bushindo

Lets restate the problem so we know where we're going. We have 10 coins, 9 of which are fair, and 1 is a 2-headed coin. You randomly pick a coin and toss it. It comes up head. You claim that you gained no information from the toss. Actually, you just gained some information, just that it isn't a lot of information. After you pick up the coin, but before the toss, you have no information whatsoever, so the probability that your coin is biased is 1/10. After one toss and the coin comes up head, the chance of it being a biased coin just increased a bit. Instead of giving calculations, let me present two examples. To argue that you gain information from 1 toss, let's look at the opposite side. Assume that you randomly pick a coin and it comes up tail upon the flip. The probability of the coin being 2-headed goes from (1/10) before the toss to 0. You just gained some information from 1 toss. Suppose you have ten 1000-sided dice. 9 of which have sides numbered from 1-1000. The last die has all sides labelled with 314. You randomly pick a die and toss it once. The number 314 comes up. What die do you think you got?

I think there is a fundamental difference in the thinking of the two camps here. The first camp assumes a priori that p, chance of a head, is fixed at 1/2. Consequently, regardless of the sequence of heads, the chance of the next toss is still 1/2. psychic_mind and others are taking the bayesian point of view, which is that p is unknown. They then estimate p given the data. In that approach, given the 20 heads, the probability of head is very near 0. Given the original wording, I'd go with the second approach. Perhaps this is a rewording of the original problem that makes it clearer. Assume that psychic_mind approaches you and proposes a game. He'll take a coin out of his pocket and flip it. If you can correctly guess the coin's orientation, he gives you 1 dollar. If you're incorrect, you give him a dollar. You figure head is equally likely as tail, so you guess tail 20 times. The coin come up head 20 times, and you're down 20 dollars. Would you guess tail on the 21 try?

Good catch. Back to the drawing board.

Bravo, CaptainEd. I agree that 4 is pretty much the minimum number of turns required for guarantee a sorted array. This is is solved way too fast, I should make them harder next time.

Yup, they do get the obligatory one night of preparation.

This topic is inspired by previous prisoners on a death row problems. There are 10 prisoners. The warden implements a simple game. He will put 100 slips, numbered from 1-100, randomly into 100 identical jar. Each prisoner will be able to open 50 jars and shuffle their contents. The prisoners will each do this individually, and in sequential order. The waiting room and the shuffling room are separate, so those prisoners waiting their turns can't see the jars being opened. Each prisoner does not have to open the 50 jar sequentially. Each prisoner can remove the 50 slips from the jar and place them back in any order he desired. However, at the end, each jar must contain exactly 1 slip and no modification can be made to the jars' appearances, placement, orientation, etc. After open and closing the 50 jars, the prisoner will be moved to a new room and have no communication with the ones who have yet to have their turn. What is the minimum number of turns it takes to guarantee a sorted array of jars? Describe the strategy.

Just thought of some more

Very nice, Geoff. I'm very impressed by your creativity. Since you mentioned food, here's another way

There is a high-rise building in front of your house. You want to measure the building's height. Using common household items, list out the strategies to measure the building's height. (And no, common household items don't include laser measurement tool or a coil of rope as long as the building's height, though I guess watches with atomic accuracy are allowed). I hope this will be more of an exercise in creativity. Members of this board have certainly surprised me with their novel and creative solutions before, and I certainly expect to see creative solutions for this one also.

OBoyOBoyle got it. Well done.

Mr. and Mrs. Smith are friends whom you haven't seem for a long time. The last time you saw them was at their wedding ceremony, where they confided that they wanted to have 2 daughters and 2 sons, and they would have as many kids as needed until they have a minimum of 2 daughters and 2 sons. (For instance, suppose they have 2 daughters in a row, then would then give birth as many times as necessary until they get 2 sons, at which point they will immediately stop having kids. Likewise, suppose their first three kids are son, daughter, and son. They would then try again until their number of daughters reach 2. ) Assume that the probability of getting a son is .5. You haven't seen the couple for years since the wedding. One day, you happened to meet Mr. Smith, who told you that his wife and he succeeded in getting the family they wanted. He then offered you a simple game. If you can correctly guess his total number of children, he would give you 100 dollars. You can only guess once. What is the best guess?

61 is actually the average number of wins for A over 10 repetition of the tournament. Dividing that over the number of games does not give the expect change of winning a single competition. The chance of A winning a single tournament is actually higher than that.

Saw this on another site. Players A and B are taking part in a competition consisting of 11 different games. Each of the 11 games is always won outright by either A or B (i.e. a draw is not possible). 1 point is scored for winning a game, 0 points for a loss. So the overall competition has to be won outright by either A or B. Whoever has a higher score at the end of 11 games wins the competition. Player A's win odds for all 11 games are known (see below), and it follows that Player B's win odds for any game n is [ 1 - P(A wins game n) ]. P(A wins game 1) : 0.9 P(A2) : 0.9 P(A3) : 0.9 P(A4) : 0.9 P(A5) : 0.7 P(A6) : 0.6 P(A7) : 0.5 P(A8) : 0.2 P(A9) : 0.2 P(A10): 0.2 P(A11): 0.1 What is the probability of A winning the overall competition (having a higher total score than B)?

This isn't a puzzle per se, but I just saw this movie scene and it's bothering me about whether it's possible in reality. In case you haven't seen Transporter 3, at the end of the movie, the protagonist encounters a scenario where his car, an Audi A8 W12, plunges into a lake and sinks to the bottom. The protagonist then nonchalantly swims to his trunk and takes out four balloon-shaped canvas sheets. He attaches those canvas to his tires, one canvas per tire, and then releases the air from the tires into the canvas, thus inflating them like balloons under water. The four canvas balloons then lift his car from the bottom of the lake to the surface. What you do think about this scenario? I feel that it's unlikely, but perhaps someone here can positively convince me so or otherwise.

Interestingly, I think CaptainEd's sort and search approach is the optimal solution for the case where the prisoners don't know each others' names. The reason HoustonHokie's strategy gives superior performance is because we are taking advantage of extra information- the fact that each prisoner knows the other 99's names. The sort-and-search approach makes does not use of that piece of information, and thus equivalent to the strategy where the prisoners don't know each other's name. Some interesting extension. Suppose that each prisoner has limited memory and can only remember the names of 49 other people. Let's say that the night before the game, the prisoners randomly divide into two groups, each with 50 members. Each prisoner then memorizes the names of all members of his group. What is the optimal strategy for this situation and what is the expected number of prisoners saved. I expect it to be bigger than 99.45 ( no one knows each others' names) but less than 99.5 ( each prisoner knows all others' names ).

If you're integrating the length of the tape as the sum of successive concentric circles, shouldn't you use the fomula for the circumfence 2*pi*r as opposed to the area of the circle pi * r2?

HoustonHokie's approach is one that I didn't consider. Very nice. However, I think the expected number for his approach will be a tiny wee bit smaller, though. My approach is along the line of CaptainEd's, though with a slight difference.

Some clarification to the original post. Each prisoner, in turn, is allowed to open and shuffle the contents of 50 jars. At the end of each prisoner's turn, he identifies one jar as the one containing his name. The jar is not removed from play, but the warden determines whether the choice is correct or not at the end of every turn, and consequently, whether that prisoner will live or die.