bushindo

I am not sure if i am within a whisker of the right answer or am on a different continent altogether. Will somebody please put me out of my misery!!! Hi ReArtemis, sorry for the misery. Your attempt, if I understand it correctly, if to encode the parity (odd, even) of every 5 jars with 1 prisoner. That is a good approach, but it doesn't completely solve the problem. If you want a solution, then you can read the previous posts, especially adiace's post. if you want a hint, here's one

Shouldn't the survival probabilities for A-E add up to 1? The game requires that only 1 person survives, so survival probabilities are mutually exclusive. I think you're missing a 0. E's chance of killing anyone is 16/20 = 80%. A's chance of killing anyone is only 4/20 = 20%. A can't be the best place to be because eventually the game comes down to a 2-people duel. Even if A is among the last two, his low chance of killing makes him lose way too often.

Very close. I have a ballpark estimate, and it's pretty close to yours. How did you do that?

I promised final and adiace a break after their exemplary effort on Prisoner on a death row 8. Well, break's over. There are 5 prisoners on a death row, call them A, B, C, D, and E. The warden gives them a chance to live. He gives them each a doctored gun and let them engage in a death match. Let's say the guns are modified so that their chances of hitting varies. A's chance of hitting and killing any other player is always 4/20. B's chance is always 7/20, C's is 10/20, D's is 13/20, and E's is 16/20. Assume that every single shot will either miss or kill. The players shoot in this order, A, B, C, D, and E. Unlike Prisoners on a death row 8, prisoners don't have a choice of shooting anyone they want. Everyone must shoot someone right after him in the sequence. For example, A must shoot B, and if B is already dead, A must shoot C if C is alive, or go on to the next person if C is dead too. The person last in the sequence must shoot someone at the start of the sequence. So for instance, let's say it's E's turn, he must shoot A, or the next alive person at the start of the sequence if A's dead. The game continues until there's only 1 person left. The last person standing earns his freedom. The warden likes you, so the night before the game he allows you to pick your position as A, B, C, D, or E. 1) What position should you pick? Super hard bonus: What is the exact chance of survival for A, B, C, D, and E. Any method is allowed, but probabilistic method like simulation is not allowed since simulation provides an estimate of the rate but not the exact numbers. I'm not after the numbers, though. I'm looking for methods that would allow us us to compute the exact chance of survival. You don't have to implement the method, but you must describe it, and it must be doable within a reasonable amount of time.

Did you count symmetries, and did you make sure not to count the order of placement (order is not important)?. This problem should have a large number of solutions, since 8 isn't the maximum number of bishops you can put on the board. The max is 13 or 14, i think. I haven't worked a solution out completely, but maybe post your solution and we'll evaluate the logic?

Just to add something, there are infinite real solutions on the real and imaginary plane. Taking the log and re-arranging, you get x/log(x) = y/log(y) Here's a graph of x/log(x) http://www93.wolframalpha.com/input/?i=x%2FLog[x] Here's the equations for the solution function http://www93.wolframalpha.com/input/?i=x%2...+%3D+y%2FLog[y] Wolfram alpha is like the nerdy search engine, sitting in the corner at prom in weird gaudy clothes, while the popular Google is in the middle of the room talking up the ladies.

yeah, sorry about that. Let's say that the rain density is 200 drops/meter3, or 200 drops per cubic meter. Thanks final.

Usain Bolt

I've always wonder whether it is better to run very fast through a constant rain, or simply to walk through it. Intuitively, running leaves you less soaked, but how much of an improvement are we talking about here. This question will settle it once and for all. Assume that all rain drops are identical in size, and that they are falling at 9m/s straight down vertically. Their density is 200 drops/meter2. You'll be walking or running through the rain. Let's make it simple and assume that you are a rectangle, and that all water that hit you will either hit in the head portion, or the chest portion. Let's say that your head portion has a surface area .06 meter2, while your chest potion has a surface area of .5 meter2. Let's say that water drops from head hits and chest hits count the same. Now, you are supposed to get across a 200 meter field, 1) Let's say you take your time and walk at 1.5 m/s, how many drop of water would hit you during your walk? 2) suppose you ran across at a speed of 8 m/s, how many drops of water would hit you during the run?

For the bishop, you're assuming that each placement of the bishop takes out 8 possible spaces on the board. However, depending on location on the board, bishops take out more than those. Consider the first placement. If the bishop is placed in the center of the board, let say at position (4,4), it would remove 14 positions from the board.

The following questions assume a 8x8 chess board. Starting with an empty chess board, 1) How many different ways can you place 8 rooks so that none can attack another? 2) how many different ways can you place 8 bishops so that none can attack another?

Copied and pasted from final's work. A shoots E first: probability of survival A-E (0.241, 0.3076, 0.30339, 0.12824, 0.01977) A shoots C first: probability of survival A-E (0.23834, 0.29741, 0.25777, 0.16702, 0.0394) Looks like if A shoots C first, he lowers his own survival rates by about 1 percent, but he will drastically improves D's survival rates (33% increase) at expense of C (16% reduction). It is probably to D's advantage to slip A a couple of cigarattes before the game to bribe him. I'll take your results over mine. I only ran the simulation only half a million times. So I guess we'll declare B the best spot to be, since it has a slightly better chance, and because it is not too sensitive to bribing, as we just discussed above. That is, unless adiace comes up with a exact analytical solution to this problem.

2 prisoners are on a death row. The warden gives them a chance to live. He shows them a gun that has a cylindrical barrel with slots for bullets (kind of like a six-shooter in old western movies except this gun has more bullet slots). He will slip a bullet into a random position in the gun barrel, and then have the prisoners take turn shooting at each other using the same gun. As you know, the first shot is most likely a blank, since there's only 1 bullet in the gun, thus the chance that the bullets fires on the first shot is low. However, as the prisoners take turn pulling the trigger, the chance that the bullet fires goes up accordingly. The person still standing after the gun fires wins his freedom. If the bullet fires, assume the prisoner the gun is aimed at will die. On the day of the game, the warden show them two guns- one with 10 bullet slots, and one with 11 bullet slots. He then says that the prisoners will pick from the following two options. If a prisoner picks one option, the other options falls to his fellow prisoner by default. A) pick a gun (10-bullet slot or 11 bullet slot ) to use during the game B) choose to go first or second. 1) Suppose that you are prisoner 1, and you are required to pick first. What option do you pick, A or B, and what choices within that option? 2) Suppose that you are prisoner 2, and you get to pick second. Your son-of-a-gun prison-mate made the same choices as the answer in part 1, what would you pick now?

i dont know where u get these problems bushindo but my sleep schedule hates u Your programming is correct. Well done. I have the following slightly different number, but that's because we're running simulations I'll tone down the computational aspect in the next couple of questions. I figure you and adiace deserve a rest. I swear you two alone have solved about 90% of my problems.

For what it's worth, I got to give you and final props for tackling this problem. These results so far looks right for the number of iterations you have, but the decision tree is pretty deep, and you probably don't want to fully extend down the entire tree. Another route is to divide and conquer. Consider the case where there are 3 players with their known shooting probabilities, and you can probably derive exact analytical expectation for their survival rates. You'll see then that your game tree becomes a lot more managable since for any node of the tree where 2 player dies,you can substitute the 3-players result in. An easier route is to do simulation, which is surprisingly easy once you apply the decision rules you derived earlier (the ones in bold). Since the decision rules apply for every single player, you can simulate the game in some short code, assuming that you're using a math-oriented language. Running the game like half a million times should tell you should player has the advantage. that's my prefered route, at least.

yes, for the bonus question, the game is entirely the same, proceeding in order from A to E. Except that you get to have your choice of position. Good work on part I, by the way.

Excellent, the answer to part I is entirely satisfactory. Good work to adiace and final. Now there's still the second part.

Assume unlimited bullets. And players will continue to get turn on the next round if they miss. Players continue shooting until there's only 1 player left.

5 prisoners are on a death row. Let's call them A, B, C, D, and E. The warden gives them a chance of living. He gives them each a doctored gun and let them engage in a death match. Let's say the guns are modified so that their chances of hitting varies. A's chance of hitting and killing any other player is always 1/5. B's chance is always 2/5, C's is 3/5, D's is 4/5, and E's is 5/5. Assume that every single shot will either miss or kill. A player must shoot someone on his turn. Each player knows his gun's accuracy rate and the others' as well. The players take turn shooting in the following order: A, B, C, D, and E. During his turn, a gun slinger can choose to shoot at anyone he wants. If a player is killed, then the order of shooting will continue in the same sequence but with the dead player skipped. Players take turn shooting until there is only 1 player remaining. Assume that each player wants to maximize their own chances of living, and that each player knows that the others will do the same. Answer the following 1) What should A do on this first turn? 2) What should B do on the second turn? 3) What should C do on the third turn? Super hard bonus: Suppose that the warden likes you, so the night before the game he allows you the chance to choose your gun. Essentially he allows you to choose your position as A, B, C, D, or E. Which position should you choose to maximize your chances of living, assuming that everyone plays optimally? Assume that the other prisoners don't know about this so they won't unduly target you out of spite.

The average gun slinger accuracy function should be defined similar to how you approached it, namely by imagining what a population of gun slinger might look like and then taking some sort of 'average'. The problem with your answer for part II earlier, I think, is that your assumption is way too pessimistic. You allowed the multiplicative constant p to vary from 0 to infinity, but that essentially says that A and B are near the absolute bottom of gun fighting skills spectrum, that piece of information, however, isn't warranted by the original post, though. In real life, this problem is a bit more intuitive. If a gunslinger has prior encounters with other gunslingers, he can probably estimate how good he is compared to an typical gunslinger. When A meets B, from A's perspective, if he knows nothing about B, he should assume that B is an average gun slinger. A then can compare his accuracy function to the guessed average function for B and decide accordingly. That way, if A is an amateur or the best in the world, he can make good use of that information.

Just a question, what if the interception is in the last 5 prisoners, how do the prisoners handle that? This approach assumes that the each prisoner knows whether their communication has been compromised, and that each prisoner can communicate with his mates afterwards. It is not what I intended but it poses a good question, so please consider it. Don't forget to work on the harder case too, which assumes that the interception is random, and each prisoner, on his turn, doesn't know whether his communication has been compromised.

There are 32 prisoners on a death row. The warden gives them a chance to live. In one room, call it room A, the warden puts 26 jars and put a unknown number of balls inside the jars. Each jar is either empty or contains 1 ball. In room B, the warden puts 26 jars, and 26 balls in a separate stack. He divides the prisoners into two groups, 1 group of 31 and 1 group of 1. The group of 31 goes inside room A, and the other prisoner goes into room B. There is a lever in room A, which is connected to two lights in room B. Depending on how one pulls on the lever, it will either make a red light or a green light goes up in room B. The game proceeds as follows. Each prisoner in room A will take turn going up and examine the 26 jars away from the sight of his fellow prisoners. He can not rearrange or modify the jar in any way, shape, or form. He then must pull the level and make either a red light or green light goes up in room B. If he attempts to communicate any other information besides those two bit of information (i.e. waiting a certain amount of time before pulling the lever, pulling the level more than once, not pull the level at all, etc. ) all prisoners will be executed immediately. The order in which the prisoners take their turn at the jar is randomly determined by the warden. If the prisoner in room B can reconstruct the arrangement of balls inside the jars in room A at the end of 31 turns, all prisoners will live. However, there's a catch. The warden will randomly intercept the communication of 1 prisoner from room A to room B and flip it. For example, suppose that the warden choose to flip prisoner 16's communication. Let's say that number 16 examines the jar and choose to make the green light goes up, the warden would intercept the electronic signal and make the red light goes up instead. Likewise, if number 16 were to choose to make the red light goes up, the warden would make the green light goes up instead. The 32 prisoners are informed of this game the night before, so they know that the communication of exactly 1 random person will be compromised during the game. They have 1 night to prepare. Is there a strategy to guarantee the survival of all prisoners? Describe it.

now part two is different if you had to hard code a method then shoot at .50 but if you can make some gut calls then thats different Well done, final. Your answer to part II is satisfactory. I also like the extension to the case where A or B thinks he's a bit better than the opponent. In general, even though A or B may not know about their opponent's accuracy function, a reasonable approach is to assume that your opponent is an average gun shooter, and base your decision from there. An interesting extension from here is that if A or B gets an impression about their opponent's skill (better, worse, equal), then they should incorporate it into their consideration, as final have suggested. This allows interesting ramifications. If A and B can communicate together, and if B thinks that he is weaker than A, it is to B's benefit to bluff and pretend to be a stronger shooter than he really is.

Sorry for the double post, i waited too long and the edit function went away. This approach is the answer to this question, "If A and B were to shoot simultaneously at step k, what is the best place for each person to shoot so that he lives and the other one dies?" Now, let me clarify my statement earlier. If A were to shoot at 44 paces, he is guaranteed minimum chance of living of 44%. If he were to shoot at 40, his minimum chance of living is 40%. See adiace's graph for the proof. I think you might be missing the key info that there is only 1 bullet in each person's gun, and that if the person who shoots first misses, the second person will have a guaranteed 100% chance of killing, since we assume that it's a closed room and there's nowhere to run.