bonanova

I was born at 11:35pm Central time in the United States on yesterday's date. The fact that I now live in the Eastern time zone might have moved my birthday to today's date. IDK. I don't know how these things work, nor in the grand scheme whether it even matters. I suspect it doesn't. The point of writing this, is the shock I had last night when friends took me out to dinner, good chicken and ribs btw, and one of them said to me, so you're __ years old now. My reaction was to say "no way" until I realized they had it right. And I thought once again about my mortality. My grandfather lived to within days of his 100th birthday, and I've always had it in the back of my mind that I would finish that journey in his honor -- even tho it might entail outliving certain government benefits. But one never knows. BrainDen community lost SuperPrismatic this year, at an age much less than mine at present. But not to be morbid. Que sera, sera, and let's celebrate the day, each day. I heard from all my children and grandchildren yesterday, and life is good.

It would be interesting to simulate this strategy against a grid of Paula numbers. I can't tell, from analysis or intuition, whether the strategy is effective.

That's a nice idea wolfgang. I'm not sure that it provides an advantage for Victor. The way the puzzle is stated, Victor is not permitted to factor into his reasoning the outcome of prior hands. Nor should we suppose that Paula will be influenced, by previous outcomes, in her choice of numbers. The solution is a strategy that Victor devises that takes into account only what Paula does, as described in the OP.

With a tip of the hat to Gavinksong's that was written by an author that he references, I give one that is of the same type, but simpler. It was included in one of Peter Winkler's books, and it is attributed to Tom Cover in a 1986 publication on open problems in communication and computation. It asks something very similar to choosing apples behind doors, and it may even have inspired that puzzle. Because I have read the solution, I won't contribute anything further in the apples thread.

The faces of two cubes are inscribed with single-digit numbers. The cubes (let's call them dice) are rolled repeatedly and the sums of their top faces are recorded. It is found empirically that p(2) = p(12) = 1/36 p(3) = p(11) = 2/36 p(4) = p(10) = 3/36 p(5) = p( 9) = 4/36 p(6) = p( 8) = 5/36 p(7) = 6/36 What can we say about the numbers inscribed on the dice?

Read more.

Comment from the sidelines: Great explanation by plasmid! Which raises the question: Which takes longer: this puzzle, or ten games of Mafia?

@DejMar Thank you for clarifying. That's what I meant to imply by "virtual certainty." Also, in your post, change "upper bound" to "lower bound." Strange to use lower bound and Graham's number in the same sentence.

Sure. Graham's number has so many digitis, it's a virtual certainty that it, and every greater number, contains at least one of each of the digits 0 1 2 3 4 5 6 7 8 and 0. But even though Graham's number is almost indescribably large, there are infinitely more larger natural numbers than smaller ones. Therefore the probability requested in the OP is 1.

Give gavinksong a green up-arrow on his OP.

BMAD, I like the puzzle. Thanks. Labels would help.

Ok, I give up

For small numbers of circles the cluster is lumpy and erratic. One would expect an answer that is pretty much ad hoc. Which is to say that for the first n circles, it would take close to an nth degree polynomial to give a good approximation to the smallest bounding radius. A table would be just as good. By the same token, I'd bet there's a fairly simple asymptotic solution for large values of n.

@DejMar (and everyone.) Thanks for your contributions. They are the life blood of the site.

With probability 1 it is irrational for circles of any radii -- integer or otherwise. The area of a unit circle is pi. The included area of three unit circles involves sin(60o).

FWIW: The median area of 0.95248 was obtained for radii of (2 2 4) inches. The next larger area 0.955875 was obtained for radii of (1 1 5) inches. The next smaller area 0.858358 was obtained for radii of (1 4 5) inches.

These questions probably have obvious answers, but just to clarify: Do the prisoners meet initially to decide on a strategy? Are the times all different? Does 1:00 - 12:00 signify a span of 12 hours or of 24 hours? That is, are all times a.m.? Or all p.m.? Or can they be either? The prisoners would need to know the number of possibilities. Finally, to escape, is it required that every prisoner inform the warden, correctly, of all seven execution schedules, prior to the first execution?