bonanova

Somewhat in the vein of a previous Professor's riddle, another physics Prof proffered this puzzle: Two cats sat on a tin roof. One slid off. Which one didn't?

The folk down at Morty's couldn't remember the last time that Alex had dropped by. Some had conjectured he had given up puzzles. A few, less well informed, to be sure, guessed he had given up drinking. Whichever was the case, or perhaps for other reasons altogether, Alex had been conspicuous by his absence for several months. But last night he was there. And his barely concealed grin put the boys on notice of a question that would put them once again to the test. Ya know how those basketball folk in the States draw numbered ping pong balls to determine the teams' draft order? Well I say that's a stone-age way of assigning variable probabilities. If they had a brain in their collective head, they would design an unfair die with faces numbered proportionately to the probability that the number shows, when the die is rolled. In fact, that's something that even you geniuses might be able to do. So here's the challenge. Suppose there are eight basketball teams, numbered 1-8, and they numbered one ball with a "1", two balls with a "2", and so on until they numbered eight balls with an "8". The number eight team would have 8 times the probability of winning as the number one team. And so on for each of the other teams. You need to get the same result; only do it by throwing an unfair 8-sided die. I'll give you a regular tetrahedron - that's one of the five Platonic solids, ya know, and tell you to slice off each of the four vertices, parallel to their opposite faces, making them into new triangular faces. The four original triangular faces would become hexagons. Now you have an 8-sided die. If you assume that the probability of a polyhedron landing on one of its plane faces is proportional to the area of that face, Can you construct in this manner an 8-sided die, with faces numbered 1-8 with the property that the number on each face is proportional to the probability of that face showing? If so, what percentage of the original tetrahedron's volume would remain? Drinks for a week to the bloke who has the answers.

Assuming that "goes" means "pulls the trigger" isn't it just the opposite?

The first has 51% chance. The second has about the same. That puts them collectively below 30% even if the last 18 are certain [they're not].

I believe that's it. Someone will come up with a calculation of the cycle probability. Nice.

A similar problem described a hypothetical village of 100 child-bearing couples. We must assume equal gender probability for each birth. If each couple produced children until a boy child is born, and then they have no more children, what is the expected gender distribution of the children in these 100 families? In the present puzzle, the man bets all day until one of two stopping conditions obtains. And the question is asked whether this strategy can overcome a negative expectation for each individual bet.

Do you mean: Neither of them can see the feathers attached his own hat.?

You're right. Good point.

The OP [seems to] say the prisoners can't communicate. If this applies only to speech, then yes, there are nonverbal communicative strategies. However, your analysis seems wrong. For example, you have not explained how, if each only looks into 50 jars, the ordering could be complete. I don't think the OP has clearly stated whether the first prisoner looks into his 51st jar before or after all the prisoners have looked into 50 jars. The OP says each, which implies to me before. Although if that's the case, the puzzle seems impossible. Also, the OP suggests the best collective result is about 30%

I'd go with James' rule number 2 above as the simplest. But no rule yet explains the minus signs, unless they were used, instead of commas, for separators.

It seems the answer is No. The OP says: the ones who have finished will not be able to communicate with anyone who is in the process of looking or is waiting to look. But this might be a cleverly worded prohibition that applies [only] for person-to-person communication. If so, my spoiler suggests a way for subsequent prisoners to have a higher than 51% chance, and there are probably even better strategies.

Read the OP again. Even number is not mentioned. The King is free to use any number of Red and Black hats totaling 20.

Good one ... ! We need details.

No patient wishes to be exposed to a glove worn by another doctor.

Gravity 101 Any two massive objects separated by a distance r share a mutually attractive gravitational force given by Fg = Gm1m2/r2. The net gravitational force on a single object is the vector sum of forces from all other objects in the universe. Whether anything [atmosphere, asteroids, vacuum] exists in the intervening space is irrelevant. Because the force behaves linearly with mass but falls off as the square of the distance, the force exerted by even large objects a great distance away can be negligible. When the distance is not that large, the force exerted by a more massive object predominates. Some familiar rock-paper-scissors comparisons: Moon beats Jupiter: The moon is tiny compared with planet Jupiter but much closer; so its gravitational force, not Jupiter's, creates tides. Sun beats Moon: Because of the Sun's extreme mass, Earth orbits it, rather than the nearer but much lighter moon. Earth beats Sun: Because Earth is close, the moon orbits Earth as the Earth-moon pair orbits the Sun.

Thanks Izzy, there's a reason why your answer makes sense ... besides your intellect, of course. I left out a crucial part of the puzzle; namely that all patients are examined by all doctors. See Edit in OP.

Yes - to both of the clarifying questions. The approach to use as regards your idea is suggested in the spoiler. Also note the omission and correction to the OP. All patients are examined by all doctors. That's an important part of the puzzle.

Here's a G-rated generalization of Rookie's condom conundrum. Take a look, if only to understand the problem being posed. Now to introduce a modicum of professionalism, we'll talk about surgical gloves, instead. d doctors wish to examine p patients p <= d] in a manner that every male over 50 is aware of. No doctor wishes to wear a glove contaminated by a patient. No patient wishes to be exposed to a glove worn by another doctor. As in the "cc" puzzle, they may be turned inside out and/or worn on top of another glove. There are three cases, increasing in complexity: d=p=2. Well, Rookie's puzzle gives this answer. p=1; d=2n+1 n is an integer] any p <= d Edit: If all dp combinations of doctor/patient examinations are performed, then ... What is the minimum number of surgical gloves required in each case? examinations and surfaces.

OK, here's another. On a can of mildew spray is the Caution: "Use only in well ventilated areas." Right. I'm not sure where your mildew grows, but ...