bonanova

That result seems to be less than 1/3 the diameter. In two dimensions you have more freedom, so shouldn't it be greater than 1/3?

You and Joe are partners at Bridge. You, having no aces, are told by a reliable informant that Joe holds an Ace. You quickly calculate the probability that Joe holds at least two Aces, which would be enough to bid game. Wait. Your informant now tells you Joe holds the Ace of Hearts. Are Joe's chances worse, better, or the same?

Guidance for discussion. Tabby and the mouse are geometrical points. Can the mouse keep the distance between the points strictly positive?

Right - it's not a variant of that approach at all.

This may not be a puzzle so much as a challenge. Prepositions and grammar may be nonsense "up with which some people will not put,"[*] but others don't mind thinking about such things. Create a paraphrasable sentence [one with meaning] that has as many prepositions as possible at the end. A word counts as a preposition if the dictionary gives at least one meaning for the word as such. Have fun. [*] See e.g. here.

Yeah, that's the "pretty simple" way. And, yes, assume you have some paper. Sorry. Now what's the other way?

OK this is a little off beat, but it points out an interesting geometrical result. The problem is to [describe how to] construct an equilateral triangle. Your tools are Pencil Straight edge A device that trisects angles I can think of two ways - one is pretty simple. Post it if you think of it, but there's another way that's perhaps unexpected. [same tools] Using the pencil as a pencil, the straight edge as a straight edge... i.e. no arcs, etc.

Perhaps also not surprisingly I am ... I recall looking through Chuck's differing approach, still satisfied mine was all right. Thanks for the vote of confidence, but I'm open to anyone showing me the error of my ways.

There is a "flavor" of puzzles that include "flavor text" which may contain clues. Some people enjoy both writing and solving them. Succinct version: Is it [mathematically] possible to individually specify the face probabilities of two cubic dice, not necessarily identically, each numbered 1-6, with the effect that the sums 2-12 occur with equal [1/11] probability?

Well you can go here for my answer. Post #15. And read d3k3's answer here. We're just prolonging the agony now by thinking about allowing numbers other than 1-6 on the faces. And yes, the OP intends to say the dice are cubic, with the numbers 1-6 on each.

I computed distance incorrectly in two previous posts. Average distance is not .664, which I took to be 1/3 the diameter. I simulate the average distance on the unit-radius disk as .9047 +/- .0027 Random triangle area is still correct: .232 +/- .002 And convex probability is still .7044 +/- .0004 I found several references for average distance. Here's one. For the unit disk it gives 64[diameter]/45pi = 128/45pi = .905414, agreeing with the simulation.

Yep. [36 comes in as combinations of cubic die faces.]

Tho it's tempting - choose face numbers and probabilities. No sums other than 2-12 only requires min[a]+min = 2 and max[a]+max = 12, allowing 0's 7's 8's, etc.

While the discussion continues .... I heard this question posed, without an answer.

True. But there are strong hints that they are: [1] the friend who perfected the art of loading dice, and [2] Alex's intention of making some bets with friends Neither would agree with a pair of dice that were not 6-sided. It actually requires equal chance for 2-12. That would mean an 11-sided die with equal chances for each face to show. Can you describe such a shape? Is there a proof of this?

That solutions fits these answers: None of the the performances of a given day are simultaneous. No group performs more than once a day. No group performs and watches on the same day.

They're fair dice, cleverly weighted so the chance of each face showing has been separately adjusted. So yes, the faces are numbered 1-6. But your question suggests another puzzle, doesn't it? For now, suppose Ian has said, and I can prove it. Would you agree with Alex, who claims to have equal probability dice [with respect to the outcomes 2, 3, 4, ..., 10, 11, 12] or with Ian, who claims it's impossible to create such a pair of dice?

Questions arise: On a given day, are all performances given simultaneously? Can a group perform more than once in a day? On a given day, can a group both perform and at another time watch the performance of a different group? Normal interpretation of the OP seems to rule out question 3, but the answers to questions 1 and 2 certainly affect the answer. To clarify regarding question 1. If Group A and Group B perform on a particular day, can Group C watch both performances? Not if they are simultaneous. To clarify regarding question 2. Could Group A perform 10 times in one day so all the other Groups could be sure to see them that day? Not if a group can perform only once per day. To clarify regarding question 3. Could all the Groups see all the other groups perform in a single day? Not if a group can only perform OR watch in a single day, or can only perform once in a day.

Read the bold type. Alex claims his dice show all outcomes with equal chance. Ian says that cant happen. Who do you agree with?

Alex and his friends Ian and Davey were enjoying some cold ones at Morty's last night when Alex pulled a small pouch out of his pocket and began to tell his story. So this fellow I met claimed he mastered the art of loading dice, and he was able to separately adjust the chances of each of the faces showing. We got to talking, and I asked him to make a pair of dice that would make all the possible results, 2, 3, 4, ... 10, 11 and 12, appear with equal chances. Here they are. And he held them up for all to see. No way, said Davey, who was no mathematician, to be sure, but this seemed beyond possibility. So you're an unbeliever, replied Alex. Then watch this; and Alex rolled the dice 5 times, getting 2, 5 12, 9 and 6. How many times would you have to roll an honest pair of dice to see snake eyes and box cars in 5 rolls? Jamie heard the commotion and strolled over to the table. Here, let me try. And Jamie rolled 3, 10, 8, 2 and 11. Well, these are no normal dice, that's for sure, he said. What are you going to do with them? Haven't decided yet, Alex replied. For now, maybe just a few little wagers with some of my, ahem, acquaintances. Ian had been thinking for a bit. They're special dice, to be sure, he said. But I'm quite certain they're not what you say they are. If you were there, who would you agree with?

Infinity gives rise to lots of counter-intuitive results - let's call them paradoxes; here are two. [1] Two times infinity equals infinity. Take the set of positive integers 1, 2, 3, 4, ... One might think that's it - there aren't any integers other then these. Ah, but there are: multiply them by 2 to get 2, 4, 6, 8, .... you can assign an integer to each even integer. But that leaves "holes" in the sequence at 1, 3, 5, 7, ... and you can assign an integer to each of these holes. So without adding any integers to the original set, you have doubled, or replicated the entire set. This is another way of saying that the cardinality of integers, even integers and odd integers is the same. I = E + O. But I, E and O are all the same size! Certainly a strange way of thinking about "size" of sets. [2] Banach-Tarski Paradox The second one is breathtaking - it bears the names of two mathematicians, Stefan Banach and Alfred Tarski. In 1924 they wrote a paper which describes how the infinite set of points {x,y,z} | x2+y2+z2<=1 can be divided into no more than 5 partitions - one of them is the point 0,0,0 - which are then translated and rotated, but not stretched, so their shape is not changed, and re-assembled into two separate sets of points, each identical to the first! Or into a sphere of larger radius than the original. As one person put it, "A pea can be split into a finite number of pieces and be re-assembled into a sphere the size of the Sun." If you follow the duplication of positive integers in [1] you can kind of appreciate how this might come about. But it's beyond simply renumbering the initial set of points. In fact, you can't do it in 1 or 2 dimensions at all. The five pieces are infinitely complex - to the extent that they are not "measurable". It's impossible to determine their volume. So when they're re-assembled, the final volume can be different from the original volume. Again, intuition is thrown out the window. BTP relies on something that most mathematicians accept [but not all agree with] called the Axiom of Choice. AOC states that if you have a collection of sets - perhaps an uncountably large collection - there exists a set called the Choice Set that contains exactly one element of each of the original sets. That is, without saying precisely how an element from each set is to be chosen, it can be done, and you can talk about a representative member of each of the sets. This is equivalent, actually, to the well-ordering theorem, which in effect says you can find and then choose the "least" or "smallest" member of any set. If AOC is removed, then partitions are always measurable, and any duplication or expansion of volume is not possible. Not surprisingly, non-measurable sets have provided a strong argument against AOC. But AOC has the virtue of providing easy proofs of important theorems. So it's generally accepted, and imponderables like the BTP remain with us. So that's it - if you mess with infinity, you may need to park your intuition at the door.

Tabby and his intended dinner, the gray mouse, are loose in a circular ballroom. There are no exits, chairs, stairs, curtains, mouse holes ... just walls. They have equal top speeds and equal endurance. So the mouse should be able to stay alive indefinitely, right?