bonanova

But it is the right idea. The minimum number (8) of adjacent steps can be maintained as small parts of the four rectangles are minimally changed to effectively stitch them together.

OK I'll have to think about it some more. It seems impossible to implement a strategy. You're blindfolded and grab one of three jars your opponent didn't pick. Now, to state that problem in a way that will allow a solution ....

Four fewer lateral moves is possible.

The other end of the stick in post 18. Nice.

Indeed. One really interesting aspect of this problem is that by removing the balls, you even out the odds of pulling one color or another as the numbers get big. For example, I ran a simulation 1,000 times where I flipped a fair coin 1,000,000 times and noted abs( H - T ). The average value was 783. But if we put 500,000 black balls and 500,000 white balls in the urn, the average number of balls left after exhausting one color is very close to 2! Coin outcomes remain 50-50. Ball outcomes readjust to b/(b+w) and w/(b+w).

Lovely, isn't it? I was going to add ... think of the string of black balls like a stick. And the white balls like cutting the stick in w random locations.

What is true of any black ball is true of every black ball. They are indistinguishable. They can't have distinct probabilities. Every black ball has a probability of 1/(w+1) of being in the urn after the last white ball has been drawn. If you like, it's precisely this awareness that constitutes the aha moment. If you want to say this black ball has this probability and that black ball has another probability, you'd have to say why this is so, and you'd have to say how you can tell them apart.

My formula made such sense to me (this is often the case until someone explains the logic that I overlook) that I posted it. Then, embarrassingly, it did not agree with simulations. The second part of the probability (after M colors appeared in M draws) appeared to be incorrect by a factor of 2. I've been chasing that factor of 2 for several days, but only part time owing to preparations for a choral performance this weekend. More, perhaps, to follow this week. plasmid, your post #7 may have found it, since you have agreement with simulation. I'll read it after I've given my approach another shot. FWIW here is my simulation of 10000 cases for N = 10. The 10 columns are for M = 1 ... 10. The 20 rows are for T = 1 ... 20 The table entries are percent, rounded to nearest integer. My equations agree for the major diagonal. But they underestimate the probabilities elsewhere. 10 MNTHIST 10000 100 0 0 0 0 0 0 0 0 0 10 90 0 0 0 0 0 0 0 0 1 27 72 0 0 0 0 0 0 0 0 6 43 50 0 0 0 0 0 0 0 1 18 51 30 0 0 0 0 0 0 0 6 34 45 15 0 0 0 0 0 0 2 18 43 31 6 0 0 0 0 0 1 8 33 40 17 2 0 0 0 0 0 4 21 41 28 6 0 0 0 0 0 2 12 36 35 14 1 0 0 0 0 1 7 28 38 22 4 0 0 0 0 0 4 20 38 29 8 1 0 0 0 0 2 15 34 35 13 1 0 0 0 0 1 10 30 38 19 3 0 0 0 0 1 6 24 39 25 5 0 0 0 0 0 4 20 39 30 7 0 0 0 0 0 3 16 37 35 10 0 0 0 0 0 2 12 34 39 13 0 0 0 0 0 1 9 32 42 17 0 0 0 0 0 1 6 28 44 22

M different colors means: exactly M different colors, not: at least M colors, right?

yeah, everything you said. the comment was to suggest the obtainable degree of diagonal moves.

A tour of the chessboard is technically a path that visits all the squares. I mean to ask for a tour that comes back to the original square - a closed path. There will be an even number (64) of moves. And the answer, no secret here, maximizes the number of diagonal moves.

Apologies to you both. I tried to think late last night, and from memory. The better (euphemism for accurate) thought solution follows: Place a set of n points pi on a circle. The basic event of the puzzle is Ei = the event that a (edit: clockwise) semicircular arc beginning at pi contains no pj where i <> j. The events for each point are disjoint, they all have obvious probabilities of (1/2)n-1 and they may be summed.to obtain p = n/2n-1.

Empirically, by simulation. Or do some area integrals. For an equilateral triangle with unit sides it's 0.364791843300... For a square with unit side it's 0.521405433.... For a circle with unit radius it's 0.905414787...