bonanova

I want to solidify the idea of using expected speeds to establish the accuracy of the (1/2)(3/4) ... formula [1] It gives an exact result. As does the formula. So if there's any difference, no matter how small, it's wrong. It must agree exactly with the formula in order to be correct. That criterion is not available to us when we compare the formula to the result of a finite number of random cases. [2] It gives the correct result for the four-bullet case, as shown here: [3] I think the six-bullet case could also be done by hand [4] I would guess the ten-bullet case could be done by computer.

I'm not sure which was fired first but either way you can see that two bullets survive. .4 .8 .6 .2 Assuming .2 was fired first - that they will travel to the right.. .6 will catch .2 (before .8 catches .6, owing to the greater speed difference between .6 and .2 ) and they annihilate. That leaves .4 and .8 to survive. Assuming .4 was fired first - that they will travel to the left .8 catches .4 and they annihilate. That leaves .6 and .2 to survive. Assigning expected speeds allows you to calculate collision times, and thus annihilation condition, once for each speed permutation, instead of a million times using random values. It may, but I have been unsuccessful, lead to a derivation of the formula (at least for small n.) Maybe it's still not clear. You just enumerate and analyze the speed permutations once, and you're done.

I disclaim authorship, since BMAD posed the "infinite bullet" version here last year, and I read the finite bullet version elsewhere, along with the conjectured solution. I share Captain Ed's amazement at it. Here is my contribution. Change random speeds into expected speeds. That eliminates need for simulation. There is a neat result that says for n random speed [0, 1] bullets the expected speeds are i/(n+1). The order they occur defines all the representative cases. And then you're done. (In this approach you know values in addition to ordering.) Although for large n you're back to millions of cases. For small n, though, maybe that formula can be derived. I haven't succeeded yet. Clarifying: The expected speed of the fastest of n bullets is n/(n+1) The expected speed of the slowest of n bullets is 1/(n+1) The expected speed of the ith-fastest bullet is i/(n+1)

There is a conjecture for the probability p0(n) where n=2m for all n bullets to be annihilated. I do not have a proof. Now that there are two programs that are giving the answers for n=4 and n=10 that agree with p0(4) and with p0(10) obtained from the conjectured formula, I think I will give the formula. It can be tested against simulations for other values of n, and its simple form may suggest a line of attack to derive it analytically.

I followed most of this on first read. Nicely done and clearly commented. Thanks. I'd like to compile and run it. What do I need for that?

dgreening points out that the same can be said if the last bullet is slowest.

I just thought that a generalization of "perimeter" to mean the length of the boundary between the interior of a shape and its exterior would give

@bubbled, Nice job. Would love to see the code. Actually I'm trying to learn python. @araver, belatedly your gold star!

Yeah, my reasoning was wrong. One can say that matches are more likely to end with one of the scores. Julie's half of those matches are then more numerous than her half of the matches that end in the other score. Nice little puzzle.

That's a good result. Curious, what language did you do your simulation in?

A 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?

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.

Yes, this one has a definitive answer. I'm not sure consensus was reached on the infinite puzzle.

A child's game uses twenty-one 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.

araver has it.

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.

A little low.

Hi dburns, and welcome to the Den.

I'm traveling until September 5. After that I'm available to play again.

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.

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.

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 4-bullet case part of the puzzle. I think it's simple enough to think through while having enough complexity to guide an analysis for the 10-bullet solution. So there are two questions now: What are pzilch(4) and pzilch(10)? Where pzilch(n) is the total-annihilation probability for n bullets.

Yes. I used turn and round to mean the same thing. Sorry for the confusion.

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 odd-man-out. (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 well-known puzzle site, which I will attribute when the best solution is found.)