bushindo

The chord, defined as the segment through the randomly chosen point and perpendicular to the line connecting the center to a randomly chosen point, should pass through the point. I include an image because a picture is worth 1000 words, plus or minus about 999 words or so. The ambiquity in the original post is similar to asking, lets say we 'randomly choose' a point in the interval (0,1). What is its average value? There are an infinite number of density distribution available for sampling, such as the uniform distribution, gaussian, cauchy, poisson, and so on. And all of them are considered 'random sampling'. Thus we see that the answer depends on the random distribution we choose. As far as choosing a random chord goes, there are an infinite number of ways to sample a 'random chord' through the circle. Some additional ways to sample a random chord are * Let the circle be drawn centered at (0,0), with radius 1. Sample the chord from the family of lines y = ax + b. The prior distribution on a and b will also change the answer accordingly. * choose two random points within the circle's area. Connect the two points with a line. That is your chord. * inscribe the circle within a square. Construct a chord by randomly choosing two points on different edges of the square and connect.

No, the antidote would still be efficacious if drunk before the poison. I might be opening myself to loop holes here, but for clarify's sake let's say that the antidote has an effectiveness window of 2 hours. Any poison in the drinker's body system will be removed during that 2 hours. Therefore, drinking the antidote on the first day does not grant immunity from poison for the entire 48 hours.

This is based on Sourabh's poison puzzle, which i must say is one of my favorites. Let's say that you have n barrels of wine, one of which is poisoned and one of which contains the antidote. If a prisoner drinks from a poisoned barrel, he dies in 24 hours. But he drinks from the poisoned barrel and the antidote barrel, he'll be fine. Drink from the antidote barrel alone has no ill effects. Let's say you have 5 prisoners that you are willing to sacrifice, and you got 48 hours. What is the maximum value of wine barrels n for which you are guaranteed to find the poisoned barrel?

As hall and bonanova mentioned, the key lies in how the 'random chord' was chosen.

I think this can save up to 243 prisoners. I don't know what's wrong with the spoiler function. This is weird. I'm sure the tags are right.

Shouldn't it be a bit less? You're probably thinking that 1/8 = .12000

You meet a guy in a pub. After a couple of beers, he offers you a chance to play a game. The game is as follows: at the beginning every game, you pay him X dollars to start. You then would pick a number from 1 to 10. He would roll five 10-sided dice, and pay you 1 dollar for every time that your number comes up. So instance, suppose that you guess the number 3, and the five rolled dice come up with the numbers (1, 9, 3, 3, 10). You would then win 2 dollars. 1) The guy in the pub offers you the chance to play the game if you pay him 0.90 dollar in the beginning. Would you play? 2) What the the maximum dollar amount X that you would pay to play the game?

Ah, I see what you're saying. All this time I thought that P1 denotes one person, while it actually denotes the first person to arrive. Mea culpa. In that case

Here's another attempt

Suppose person A plans to arrive exactly 12:30, but he is delayed for 39 seconds by some bum asking for change. When he arrived the hour is 12 o'clock 30 minutes and 39 seconds, what is his chance of meeting B then?

There is a flaw because a picture is worth 1000 words [spoiler=]Probability of meeting as a function of P1 arrival time

I agree with simulacric_bro

I remember long ago McDonald used to run an ad campaign after every major movie. They would make a set of, say, 10 different figurine from a specific movie, and include supposedly a random figurine in every kid's meal. The exciting encouragement after the commercial is always "Collect all 10!". So, here's the puzzle. Suppose that you are a kid who wants to collect all 10. Assume that every time you go to McDonald, they pick a random figurine from the set of 10 and include it in your meal. How many times, on average, do you expect to whine and drag your parents to McDonalds before you get a complete set of 10?

You can flip whichever coin you like at any turn. The goal is to be able to identify any coin with 95% correct probability. Obviously, if you happen to start with the 1/5 or 4/5 coin, it would take less flips to be certain of the coin's identiy. The interesting part here is whether you should just pick a coin and continue flipping until it's identity is clear, or should you flip each coin once or twice first to get some information about their identities.

The interpretation in part a) is correct.

You have 4 biased coins. They look identical. They have the following respective probability to land up head: 1/5, 2/5, 3/5, 4/5. You accidentally jumbled the four coins together and now you don't know which is which. Lets say that you have to determine the coin's identities by flipping them and observe the outcomes. Assume you can flip as many times as you like, and you can choose any coin at any flip. 1) what's lowest expect number of flip required to identify 1 coin with 95% certainty? that is, you want the lowest expected number of flips to correctly identify 1 coin 95% of the time. Hard bonus. 2) What is the lowest expected number of flips required to correctly identify all 4 with 95% certainty?

This applet really helps. Make sure that you overlaid a grid on the applet, and drag the points around to see how the area varies. Put your mouse over a point until the icon changes into a cross-hair, then you can click and hold to drag the point around. http://www.voronoigame.com/VoronoiGameApplet.html

Four friends are playing a game. They have a pool of 100 dollars. Let's say that they have a unit square, with the corners located at (0,0),(1,0), (1,1), and (0, 1). Each player would sequentially take turn marking a point on the unit square. After the 4 markers are placed, the square is divided into 4 region, each centered at one of the markers. The rule for region division is that if any point on the square has the smallest distance to marker i, then it belongs to marker i. Apply that to the infinite number of points consisting the square, and you have the area associated with each marker. At this stage, each player wins an amount of the $100 that is proportional to their polygon's area. For instance, if player A mark a point so that he has 50% of the area, he wins $50 bucks. The remaining money are divided proportionally between the other 3. Here is an example. Here, there are four point marked on the square, and the square is divided into 4 regions by the rule defined above. Note that boundaries between any two regions are actually lines that are equidistant from the two marked locations. The pink point has the largest area (about 35%), so the player who placed that point wins about $35. 1) Suppose that you are playing this game, and you have to go first. Where would you mark your point? Assume your opponents will each try to maximize their winnings, with no collusion between any two party. 2) suppose that you are playing this game, and you're offered the choice to go first, second, third, or last, which would you pick to ensure the highest expected winning? Assume your opponents will each try to maximize their winnings, with no collusion between any two party.

I think I have an edge in interpreting final's solution because it is along the line of what I was thinking when I posted the puzzle. It's easier to understand someone's instruction when you have a similar roadmap. I think what final was trying to communicate was the following: Let's label all the prisoners 0-19. Divide the prisoners into two groups: the first group with prisoner 0 and second group with prisoners 1-19. To win the game, the second group need to communicate Prisoner 0's number to him, and prisoner 0 needs to communicate X to the second group, where X = modulo( P1's hat + P2's hat ... P19's hat , 20 ). For any prisoner in the group P1-P19, if he knows the remainder of (sum of hat 1 to hat 19) /20, then he can figure out his own hat number, since he can see the other 18 in the group. final's strategy is as follows 1) starting from P1 and going to P19, let each prisoner raise his hand in order, allowing a small pause in between. 2) If any prisoner's index is the same as prisoner 0's hat, then he won't raise his hand at all. So we are guaranteed to have at least 18 prisoners raise their hand. If all 19 raises his hand, then prisoner 0 knows that his number is 20. Otherwise, prisoner 0 can get his hat number from the position of the guy who didn't raise his hand. 3) Prisoner 0, at the same time, would compute X, the sum of hat 1 to hat 19 modulo 20. Prisoner 0 then would communicate X to the rest of the group by raising his hand at the same time as the prisoner whose index correspond to X. In his example, if X = 2, then prisoner 0 would wait until P2's turn, and then raise his hand at the same time as P2.

Technically, at any flip, it is guaranteed that your number will either be greater than 1/pi or less than 1/pi. The proper early stopping conditions should be if your binary number X is greater than 1/pi, or if X is less than 1/pi regardless of subsequent flips. The second condition makes it messy to calculate the average required number of flips, so Im just going to go out on a limb and estimate the expected number of flips as 1 + 1/2 + 1/4 + 1/8 ... = 2 flips.

This will guarantee any desired degree of accuracy

so something like (20/21)^20 everyone lives or 20/21 individually I like this answer. This sounds like it would save all the prisoners 100% of the time. Well done.

The use of the time variable is a reasonable solution, but that wasn't the intention of the puzzle. Assume that the prisoners do not possess watches, so they can't synchronize their mental clock down to the seconds, or intervals thereof. Let's say that the game is structured so that each prisoner only has access to the following information: the other 19's hat numbers, and the ordinal sequence ( with the time duration between any two elements stripped) of the prisoners getting ready.

Twenty prisoners are on a death row and are allowed a chance to live through a game. The game is as follows. Each prisoner, as they enter the room on the day of the game, is given a hat with a random number ranging from 1 to 20. The numbers of the hats are independent of one another. The prisoners then stand in a circle, and each one can see what the other nineteen have on their hat, but not his own hat. After the prisoners stand in a circle, the prisoners are allowed 3 minutes to think of a guess for the hat they are wearing. At any time that a prisoner is ready with his answer, he can raise his hand to indicate that he is ready. The other 19 prisoners, being in the same room, can see whenever someone raises his hand. Assume the prisoners can not communicate with each other in any other ways such as raising his hand with different number of fingers extended, raise it with different speed, make his fist into a prearranged shape, etc. The only information available to each prisoner is the other 19 numbers, and the order in which his fellow prisoners indicate they are ready. After 3 minutes, or after all 20 indicate that they are ready, whichever is sooner, each prisoner is required to write his guess for his hat number without being seen by the other 19. All prisoners write down their guess simultaneously. If all 20 guesses are correct, the prisoners win and all are released. If 1 or more is incorrect, all 20 will die. With optimal strategy, what is the chance of winning for the prisoners? Describe the strategy.