Logophobic

@BMAD What sequence of four folds yields a four-sided figure? It appears to me that the least number of sides after four folds is six. @tojo928 Barring a four-fold solution from BMAD, a six-fold solution is best.

Simple: Alternately: Question:

Except that there is one result (b=c) that you did not test:

Counting paths on Y instead of K:

OEIS: A097227

Your presumption is wrong. Here is a counterexample: AB/BC = 0.75 AD/DC = 0.5625

1+6+9+5+12 = 5+12+10+4+2, as required: A+B+C+D+E = D+E+F+G+H Of course, A+B+C+D+E = F+G+H+I+J does seem more appropriate, but I was working with what you posted.

I think DejMar was right, though I have not yet proven it. Here is my solution for N=3 (100):

Hint: Solution:

We cannot assume that the general letter frequency of the English language applies to this collection of DVDs. The only data we have, one occurrence of each letter in a collection of 26 titles, suggests a uniform distribution.

I was looking a this question yesterday, and while I understood this hint, I did not know how to apply it. Well, today while browsing another site, I stumbled upon a similar question concerning (1+√3)2015. The solution to that provided all the information needed to solve this one. Interesting to note: there are only 44 possible results, recurring in a cycle of length 500. We can index these results and use (n mod 500) to find the result for any n.

It certainly is provable, though a brute force approach would be time-consuming. There are 11 * 2 * 9! = 7,983,360 unique arrangements of pegs. And, if I counted correctly, there are 68,600 unique solutions for peg #1, with five rotations of each solution for each arrangement of pegs. That's 343,000 positions to check for each arrangement of pegs. More than 2.7 * 1012 positions to check, and we still have 15 more triangles to place. Should we go ahead and reserve time on some super-computer? BTW, the 12th triangular number is 78, not 45.

Ah. In that case my simulations agree with yours. I noticed a certain relation between N and EV, so I ran a few more simulations. The first number is N, the second is EV based on simulation. The third is a very simple calculation. (See below.) 100 88.21 87.50 200 182.96 182.32 300 278.84 278.35 400 375.62 375.00 500 472.53 472.05 600 569.96 569.38 700 667.59 666.93 800 765.54 764.64 900 863.13 862.50 1000 961.13 960.47 If that wasn't evidence enough: 1000 960.85 960.47 2000 1944.74 1944.10 3000 2932.26 2931.53 4000 3921.39 3920.94 5000 4912.13 4911.61 6000 5903.97 5903.18 7000 6896.78 6895.42 8000 7888.41 7888.20 9000 8881.87 8881.41 10000 9875.06 9875.00 The very simple calculation?

OK, I realize that calculating the sums is not that simple. However, calculating the probability of card M of N being the highest card in the left pile as X = N - M; PX = X / (N-1) * (1 - PX-1) gives a close enough approximation. My simulations show that half of the time the highest card will be greater than or equal to 963. My simulations also show that the card with the highest probability of being the highest card in the left pile is 968. Which of these is the expected value?

I'll bring this back up because I believe I have a better solution.

I realized an error in my code for calculating the four-person probabilities. For the sake of correctness, here are the true (and extended) results: