I'll be seeing a "Project Steve" on that any time soon.

Sorry, but I never said that. On the contrary, I (and others) state that it IS different. Hence the difference between 50% and 33.33%, according to the (different) available info.

Note: what is "Project Steve"?
Quote: 24Bingo

But that kid could still be as easily a boy as a girl.

Evidently, it is you here who are saying both situations are the same (namely, 50%).
Reperiet qui quaesiverit
24Bingo
Joined: Jul 4, 2012
• Posts: 1348
January 17th, 2014 at 11:08:51 AM permalink
Quote: kubikulann

Sorry, but I never said that. On the contrary, I (and others) state that it IS different. Hence the difference between 50% and 33.33%, according to the (different) available info.

No, you're being confronted with the latter and saying it's the former, again and again, because I don't know whether the coin I saw was flipped first or second. That's exactly what you're saying.

EDIT: Actually, looking at the asymptotic behavior of those integrands and their derivatives, I'm starting to think the answer probably involves power towers. I'll work it out in more detail when I'm not typing on a phone, but right now I just don't want to lose bragging rights in case someone else figures it out before then.
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.
dblanch256
Joined: Jan 9, 2014
• Posts: 92
January 17th, 2014 at 4:02:58 PM permalink
Quote: kubikulann

Quite so. Now I have a dilemma on my hands. Several people in this thread have made critically important contributions to the solution, but independently! I think if someone could combine the essentials of what's already been said, a complete and correct answer would result. [Again, harder than it sounds.]

I'm tempted to offer another hint, but I don't want to spoil the fun, unless the fun has already dissipated for most people! [If so, no harm no foul. If this were an easy problem I'm pretty sure it would have been solved days ago.]

OK, for those still determined enough to pursue it, I'll cite below what I believe to be facets of the solution, trying not to "give away the store" in the process.

This is a different problem, but its solution provides significant hints for solving the original (harder) problem.

"You collect ten wizards in a room, each (you guessed it) wearing a randomly chosen white or black hat. Again, all wizards must simultaneously either (a) declare their own hat color, or (b) remain silent. Thousands of rounds are conducted, but only rounds in which one or more wizards declare are counted toward the final score.

I claim they can collectively win 90% of the "qualifying rounds" using this strategy, based on the first two entries of the tenth row of Pascal's triangle (which is simply and easy source of binomial coefficients).

WARNING -- READ FURTHER ONLY IF YOU WISH TO SEE THE SUGGESTED GROUP STRATEGY

Only when all of his neighbors' hats are the same color, should a wizard declare the opposite color for his own hat, otherwise he should abstain. Clearly there will be many rounds for which nobody declares anything, but as mentioned earlier, those "silent" rounds do not count in the scoring.

• Wizards should work in groups with a single "spokes-Wizard" for each group
• Nothing prevents the wizards from forming a single infinite line corresponding to the natural numbers
• Failure is always an option. Only a 90% success rate is required for each "round". This is important.
• Imagine how one might calculate the failure rate of a finite group of wizards. See solution to easier (subset) problem button above.
• Try to extrapolate the above to more wizard groups (not necessarily the same size) whose successive failure rates drop precipitously
• Might the collective failure rate of 10% comprise the limit of an infinite series?
• David C Blanchard
kubikulann
Joined: Jun 28, 2011
• Posts: 905
January 17th, 2014 at 4:33:58 PM permalink
Quote: dblanch256

<item>Wizards should work in groups with a single "spokes-Wizard" for each group</item>
<item>Nothing prevents the wizards from forming a single infinite line corresponding to the natural numbers</item>
etc.

Let me guess... You are a frequent contibutor to Wikipedia? I used to mix up their < > with the [ ] of this forum, too.
Reperiet qui quaesiverit
24Bingo
Joined: Jul 4, 2012
• Posts: 1348
January 17th, 2014 at 9:48:46 PM permalink
I think I may have figured out the general form of the answer, although I can't prove it rigorously (never was too great with series), and I'd have to work out the size of the first group (at least 10, though). Are the groups such that...

...the size of each group is the size of the previous - call it N - plus ⌊2N/N3⌋?

e.g., although I don't know where it would start, 10, 11, 12, 14, 19, 95, 46203914573123975852867, and so on...

...yeah, that seems a bit random, doesn't it? My reasoning is so sloppy I was reluctant to share it, but I guess I probably should.

The reason I thought it was a power tower was looking at those integrands above. I wanted the convergent one to asymptotically approach 1/x2 to make it as likely as possible that the divergent one should diverge. To that end, I worked out the derivatives of the integrands, for the convergent one, ln 2 * f'(x)/(2f(x)-1-1), for the divergent one (ln 2 * f'(x) * f(x) - f'(x))/(2f(x)-1-f(x)). Seeing the second, I thought I probably didn't have to worry about it, as long as the first was reasonably close, so... yeah.
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.
dblanch256
Joined: Jan 9, 2014
• Posts: 92
January 18th, 2014 at 3:48:13 AM permalink
Quote: kubikulann

Let me guess... You are a frequent contibutor to Wikipedia? I used to mix up their < > with the [ ] of this forum, too.

Sadly no. I wish I were, but all I do is send them small donations in exchange for the unspeakable volumes of data I shamelessly suck down from them.

I rushed this reply ... so I need to fix that. Meanwhile, I thought of an easier, but very related problem that might be a good stepping stone. Give me an hour or so, and advise whether you think it's the "right kind of help".
David C Blanchard