Logophobic

A bit more probabilities for at least three persons sharing a birthday. And some probabilities for at least four persons sharing a birthday. Because: Why not?

For further information regarding this puzzle and its origin, see http://gurmeet.net/puzzles/tiling-with-calissons/ " Source: On the back cover of Mathematical Puzzles: A Connoisseur's Collection(163 pages, 2003) by Peter Winkler. Originally, the problem appeared in "The Problem of Calissons" by G David and C Tomei, American Mathematical Monthly, Vol 96(5), May 1989, p. 429 - 431."

I didn't fully trust the results of my first simulation because I was using a pRNG with has a relatively short cycle. On Saturday, I dug up code for a better pRNG and repeated the simulation. The result was far more accurate, giving a probability of 49.95% (leap day excluded). I then began an iterative sum approach similar to plasmid's above. I manually calculated the probabilities for up to 8 persons, then began to code a routine to sum the probabilities for up to 99 persons. I didn't have time to finish, so plasmid beat me to the post. The results, of course, were exactly the same: The code (Excel VBA)

Based on simulated trials, the answer is... I recorded 100 runs of 100,000 with this number. The minimum was 50,046. The maximum was 50,316. The average was 50,177.64. Final result: 50.18% Then I considered February 29, which was conveniently omitted in those trials. I repeated the process, allowing for leap day in its proper ratio. The minimum was 49,891. The maximum was 50,068. The average was 49,999.04. Final result: 50.00% Very interesting. Given this number of people, the probability that at least three share a birthday might be ever-so-slightly less than 50%.

Are you sure about that?

Let me point out that we don't need two separate formula for Iceberg 1 and Iceberg 2. The formula given for Iceberg 2 is valid for Iceberg 1. As for Vsw decreasing for Iceberg 2, this is only true when θ <= 30 (base angle not greater than 60.)

Not so: 50% more is 'half as much' more, so 50% more than 50% is 75%. That said, I would interpret 'heads 50% more of the time' to be 'heads 50% more often than tails', which would be 60% heads, 40% tails.

You did not need to solve the integral. The submerged volume is a spherical cap, the volume of which is pi*h2*(3r-h)/3, where h is the height of the sphere from the intersecting plane. In your diagram, this is (R+s). For the more general case this is 5 - (r/sin(t) - r), or (1-1/sin(t))r + 5. Now, after some review of the derivative, which I was barely introduced to some 14 years ago, I have a complete solution for the question as posted. f(r) = pi/3 * h2 * (3r-h) h = (1-1/sin(15)*r + 5; letting n be the constant value (1-1/sin(15)), we have h=nr+5, then: f(r) = pi/3 * (nr+5)2 * (3r-nr-5) f(r) = pi/3 * ((3n2-n3)r3 + (30n-15n2)r2 + (75-75n)r - 125) f'(r) = pi/3 * (3(3n2-n3)r2 + 2(30n-15n2)r + (75-75n)) f'(r) = pi * ((3n2-n3)r2 + (20n-10n2)r + (25-25n)) (3n2-n3)r2 + (20n-10n2)r + (25-25n) = 0 The radical cleans up nicely, giving us (-b + 10n)/2a and (-b - 10n)/2a (-b + 10n)/2a evaluates to 1.74599, which is 5 * sin(15) / (1- sin(15)), where the sphere just touches the surface of the water, giving a submerged volume of 0. (-b - 10n)/2a evaluates to 5*sin(15)/(1+sin(15)+2*sin2(15)) = 1.150465939351, giving a submerged volume of 5.3177342075568337849950 cubic inches.

A simple exhaustive search shows that of 46656 possible outcomes, there are 1860 that end at the starting point. 1860/46656 = 0.03986625514403292181069958847737

I like your solution. I attempted to come up with a logical proof by adding/subtracting a constant value, but couldn't sort it out.

I have no idea how to find the solution algebraically, but through recursive approximation...

Your example is not a counter to my second point because it is in opposition to my first point: it contains both even- and odd-weight gears. The validity of my second point is implied by the validity of my first point. There does not exist a counter-example to the second point that does not contradict the first. The first (k=0) and second (k=1) together are a proof by induction of the third.

You have overlooked the integral weight requirement, but your argument against my 'simple proof' still holds. I was trying to simplify a more complete proof that I had in mind:

kbrdsk is correct: jasen's reply has error because more than two students would have spoken wrongly in each case: (Given: Exactly two students spoke wrongly.)

A simple proof:

After further consideration, a larger sphere, while not displacing its total volume, could possible displace a greater volume than that which I gave at first.

Largest sphere to fit inside given cone?