bushindo

Link to detailed answers by people with physics degrees

Haven't tested it out to see how many would say the right color, but that's my first idea. But what about the Red hats? Here's another method

That was a good solution. Serves me right for not thinking outside the box.

By the way, if the kid takes 10 minutes to write a program to "to print all the even numbers till 500", he's not a smart kid, as you claimed.

Not to beat a dead horse, but even assuming the above scenario there is a way to convey information between the prisoners. Let's say that when each prisoner is ready, he presses a button, but no other prisoners can see the act. The light will dim when all buttons are pressed to let the prisoners know that all buttons have been pressed. Let all prisoners except the first prisoner (P1) immediately press the button to indicate that he is ready when the game just starts. P1 then looks at all the hats he sees and convert that into a base-N number, call it x. Wait x seconds before indicating that he is ready, thus dimming the lights. All prisoners except P1 can then factor his own number from x. Another possible way to solve this without using the time variable is to assume that each prisoner can see the others indicating whether they are ready or not. The order in which the prisoners indicate they are ready may convey some information also. But not as much. There probably is a flaw in the logic or wording of this puzzle, as final have observed. With absolutely no communication between the prisoners during the game, which the original post seems to suggest, there is no way to improve the (1/n)^n odds.

This seems to indicate that each prisoner may, at any time he chooses, indicate to the warden or the official that he is ready. I'm making the reasonable assumption that since everyone is in the same room, all prisoners can see when one of their own indicates that he is ready.

This part isn't clear to me. I assume that the prisoners don't go in sequential order and call out their guesses like the typical prisoners and hat puzzle. Also, do the prisoners get to pick when to call out their guesses, or is there a set time limit to consider their guess?

I cheated on this problem...

Here's part B of the question. What is the chance that we would get the above data if we assume B) That the chance that a drug is stolen is the same for the on-schedule days as it is for the off-schedule days for employee A? C) That the employee's on-off schedule days are independent of the stealing-activity days?

Here's problem from real life. Suppose that at a certain company, valuable products are stolen over a 4-months periods, presumably by some employee. In the 114 days period, products were stolen in 45 different days. The company keeps a log file of the days the products were stolen. They then looked at the records to see who worked on what day. The following is the contingency table in terms of days for employee A Onschedule Off-schedule Stolen 45 0 Non-Stolen 32 37 The evidence looks pretty damning. But you need to convince a judge that it is more than chance. Assume that the chance of product being stolen is binomial and independent for each day with p = 45/114. That is, each day has a 45/114 chance of having product stolen. Calculate the likelihood that A's work days would have the same pattern as above, given the binomial loss assumption. Show your math.

It'll be a cold day in hell when I can one-up CaptainEd. Anyways, I made a mistake earlier when computing the possible number of ways to distribute 3 right switches among 32. For some reason, I entered Combination( 15,3) into my computer. Don't ask where the mental disconnect comes in. Anyways, I'm revising the estimate. 9 turns, and possibly even 8, is required to solve part I.

Can you find the solution to part I in 6 turns or less? phillip1882's method is a good point to start but needs refinement. For part II, it is possible to win in less than 32 turns.

I'm seeing some interesting solutions here. The issue is that most of these solutions require way more turns than necessary. Can you find a solution to part I in less than 8 turns? I think 8 turn is not the minimum required, it may be even less. I'll get back to you on that. phillip1882's post is interesting. I'll need to take a longer look before I'm convinced.

This is inspired by bonanova's light switch problem. Suppose that there are 32 light switches in room A, which corresponds to 32 light bulbs in room B. Let's say that each switch can either be set in an "Up" or "Down" position. Which position corresponds to the "On" status in the next room is unknown for every switch. The game consists of the following terms. At each turn, you are allowed to set all 32 switches in any configuration you pick. You can not see the light bulbs in room B, but an official will tell you how many light bulbs are on at the end of every turn. The game ends when you can turn all 32 lights on. 1) Suppose that on your first turn, all the switches were in a random up down configuration, and the official tells you that 3 lights are on. What is the minimum number of turns required to turn all the lights on, even in the worst case ? 2) Suppose that the switches are set in random positions. What is the minimum number of turns required before you are guaranteed to win?

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Good job to manwe and final. You got it. I agree with final that the chance of a colony living indefinitely is the complement of the chance of extinction. So, in this case, the two bacteria and their possible descendant will go extinct with probability 0.618034^2 = 0.381966. That means they'll exist indefinitely with probability 1- 0.381966 = 0.618034. It is also interesting to note that every generation with k descendants will have an extinction chance of 0.618034^k. Of course, after a certain k, that chance is trivial.

In a certain bacterial colony, all bacteria reproduce by asexual reproduction. That is, every bacterium over the span of its life has a 1/2 chance of splitting into 3 identical clones. The clones then have the same reproductive mechanism as their parent. If a bacterium doesn't divide (1/2 chance), it will die eventually. Assuming that the bacterial company starts with just two bacteria. Assuming that the division process for the bacteria are independent of each other, what is the chance that the colony would go extinct, i.e. having all the bacteria dead at some point?

This logic checks out for me. I verified that the code does examine all the possible 64C8 = 4426165368 bishop placements. The subroutine that checks for attack lines seems to be correctly used. Unless someone contests this, I declare Phatfinger's the most satisfactory. Well done, Phatfinger. Adding the bishop one at a time and using next is much more optimal than generating 8 placement locations and check them for conflict.

You didn't happen to meet your wife 40 years ago, did you?

I feel the same way you do

Congratulations to bonanova and EventHorizon for getting part I and the bonus, respectively. Well done.

My math professor gave out this problem as a bonus during my first year of college. I bet the den will solve this in a jiffy. Here is a picture of a 12-faces regular polyhedron. Each face is a regular pentagon, where all sides are equal and all interior angles are equal. 1) Find the angle between any two adjacent faces. Bonus: Give the answer to number 1 in its exact form (ie. an expression containing square roots, trig functions, and whole numbers instead of giving the decimal form)

I meant for the puzzle to be a statistical exercise (i.e. chance of dying the same for A and B on 10-shot gun). But good job for thinking outside the box. I know much more about statistics than I do about gun, as your post just showed.

Good work, CaptainEd. I didn't actually expect you to actually implement the method, so your solution makes it all the more impressive. Well done.

This is counter-balanced by the fact that C has a 1/2 chance of killing D on everyshot. Most of the time, C survives B's shot, so he gets to shoot D. If A kills B, that's even better for C. In this game, killing someone above you is not as important, or urgent, as having someone with low accuracy below you.