plasmid

Alternatively, you could just do what biologists do. Calculate a p value, and if it's less than 0.05 then consider that to be proof that the coin is biased. I think p<0.05 for n>7 based on bonanova's formula.

In that case, I'll go ahead and just say what I consider the answer to be.

It looks like it's heading towards DrXP's solution, but I can't tell for sure.

A general case could in principle be solvable with an integral, but in practice it would be so messy that simulation would still be more feasible. Edit: It's actually more complicated than that hideous monstrosity... I left out that you would need to normalize the probabilities to sum to 1 instead of using raw squares of distances from the center.

This could be turned into a math problem by asking the following questions

I chose to define "best path" as the path which gets Santa from the North pole to the South pole with the lowest possible top speed at any point in the path, neglecting the tilt of the Earth (and hence possibility of one pole being in constant daylight). This seemed to be the definition that makes the problem non-trivial. If you consider the way it's phrased in post #16, asking what the longest duration of travel Santa could make while staying in darkness, then I agree with harey's answer, but I suspect that BMAD meant something else. Comments are within the spoiler, although I somehow get the feeling that maybe you were trying to make a point that I'm not picking up on?

If we continue to consider the question to be asking "what value for the third (or Nth) number from smallest to largest is most likely to occur" then If we instead use the definition of "expected value" as the average value after a large number of trials then

I agree with your answer (if the problem is asking "what value is most likely to occur for the third number", which might be different from the expected value if it's defined as "the average outcome if you were to run a large number of trials") but that formula makes me think you took a different approach than I did. How'd you solve it to end up with that formula?

Solutions to the equation n x 2a + 1 = m2 using the notation I've been working with should give you all anomalous numbers if you can solve it... or to be precise: if you can come up with a way of recursively coming up with all solutions for m given a set of n's, and then adding those m's to the list of n's and coming up with more m's, etc. In my lists of pairs of integers (n, m) that satisfy that formula, I noticed that all numbers of the form 2b-1 and 2b+1 were solutions. Now that I know my solutions have that form, I can prove that all numbers of those forms are solutions using my approach, which makes our approaches in practice look much more similar. By modifying my perl code a bit I made it much more efficient and have it outputting only m's that have a preceding n in the solution set, so it can look for anomalous numbers more efficiently with output that's easier to digest. The only anomalous numbers on the list it creates are the same ones you have 11 (arising from n=15 (24-1), a=3) 23 (arising from n=33 (25+1), a=4) 181 (arising from n=4095 (212-1), a=3) And it searches up to about 1014 (about 247) so it's maybe covering them all. If I try increasing the upper limit on the exponent $a, then it starts spitting out numbers that arise from rounding error (confirmed to just be rounding error by a separate program). But I cannot prove that those are the only anomalous solutions.

How'd you come to that answer?

I went back and noticed that my approach didn't miss 47, 93, and 185 -- I just missed transcribing that (33 x 2a + 1) has 529 (m=23) as a solution, so 23 should have been included on my list and 23 x 22 + 1 = 47, 23 x 23 + 1 = 93, and 23 x 24 + 1 = 185. So all's well there. At first I thought my algorithm was also going to miss 363 because solving (n x 2a + 1) = 363 would mean 181 must be on the list, and it initially looked like it wasn't because values for n were growing past 181 and looked like they had settled into a pattern. But there was a break in the pattern when I started looking at larger values for n and m. The set of n's and what m's they would produce goes 1 -> 3 3 -> 5, 7 5 -> 9 7 -> 15 9 -> 17 15 -> 11, 31 17 -> 33 11 -> nothing 31 -> 63 33 -> 23 33 -> 65 63 -> 127 23 -> nothing 65 -> 129 127 -> 255 129 -> 257 255 -> 511 257 -> 513 511 -> 1023 513 -> 1025 1023 -> 2047 1025 -> 2049 2047 -> 4095 2049 -> 4097 4095 -> 181 4095 -> 8191 Incidentally, 181 also -> nothing I have not yet had a chance to think through why my algorithm is generating n's that don't follow the pattern of being 2a +/- 1 for exactly the same set of numbers that you consider anomalous, and I will not have such a chance before going to sleep, but it seems unlikely to be coincidence.

The thing I still don't understand is whether or not you can prove that you've got a set of generating functions that will cover all numbers that will converge. That's why I took the different approach of starting at the number 1 and applying the algorithm backwards -- if you can cover all cases that can be reached by working from 1 backwards then you know you've covered every number that converges, so any number not on that list can't converge. I just can't come up with an adequate way to show definitively that I've covered all generating functions of the form (n x 2a + 1) x 2b by finding all values of n using the function sqrt(n x 2a + 1) = next value of n In fact I know I'm missing some numbers that do converge, maybe because the generating function behaves strangely at higher values or something, because my approach didn't pick up that 47, 93, and 185 converge using the list of numbers generated in the previous perl program. (Either that or maybe rounding errors are making my computer miss whether it's dealing with integers or not after taking log base 2.) If you want to make a list of numbers that your functions cover, you could compare it to the list of the numbers from 1 to 200 that appear to diverge: 27, 39, 43, 51, 53, 55, 59, 71, 45, 77, 79, 83, 85, 87, 91, 95, 99, 101, 103, 105, 107, 109, 111, 115, 117, 119, 123, 135, 139, 141, 143, 147, 149, 151, 153, 155, 157, 159, 163, 165, 167, 169, 171, 173, 175, 179, 181, 183, 187, 189, 191, 195, 197, 199 and multiples of powers of two for each of those.

The following perl program brute-forces an answer to my earlier question, which I think gives a solution, although I'd rather find a way to solve it without brute force.

Interesting, but I'm not sure that those formulas cover all numbers that converge, so I don't think you could conclude that a number doesn't converge based on the fact that it doesn't fit any of the described cases that do converge. As an example, 23 converges but I don't think it fits any of those cases.

Still not a solution, but if I could get a better handle on the function (n x 2a + 1) = m2 it might lead to one.

Not exactly what I had in mind, but so close that I'll call this won.

I'm afraid I know nothing of the sport of curling, other than that it somehow involves sliding a stone along ice as broomers sweep a path. So it's of course not what I've intended, and I'm at a loss to tell how well it would fit.

King of the mount, I'm venturing out To one of the four nearby Straight to the knight in armor I sight While tempting the fates I fly Foe in my path assails with wrath And dare I not land a blow Forced to retreat in shameful defeat My allies won't let me go

If the pieces of cake must have the same size and shape, then I get the feeling this isn't one of those questions of the "one person cuts and another chooses" variety. It's asking how a certain number of slices can result in equally shaped pieces if the guests can cut with absolute precision.