bonanova

Mitch is quite a swimmer - do you want him running twice as fast, instead? Either way, there is an answer.

That will work. I will leave the puzzle open for other approaches if there are any.

Good point. It must be a yes/no question. Thanks.

With a nod to itsmee's 999 puzzle. You fall down Alice's rabbit hole and are faced with three doors, each labeled This Way Out. But you've been around this block before. You know only one door will get you home, And you don't even want to imagine what's behind the other two. Mad Hatter volunteers to help. edit He will answer one yes/no question truthfully, if he can. If he cannot, he will disappear without uttering a word. Let's say the doors are A B C and you've got tickets for the Grateful Dead. What's the question that will get you to the show?

Trying to resolve this puzzle assuming a rectangular house. There are four places the dog can be tethered, between which the area changes monotonically. Therefore, their respective areas contain the extremal area values.

Can't you prove it just by expanding the product? All the steps are reversible, so going either direction accomplishes the same thing.

This is the method I had in mind. It follows from a previous discussion of simulating an arbitrarily biased outcome using a fair coin. But let the discussion continue.

I agree with Prime provided the OP is worded, What fraction of integer-sided triangles whose perimeter is 12 are isosceles?

Конечно ... Веревка подбежал шкив, на одном конце была обезьяна, на другом конце вес.Два оставшихся в равновесии.Вес каната было 4 унции / футбол, а в возрасте от обезьяны и мать обезьяны составил четыре года.Вес обезьяна была много фунтов, как мать обезьяны было лет, и весом весом и весом каната вместе были в полтора раза больше, поскольку вес обезьян. Вес вес превышал вес веревки, как многие фунтов, как обезьяна лет, когда мать обезьяны был вдвое старше брата обезьяны было, когда мать обезьяны был вдвое старше брата обезьяна будет, когда брат обезьяны в три раза стара, как мать обезьяны Когда мама была обезьяна была в три раза стара, как обезьяна в пункте 1. Мать обезьяны был вдвое старше обезьяна Когда мать обезьяны был наполовину стара, как обезьяна будет Когда обезьяна в три раза стара, как мать обезьяны было, когда мать обезьяны в три раза стара, как обезьяна в пункте 1. Возраст матери обезьяны превысил возраст брата обезьяны на такую ​​же сумму, как возраст брата обезьяны превысил возраст обезьяны. Какова была длина веревки? So the rope has legs and plays footbal?! And monkey's mother is a male?! And age is money?!Was it Google translation? I guess, it'll be awhile before computers can understand and translate natural language.Although it's worth noting that about the only bit translated correctly, is that very thing, which I misunderstood. The translation states, the weight of the rope together with the weight is 1.5 times the weight of the monkey, as the problem statement intended. Whereas I understood it as half weight of the monkey. I was going to add the caveat that I cannot verify the usefulness of google translate. But yes, I thought all monkeys played football. Actually now I am convinced the puzzle is broken in paragraph 2. Trying to fix it, knowing the answer, but I may just close the topic. Thanks for working on it.

I'll take that. Very nice.

The puzzle may have run its course. Probability is ratio of areas, and the triangle area is an hellacious multiple integral. The problem is known as Sylvester's four point problem. I did find the number, as RG gave it, from simulation, not calculation. Uniform point-picking, necessary to form uniformly distributed triangles was an issue. Equally likely radius and angle gives preference to central points. What was interesting to me when I simulated was that I was off by a factor of four that the four points would not be convex. That is, and should be 4 time the probability the 4th dart lands inside the triangle. There are three other areas the 4th dart can land that make a non-convex configuration, and those areas, on average, are equal to the triangle size. Not an intuitive result until you realize any of the darts can be taken to be the 4th dart. So my last question now does not need to be asked, namely, to compare the probabilities of the two cases in the OP. also, that question has been answered in this thread. Also, the mean triangle area was (to me) surprisingly small.

OK, that works. Any others? What if n was large? Like 100.

You're out with friends at Chuck's Steak House and decide to flip a coin to select one person get a free dinner. The bill will be split n-1 ways instead of n ways. Since I was not invited, I don't know how many are in the group. (Maybe next time you'll include me; I love Chuck's place.) So anyway, your selection method has to work for an arbitrary numbers of participants. You have only a fair coin, and the method has to treat everyone equally. It must be absolutely fair and unbiased. There might be many ways; bonus points await methods with originality, flair, and minimization of flips. Pick one person out of n, fairly, with a sequence of fair coin tosses.

this is what even i think how can you split 45 into two equal parts...P.S- he noted that we cant use 9 as 6... Ah, I misread that. Good catch.

Most people attribute the "truel" puzzle to Martin Gardner. Its charm, like many of his puzzles, it that doing the thinking and the math rewards you with a surprise. The surprise here is that the best shooter can end up the underdog, and the worst shooter can end up the favorite to win. If Alex shoots first we'll bet on him all day long. Having to shoot last is clearly the reason why he might be least likely to win. And that is not a greatly surprising result. But it is not because Cole may shoot first that he can be the most likely to win: it's because he is able to "volunteer" to shoot last -- along with his being the least desirable target. That is the genius of Gardner. I can't add anything to what you guys have said. This was enjoyable.

What I tried to convey by modulo 100 simply was 2013 becomes 13, which then is the straight product of 1 and 13. Being prime and all that is not a requirement, just dates like 5/12/60. You've opened up a richer set of possibilities, which I will now go back and work through. Thanks.