HoustonHokie

You're right. I did try it before I submitted the post and it failed. My circles were too small and didn't touch each other. One problem is that it is almost impossible to keep the compass perpendicular to the paper. The other, of course, is being able to pick out the tangent point of a circle by eyeballing it. This solution relies on both perfect circles and perfect tangents. If it were one or the other, it would probably have worked. Almost pulled the post, but I figured it was still closer to the solution you were looking for. I love a good geometry problem, so I'll keep looking.

Assuming the pencil lead in my compass could draw on heavy gauge steel, this is another possibility.

That's funny - the high-powered one I got for my college drafting courses doesn't have numbers! Maybe I should have gotten the cheap kind!

Yes, but don't you need a straight edge to complete the bisection of an angle? As I recall, the process goes like this: 1. With compass point on angle vertex, swing an arc from side to side of angle. 2. With compass point on intersection of arc and side of angle, swing an arc with radius of at least half the distance between the two intersections interior to the arc. Repeat on the other side. 3. The intersections of the two new arcs will be on the bisecting line, which you can connect to the original angle vertex with a straight edge. And the OP doesn't say that we have a straight edge... So I folded instead. Of course, the OP doesn't say that our triangle is on paper, either - but it does say that we got the triangle from our work on bonanova's topic, which did specify paper (and allowed use of a straight edge, but I think that tool has been taken away). If I had a way to guarantee that my bisection point was at the center of the desired circles, then I'd say that bisection would work. Otherwise, I'm going to fold.

Very carefully... Basically I performed multiplication one digit at a time and kept expanding the candidate number based on the results of my previous multiplications. Because that sentence makes almost no sense to me, here's a better look at what I did:

I've never had to automate multiplication digit by digit before, but my spreadsheet gave out at 15 digits. It was an interesting process to set up formulas to multiply two numbers, add the "carryover" number, figure out the 0 and 10 place, and report those in various locations. Then it was a matter of carrying it out place after place and looking for a spot where the product began with "71". First time for everything...

oops... I forgot there were 117 other possible numbers to share with Obviously, you remembered the results from the birthday thread.

whew - a lot less calculations

Actually, there are 10 possibilities (0-9) & 10^4 = 10000. But since 0000 is disallowed, it's actually only 9999.

Strictly speaking, it seems to me that this puzzle might be dependent upon the rules the SSA uses for creating SSN's. Rules for SSN's: 1. Only numbers 0001 - 9999 are allowed for the last 4 digits (0000 is not allowed). 2. One set of 4320-4329 has been removed for advertisements. 3. Each SSN has 3 parts in the form AAA-GG-SSSS (area-group-serial). The serial numbers (last 4 digits) are assigned sequentially for each area and group combo and are therefore not random. This has the effect of making 0001 more likely than 0002 which is more likely than 0003, 0004, ..., 9998, 9999. That being said, a fair number of the low numbers were issued when the SSA was formed in the 1930s and maybe the professor's students would be more likely to have higher numbers (I'm assuming you don't have many octogenarians in your classes). It's probably too difficult to figure out how much more likely some potential serial numbers are than others because I doubt we could get statistics on their distribution. Should we make the assumption that the numbers are sufficiently mixed in your student population that all serial numbers from 0001 - 9999 are equally likely?

Thanks And I haven't quite put everything to rest in my mind. renan proposed an interesting variation and I'm trying to figure out if that one can be done. Problem is, I may not have the right picture in mind. renan's post mentioned an oval, but I don't see a need for one if it looks like this: If it looks like that, each circle affects a different number of coins - the outside circles affect 3 coins each, and the middle circle affects 5 coins. Makes it interesting. If I've got a little more time, I might try to work that one out.

Question: if he [expects to receive a clue] is that the same as [receiving a clue]? I think my truth tables showed that as long as his expectation pans out and he receives a clue of the same form as the others, he can say his probability is 2/11 without even knowing the specific value to be revealed. Otherwise - either because you can argue that the form of or future presence of a clue is not guaranteed until actually provided (or at least promised) - he would have to say 1/6 until receiving the actual clue (or the promise of one). In other words, if the magician had stated up front that everyone would be told one of the two values of their dice, they would all be able to say 2/11 and we'd only be arguing the point from the boy/girl problem. No such promise was made, so the argument we must also wrestle with is whether it is reasonable for subject 7 to assume that the pattern of clue-giving would continue. I think should be obvious that if he had [not heard anything] he would have had no expectation of receiving a clue and could not venture any guess except 1/6. Similarly, if the magician had told each of the previous 6 subjects that the value of a particular die was n, he (like they) would be forced to say the probability is 1/6. I realize that I am probably arguing out of both sides of my mouth but that's because I think we've left the realm of probability/statistics and moved into human behavior with the subleties of whether and how a clue would be provided, which, in my opinion, is the determining factor in choosing between 1/6 and 2/11 as the appropriate answer for subject 7. That depends on a prediction of the behavior of the magician, and we are given limited data upon which to base our assessment of his future behavior.

It is an interesting problem. I wonder how the distribution of passengers' arrivals is assumed to be - regular or random? Does it matter?

So it's the being told, but not actually seeing, that makes the difference...

Oh, yeah - I could program it - and pretty easily in a spreadsheet. So I did. And I got the same answers in about 30 minutes that it took forever to draw, plus I confirmed that my groups were right. As far as the problem cases go, well, I haven't discovered much beyond the fact that they're problems. I do know that none of them can be created with a reset, so they have to be created with flips, and they're just flips of each other. If you introduce a reset to one of the problem cases, you exit them and can't get back.