bonanova

Hmm do they consider the hypotenuse a leg in America? No, they don't.

Honorable Mention: Anza Power had the right analysis for inclusion of the original person.

This calls for a distinction (is there one?) between non-intersecting and non-overlapping. Hmm, yeah, I guess there is. TSLF, care to comment?

Rookie's classic (with some assumptions not made in the OP).

Anza Power has it.

The cases of 3 and 4 both match my answer above, in case of four people you and the person right in front of you have 1/3 chance each and the two others who are beside you have 1/6 chance each... I didn't read your answer carefully enough to see that you included the initial flipper. So that degrades the chances of the adjacent persons. Do you get the expected result now if you exclude him? The process is supposed to find "another" person. I.e. a person other than the initial flipper. The one who buys is the one who never touches the coin.

Someone has the right answer, and someone is close. To decide, and get a clearer view of the processes, conside a simpler case. The case of three people (two others) is trivial. Four people is rich enough to be instructive, and still tractable by hand. It's an interesting result.

You're having coffee with friends sitting around a table on a work break. You paid for everyone's coffee the last time, and so you propose a method for randomly assigning that task to another person this time. Your proposal involves flipping a fair coin until a particular event happens. Here are the details: You flip first. If it's heads, you pass the coin to your left; tails, you pass the coin to your right. The person receiving the coin flips it and passes it left or right according to the same algorithm. And so on. As the process continues, more and more of the persons will receive the coin. At some point, there will be just one person who has not received the coin. That person must pay the bill. Is this a fair method?

Bad thing for a moderator to say, perhaps, but I think this puzzle is worth looking at, even if it may have been posted before. So if you happen to know the answer from a previous posting, let it ride for a few days. There are three pairs of balls - red, white, and blue. In each pair one ball is a little bit heavier than another one. All the heavy balls weigh the same, and all the light balls weigh the same. Also you have a balance scale with exactly three outcomes: <, = and >. You need to identify the three heavy balls. How many times do you need to use the scale?

Kudos to DeGe. I started a recursive / summation approach, and ran quickly out of steam (read: lacked ability to think through to conclusion.) If you're familiar with the problem (a classic that I shared a while back), when I was first asked about it, I ran away from the calculus and made the assumption that gave the answer immediately. And I impressed the female colleague who posed it to me. I looked for the easy way out here as well.

From what I remember while solving it, there were no unresolved elements and each piece fitted in one place only (I have now deleted the file that I used to work this out). The interchangeability between Mike/North and Paul/Fisher is also resolved. I don't remember how exactly I resolved it but it had something to do with there being only 3 girls names... Grace, Karen, and Dot. In case of interchanging Paul/Fisher and Mike/North, some other condition was being violated. I hadn't thought of that, assuming Ali was short for Allison, who, in my social experience, is decidedly female. I guess that wraps it up. Nice solve. And since chocchief doesn't seem to be around, I'll mark it.

It seems the OP allows all three of the above solutions. Barcallica's Paul/Bill uncertainty is resolved using clue 7. Nevertheless, Mills and Ross seem interchangeable. DeGe's solution shows that {Mike and the North girl} can be interchanged with {Paul and the Fisher youngster}, whose wrong answers can be either {7 0} and {3 4} or v.v, respectively. Sometimes the flavor text hides a clue, but not this time. Great puzzle otherwise.

[spoiler=It has to be different in the following way]In your sketch, the top row circles are 1-1 with the bottom row. It has to be the case that two extra top row circles appear. That won't happen unless at some point (at least in two cases) a top row circle is directly above a bottom row circle. So the top row circles must move up and down in groups. Start with a group of three top row circles nestled down in triangular packing position. Then a group of three top row circles packed against the top of the box. Then three more down, then three more up. Do that 66+ times, to the end of the box. It seems like that permits the top row to be compacted in length. That may be enough to fit two extra (199 = 2 = 201) top row balls.

Nice, but not it.