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bonanova added a question in New Logic/Math Puzzles
Bent octagonA convex octagon completely contains all 28 of its diagonals. But if you move the vertices around you can make portions of some, but not all, of them pass through its exterior. What is the largest number of such diagonals? Equivalently, how many diagonals of an octagon must remain totally in its interior?
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bonanova added an answer to a question Minimize the perimeter
Thinking about this ... One might make the case for three different values of perimeter for three different placements of the "hole," yielding a distinct best answer.

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bonanova added an answer to a question Random bullets
Yes, this one has a definitive answer. I'm not sure consensus was reached on the infinite puzzle.

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bonanova added a question in New Logic/Math Puzzles
Minimize the perimeterA child's game uses twentyone unique shapes that comprise from one to five squares. This puzzle asks how tightly can the shapes be packed without overlap so as to achieve a figure with the smallest perimeter? The individual shapes may be rotated and turned over (reflected) as desired.
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bonanova added an answer to a question Double up!
araver has it.

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bonanova added an answer to a question dividing infinity by 17
I was going to ask which direction the exponentials are evaluated.
But obviously 3 is first raised to the 4th power; that result is then raised to the 5th power; and so on.
Only finite power towers can be evaluated right to left.

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bonanova added an answer to a question Random bullets
A little low.

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bonanova added an answer to a question Badminton

bonanova added an answer to a question Random bullets
Hi dburns, and welcome to the Den.

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bonanova added a post in a topic Rest in Peace, Games Forum
I'm traveling until September 5. After that I'm available to play again.

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bonanova added an answer to a question Double up!
I don't like the default sort by rating thing either. Site Admin is working on changing it to chronological.
araver, that is a valid assumption because the goal is to *assure* a given number of rounds.
If two are equal at the end of a round, they *could* be the active players on the next round.

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bonanova added an answer to a question Double up!
Yes. You understand, and now everyone does.
Also, you hold the (for now) record of N = 4. And there is now a target on your back.

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bonanova added an answer to a question Random bullets
You are correct that there was a similar previous problem. There, the gun was never turned off, and the question was the probability of at least one bullet not being annihilated. That thread had a lot of interesting discussion and simulations. Reading it now would not be considered cheating since for this problem a closed form is asked for. I tried searching once, without success, but I'm sure it's there.
Your observations are correct so far.
I am going to make the 4bullet case part of the puzzle. I think it's simple enough to think through while having enough complexity to guide an analysis for the 10bullet solution. So there are two questions now: What are pzilch(4) and pzilch(10)? Where pzilch(n) is the totalannihilation probability for n bullets.

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bonanova added an answer to a question Double up!
Yes.
I used turn and round to mean the same thing. Sorry for the confusion.

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bonanova added a question in New Logic/Math Puzzles
Double up!Three people play a money game with (perhaps differing) initial stakes in the range [$1, $255]. On each turn they flip a fair coin to determine an oddmanout. (If there are three heads or three tails they flip again.) The poorer of the remaining players then "doubles up" at the expense of the richer player. For example: If the initial stakes are ($15, $70, $150) then depending on which player sits out, the stakes at the end of the turn would be ($30, $55, $150), ( $30, $70, $135) or ($15, $140, $80).
The game ends when the two active players in any round have exactly the same stake.
For each set of initial stakes there is a minimum number of turns that must be played. In the above example, clearly two rounds must be played (more than two, actually.) Starting with ($1, $4, $6,) at least three turns must be played.
Your task is to improve on that. Find initial stakes for which the game cannot end before N turns have been played, where N is the greatest number you can find. If you can beat N=3, post your solution and let others take a shot at beating it. Gold star will be awarded if the best possible N is found.
(This was adapted from a puzzle in a wellknown puzzle site, which I will attribute when the best solution is found.)
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