bushindo

Well done, plasmid. Your logic is spot on. I like your shortcut, it is a novel way to think about probability, at least for me. Posts like yours are the reason why I come here.

I'll just make a point that if A were to shoot at 40 paces,his chance of surviving is 40%, regardless of where B shoots. If he were to shoot at 44 paces, his chances of living is now 44% percent, regardless of B's strategy. So an improvement is possible there. I like the way you think,adice, but there must be a math error somewhere. Your method is roughly equivalent to guessing what your opponent's accuracy function would be using some guesses about the population of gun fighters. Obviously if A were to shoot on the 5th step, his chance of hitting is only 5%. Normally,you would only do that if you think that if your opponent is vastly superior to you in accuracy, and that in 5 paces his accuracy already surpassed 95%.

Good job, adiace. Your response for part I is entirely satisfactory. I would like a bit more justification and elaboration for part II, though.

Final's understanding is correct. There are 200 paces between the two men. After 100 paces from each men, the two will be right next to each other.

Two gunmen are standing 200 paces apart. Each man has only 1 bullet in his gun. Every second, the two men take 1 step towards each other. If the first man (call him A) were to shoot, his chance of killing B is k/100, where k is the number of steps he already took. So when k = 0, the two men are 200 paces apart, A's chance of killing is 0. When k=100, which is when the two men are right next to each other, A's chance of killing is 1. B's chance of killing is 1.25*k/100 when k is less than 80. If k is more than 80, B's chance of killing is 1. Essentially, B's chance of killing increases between 0 and 80 paces, and stays constant at 1 after 80 paces. Assume that the two men are in a locked room, and there's nowhere to run. Therefore, if a man were to shoot first and miss, the other man would kill him for sure. 1) Let's say that A and B know about their accuracy functions as well as their opponent's, at which step should A shoot for an optimal chance of survival? At what step should B shoot? 2) Let's say that A and B know their accuracy function, but not their opponent's? Where is the optimal shooting step for A? Where is the optimal step for B?

Here's a question inspired by the NBA finals. While waiting for the Lakers game to start, I made up this problem for fun. Assume that there are four teams in semi-finals, Lakers, Nuggets, Cavaliers, and Magic. Lakers would play with the Nuggets in a best of 7 series. Cavaliers would also play with Magic in a best of 7. The winner of those semi-final rounds would advance to the final and compete in a best of 7 for the championship. Lets say that each game played is a binomial event, with the following probability distribution per game for each team match-up P( Laker win over Nugget ) = .6 P( Laker win over Magic ) = .65 P( Laker win over Cavalier) = .55 P( Nugget win over Magic ) = .45 P( Nugget win over Cavalier ) = .4 P( Cavalier win over Magic ) = .4 1) Lets say that the semi-finals just started. What is the chance that Lakers will win the championship? ( Bonus: give the championship chance for all other teams as well) 2) Let's start from the current team score, where Lakers are up 3-2 versus Nuggets, and Magic up 3-2 versus Cavalier. What is the chance that Lakers will win the championship? Chance that Cavaliers will win the championship? (Bonus: give the championship chance for the other two teams too)

Please ignore the fact that there are two images in the OP. I had two images of the same puzzles, and added 1 to the text of the post. I didn't realize that the other attachment would show up too until it is too late to edit. Work on whichever puzzle image you like. They are both the same.

No credit to me whatsoever, James8421 brought my attention to this puzzle, which was made by sam loyd. I haven't solved this one yet, so ask James if you need hints. Euler, the great mathematician, discovered a rule for solving all manner of maze puzzles, which, as all good puzzlists know, depends chiefly upon working backwards. The accompanying puzzle, however, was built purposely to defeat Euler's rule, and out of the many attempts is probably the only one that thwarts his method. Start from the heart in the center. Go three steps in a straight line in any one of the eight directions, north, south, east, west, northeast, northwest, southeast, or southwest. When you have gone three steps in a straight line you will reach a square with a number on it, which indicates the second day's journey, as many steps as it tells, in a straight line in any one of the eight directions. From this new point, march on again according to the number indicated, and continue on in this manner until you come upon a square with a number which will carry you just one step beyond the border, thus solving the puzzle. To post your solution, just list out the successive list of directions you need to take, using the 8 directions north, south, east, west, northeast, northwest, southeast, and southwest.

Ah, there it is. I did a search for "cheating husbands" and found nothing in the first 2 pages, so I assumed that the problem wasn't posted. Turns out I should have searched for "unfaithful wives". Good job to joe, james, and adiace. Anyways, the second question still stands. Let me clarify the second one. Every wives already knew of some cheating husbands, so the missionary didn't tell them what they didn't know. Explain where the extra information entered the system. Adiace made a stab at the second question, but I feel it is unsatisfactory.

Searched for this in the archive but didn't find anything. On an remote island there lives a tribe. The tribe has a particular custom. If a wife discovers that her husband has been cheating on her, she would shave his head that very night and paint it red. There are 9 cheating husbands on the island. The wives gossip among themselves, but they never tell any wife the status of her husband. As a result, each wife (cheated and non-cheated alike) knows the cheating status of every man except her own husband. One Monday, a missionary arrived and went drinking in a bar with some of the tribe's men. Under the influence of alcohol, some men let it slip that there are a few cheating husbands (but didn't say which ones) on the island. Before the missionary left, he assembled the tribe and told them "There are some cheating husbands among you". A couple of days passed and no shaved men appear in the morning. Then one day, exactly all 9 cheating men appeared with shaved and painted heads. 1) What day was it when the 9 cheating husbands appeared with the shaved heads? 2) This is more of an information theory question. Each wife already knows about the other cheating husbands. The missionary told them what they already knew. Why then did this happen?

I see now, that would take advantage of the prior knowledge that there are only 10 balls. Well done.

But what if the last ball is in the last 4-jar sequence?

Here's a joke I heard from a long time ago. In a certain family, a boy lived with his mom, dad, and grandparents. One night, the boy had a violent dream and screamed in his sleep, "Grandpa, don't go!". The next morning, the boy's grandfather died at the break of dawn. The father was absolutely dejected over losing his dad, but he dismissed the boy's midnight scream as a coincidence. A while later, the boy again cried out in his sleep, "Grandma, don't go!". Sure enough, the very next morning the grandma died at the break of dawn. The family was disconsolate. A while later, the boy again cried out in his sleep, "Daddy, don't go!". The dad was absolutely dejected, and started making his will that very night. He spent a tearful night saying good bye to his wife and the boy, and as morning came, laid down to prepare to meet his Maker. To his delight, morning came and he was alive and well. He ran down the street, crying "It's good to be alive", only to be yelled at his neighbor's sobbing wife, "Would you quiet down? My husband just died this morning." (Wait for laughter) Here's the brain teaser, is there something odd about this joke?

I nominate this solution to the Hall of Fame.

Incidentally, when I was saw the spoiler title, I thought that CaptainEd was claiming that no one had a solution by the time he posted it. Anyways, good job to nobody. It is very impressive that you were able to solve it without prior knowledge of graph theory. If you were born before euler, there would be a theorem called "nobody's graph theorem" by now.

Phatfinger got two solutions, both of which are good. Good job to kidsrange, TerriDawn21, and Nitin Agrawal, who posted about the moving knife method, that's a very slick way.

If A, B, and C agrees to a schemes where A cuts, C picks first, B second, and A last. It is possible for A and C to collude to get the entire cake.

How would they divide those 27 pieces? Who gets to pick first, and then how would it go from there?

I assume you meant why would A conspire with C in order to get the smaller piece. The assumption here is that if these two were really colluding, they would share their collective pieces after B goes away.

The cuts don't have to be straight cuts, and there don't need to be only 3 cuts. If the first decide to cut the cake into many pieces, how would the division process go?

3 people are dividing a cake. Each person wants as much cake as possible for himself, but each person thinks that the other two might be colluding. Lets call the three A, B, and C. A suggests that he will cut the cake into 3 portions, and then C will pick his piece from the 3, and B then would pick his choice from the 2 remaining. and A takes the last piece. B objects to this scheme, saying that A can cut the cake into 1 large piece, and 2 smaller equal pieces. C then would have the largest piece, and B would have to take the smaller piece. In fact, B claims, If A and C were really colluding, this method would allowing them to get the entire cake to share among them two. Given this distrusting atmosphere, is there a way to divide the cake so that each person is satisfied? Any proposed method would have to convince each person that he would get his fair share even if the other two were colluding. I know of two solutions to this, but I wouldn't be surprised if the den can come up with a couple extra solutions.

now bushindo your answer i think is if the number of locks needed to unlock increased on 4 missed regardless if there was one unlocked the real average of that im pretty sure is quite lower. Im sure you could prove me wrong tho. Also Ive forgoten how to set up that matrix equation (or i never knew i think i remember it tho) if you could explain it quickly;y please do so if its hard tho you dont have to bother Sure thing