CaptainEd

oops, another senior moment... Except I admire the problem and Phil's solution!

False alarm, sorry. It's easy to do what I suggested. There is no light shed on the overall problem. Move along, folks, there's nothing to see here...

Here's a slightly reduced form of the problem, does it shed light on the impossibility of 18? Make the partitions of size 1, 2, 3, 6, 9, 12, and 18. That is, place the first three points as before; place points 4, 5, and 6 in a way so that the six points are in separate 6ths of the line, place points 7,8,9 so that the 9 points are in separate 9ths; place 10, 11, 12 so that the 12 points are in separate 12ths, and place the last 6 so that the 18 fit into separate 18ths. In other words, remove the constraints on placements 4,5,7,8,10,11, 13, 14, 15, 16, and 17

I'm clueless. I was hoping to see some sort of kernel for an induction step--if I remove this card, I get the previously solved problem for N-1. That's one reason I like your observation about a most expensive starting point- it consists of (N-1) and (1) bracketing the fixed point from the previous round. I don't see how that helps, but it is at least a simple pattern such as you like to find in induction...

Superprismatic, that's a pretty and nicely organized template for a starting configuration! This whole problem and its solution are provocative.

Jim, you harvest one from each of the piles, destroying them. However, your harvest becomes another pile, so you wind up with one pile with 45 cards. Then curr3nt has analyzed what happens there.

Thanks for the puzzle(s) as always!

Oh, you want the minimum flip solution...sorry

BIT by Bayes, Bushindo! Thanks for the citation

I see, so when you compare with the middle number, you append the new number on the OUTSIDE (front or back) of the whole string, rather than Left or Right of the middle number. I answered a wrong question. This IS an interesting puzzle, thank you Wolfgang

WHen you say "choose a number", do you mean that I must grab one randomly, or do you mean that I can decide on which number I want to grab?

Durn, bit by Bayes again...

Edit: looks like I got here late. Gotta remember to refresh the screen before posting (note to self...)

Thanks for the pretty puzzle!

And so it remains a team effort. But your arrangement wins by one test, I believe. Your 13th test distinguishes the last pair, so no 14th test, agreed. My 13th test distinguishes my only pair, so no 14th test. Unfortunately, my 12th test touches 3, so it takes two more to find which one for sure. This is why your solution ending in 3 pairs is better than mine, ending in a trio and a pair. Good job, Y-san. Hontoo ni utsukushii desu!

I think the second player could choose one of the cards chosen by the first player, as the first player's choices are secret.

Now that Y-san has the same max, I feel more confidence. Arigatoo gozaimasu! I admire that you were able to get a larger number of full crosses early, so your Mean Time To First Hot is lower by a fraction. Now I see that, if we ignore the geometry, and just get to grab 5 cards each time, it would still take 11 tests to be sure of getting a Hot, so our taking 13 doesn't feel all so wasteful.