bushindo

5^4 means that are 625 possible combinations. To convey those to the prisoner in two words, simply assign those 625 combinations to 2-alphabet-letter combinations. Since there are 26 letters in the alphabet, we should be okay.

Sorry, made an error with some matrix index before

This can be modeled by time to first success

You guys are hilarious. I didn't think of the 3 dimensional thing, so good job.

you missed the top vertical line. Looks like this is your kryptonite

Hilarious. I hope that wasn't you.

Here's a picture. There are 16 connectors connecting the dots. Find a way to draw a single line so that it crosses through every single connector once and only once. You can start from any point on the image, and you can finish at any point. The line must be continuous, and crosses every connector once each. Here's an attempt. For ease of discussion, I label the edges from 1-16. The lines goes through connectors (2, 1, 7, 5, 4, 14, 15, 13,12,10, 9, 11, 8). Note that this doesn't satisfy the requirement above because the connectors 6 and 16 haven't been crossed. Have fun.

Yours work too. Well done

It works. Well done.

Good try, everyone. I haven't had time to read everyone's post, so I'll just comments on some of the ones that jumped at me brshravan in post 17 and Palascosas in post 25 had answers that will surely save the prisoners. Good try here, adiace, but there's a slight problem.

Easy, we covered this in the first day of zen class

I've seen some good answers, and some that are really inventive. The flurry of activity seems to be strong, so I'll leave it be for now. I've seen some solution introduce extra information by choosing the order in which the prisoners take their turn. The problem with that is each prisoner only gets to see the configuration once it is his turn. Those who have yet to take their turn don't know what the ball configuration is, and thus can't choose who to go next. Besides, introducing the new information makes this problem too easy. For this problem, assume that at the beginning of every turn, the warden randomly chooses a prisoner to go up.

I would like some detail filled

55 prisoners are on the deathrow. The warden gives them a chance to live. He puts 100 empty jars into room A and randomly puts 10 balls under the 100 jars. Each jar is either empty or contains 1 ball. In the other room, call it room B, the warden puts 100 empty jars, and a stack of 10 balls. The warden then divides the group of 55 into two groups of 54 and 1. The group of 54 he puts into room A. The last prisoner goes into room B. Each prisoner from room A will take turn looking under the entire 100 jars, but can not move or rearrange the contents. He then can go to room B and must say one of two possible words to the prisoner there. The 2 possible words are Lakers and Rule. Assume he can not convey any other information besides that word (so no facial expression, tone, body language, hand gestures, etc. ). Any attempt to convey extra information like remaining silent, concatenating words, or walking a certain number of paces before stopping in room B will get all prisoners killed immediately. The prisoner in room B will then have to reconstruct the permutation of the balls in room A. If the prisoner in room B can successfully reconstruct the permutation in room A after the 54 turns, all 55 will live. Otherwise they will die. The night before, the warden tells the prisoners this scheme, so the prisoners know that there will be exactly 10 balls under the 100 jars. They have 1 night to discuss a strategy. They are not allowed to bring any mechanical computational aid to the game (yes, abacus are out too). Assume that each prisoner has the mental computational skills of a reasonable average person. 1) What strategy would give the prisoners the best chance to live? Describe the strategy.

The part about each guest taking room (2*i - 1) was the instruction for night 1, when guests bearing numbers 1,2,3,4,5,... arrived. The guests with shirts bearing fractions arrive on the second night, and we're supposed to find the solution for that case.

You might be a pan-dimensional being, but you're still a business man first. Giving an infinite number of guests 1 keg each can really cut into your profit, you know. Although you can probably recoup the cost by giving each drunken guest a typewriter and have them churn out shakespeares and stuffs.

adiace and final got it. Well done.

You're Q, a pan-dimensional being living in the Q Continuum. Unlike other beings on the Q Continuum who have nothing better to do but going to lower dimensions and bothering mortals on spaceships, you actually have a day job. Your job is to manage the Infinity Inn, which is a hotel with, you guessed it, an infinite number of rooms. Let's say the rooms are numbered from 1 to infinity. One day, your Infinity Inn is fully booked, but an infinite number of guests arrive. The new guests each wear a shirt with a unique whole number (ie. 1, 2, 3, 4, ... ) . Not one to miss a beat, you ask every currently settled guest to move into a room twice their current number. So the guest currently in room 2 moves into room 4, the guest in room 3 moves into room 6, and so on. The new guests then simply move into the now vacated room. For convenience, you simply ask the new guests to move into room (i*2 - 1), where i is the number on their shirt. The next day, a family called the Fractions arrived. Each member of the family wears a number of their shirt. For every possible two whole numbers a and b, there is a unique family member who wears a shirt with the number "a/b". So for instance, pick a = 2, and b = 4, there's a guest with 2/4 on hist shirt. Pick a =1001, and b = 23, there's a guy with 1001/23 on his shirt too. The head of the Fraction family look at your Inn and said that it is too small for his family, not to mention the fact that it is already full. The family was about to leave you said "Wait....". 1 ) Is there a way to fit the whole family inside your inn? Remember, you just fully booked it the night before. If so, describe a way to fit everyone in.

I like James' answer the best. Very nice work.

Interesting concept. This may very well work, with some refining on the details. This is a reasonable compression scheme, and since having 39 prisoners transmitting 5 bits of information each is equivalent to having a 90-digit binary string (compared to a 100-digit binary string we are supposed to encode), this may work a a reasonable number of cases.

Even if my method is correct (which it may well not be) I may have done the calculation wrong so it would be good if someone could check. psychic_mind got problem 1. Onto problem 2.

This problem is inspired by a recent viewing of the famous Press Your Luck show where Michael Larson memorized the lights patterns and broke CBS's bank. Assume that a contestant on a game show has 10 spins. Each spin of the wheel might land on a one of these outcomes with equal probability. The possible outcomes on the spins are (Whammy, 1000 dollars, 3000 dollars, 500 dollars, 2000 dollars ) Each contestant starts with 0 dollars, and accumulates money depending on what outcome they land on. Landing on a Whammy means that the current money is wiped out, and the amount is reset to 0. Landing on a Whammy does not affect the number of spins remaining for the player. 1) What is the expected winnings after 10 spins ? 2) In this scenario, the wheel is modified to contain 1 additional outcome. The chance of landing on any outcome is 1/6 (Whammy, 1000 dollars, 3000 dollars, 500 dollars, 2000, 1000 dollars + 1 extra spin ) Landing on the last outcome to add 1000 dollars and 1 extra spin to the player's account. If a player starts with 10 spins, and continue until all spins are gone, what is the expected earnings?

final, adiace, and psychic_mid got it. Well done.

Your math for the first three situations are right, but the deuce part isn't. Thelogical1, your approximation is quite close. Care to elaborate on your method?