BMAD

take naked pictures of yourself put them as your screensaver, make the screensaver password sensitive leave used tissues (or at least they need to appear that way) around the couch and remote control ensuring to coat the remote in something sticky apologize for mistakenly using their toothbrush fart and demand your roommate say 'excuse me' brag incessantly about simple accomplishments, "oh, yeah! I tied my shoe." "oh, yeah! i turned my computer on." "oh, yeah! I found my pencil" put mayonnaise on your pepperoni pizza, heat in microwave, sprinkle on garlic, enjoy....talk with husky voice for the rest of the day.

Q: How does every Russian joke start? A: By looking over your shoulder. Q: What is 150 yards long and eats potatoes? A: A Moscow queue waiting to buy meat. Q: What did the Russian people light their houses with before they started using candles? A: Electricity.

selling an apple that is 2/3rd eaten would be difficult But if you insist, add 1/3 to my previous answer that bellies the assumption that he must have apples to drive. which is not the case. he only eats apples while driving if they are available. ps. you are also answering a different question.

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Do we have to have that many players? I wouldn't mind having two characters. Being new at the game, I am sure to die often

on the right track, but no sure answer yet

Computer's are not confident, they either compute an answer or they don't. It would play the probabilities of the human selection.

You have a point. The trick in the OP wouldn't really work since saying "no" is pretty much a confession. But changing the coin setup in the way you described wouldn't work either. The only honest way to make it work would be to tell the surveyee to lie or tell the truth based on how the coin lands, but that wouldn't give you any useful data. I disagree. This is actually a classic setup that many psychologist use. Since the participant has the security of the coin flip (if it is secretive toss) the psychologist can calculate using some simple statistics the rate at which one shoplifts (in this case)

remember right angled is not considered acute in the game's directions

what would your array be?

if the computer is right that you would pick both, you pay your uncle 1010 not 10.

I think the amount you owe your uncle if the computer predicts correctly is the value of the money in both envelopes, not just envelope A - if I understood the problem correctly. Also, even if the computer is wrong, you're not guaranteed an additional 10,000 since it might still be empty. The computer was designed for/by the uncle to work in his benefit not yours, so it would do what it can to ensure your uncle makes his money.

yes. i still have more listed.

Five years ago, your rich uncle secretly commissioned the construction of a supercomputer! The purpose of this binary behemoth (the "Very Good Predictor") was to predict, as accurately as possible, every action you would take over the course of these five years. It has managed to do so with 100% accuracy for its entire runtime. Now, your uncle has approached you with this information, and an offer to participate in a game involving the supercomputer's final prediction. He has placed two envelopes in a room: Envelope A and Envelope B. Envelope B contains exactly 1,000 dollars. Envelope A either contains 10,000, 10, or 0 dollars. If the supercomputer correctly predicted your selection of Envelope A alone, it will contain $0. If the supercomputer correctly predicted your selection of both A and B then Envelope A contains only $10. Otherwise the supercomputer predicted that you would not choose A and told your uncle that he could safely put $10,000 in the envelope (which he would oblige). Now since this is essentially a contest between you and the supercomputer, if the supercomputer correctly predicts your choice, the amount selected is the amount you owe your uncle, otherwise you keep the amount in the envelopes (both). The values in both envelopes is revealed once a selection is made. What should you do?

BMAD has a defective clock in his room. The clock is digital, and some of the seven bars on its units digit are broken. A bar that works is on when it should be and off when it should be, whereas a broken bar is off regardless of the time. a) BMAD enters his room and glances at his clock. He knows how many bars are broken but he doesn't know which ones they are. He immediately knows for sure what the time is. b) BMAD enters his room and glances at his clock. He knows which bars are broken and immediately knows for sure what the time is. c) BMAD enters his room and knows which bars are broken. He stares at his clock for 60 seconds. Now he knows for sure what time it is. d) BMAD enters his room. He knows how many bars are broken but he doesn't know which ones they are. He stares at his clock for 60 seconds. Now he knows for sure what time it is. In each of these four scenarios, what is the maximum number of broken bars, and which might it/they be?

In a rectangle that's 2 by 200 units long, it's trivial to draw 400 non-overlapping unit-diameter circles. But in the same rectangle, can you draw 401 circles?

Balcony seating at the opera is by ticket only, and all the tickets are sold and all the ticketholders are in line to enter. But it turns out that the first person through the door is rather inebriated by the time the hall is open for seating and, instead of taking the properly assigned seat, chooses a chair at random (presumably the one with the lowest-seeming relative velocity to his or her person). The next people come in one at a time—and if their seat is available, they take it. But if someone is already sitting there, rather than disrupt things, they just pick some other random seat. Which raises the question, "what is the probability that the last opera lover ends up in his or her assigned seat?"

x x x x x x x x x x x x x x x Distribute all the numbers 1 to 15 on the x's so that the number on the x, shows the difference between the two x's on which it sits.

why?

There is a coin at each vertex of a regular 10-gon. Alice and Bob take turns removing one coin, with Alice going first. A coin can be removed only if there is an acute-angled triangle between the coin one wants to remove and two other coins [the center of the shape is the center of the angle]. A player who cannot move loses. (Note: A 90 angle is not acute.) Who has a winning strategy?

If you drive to the store at 20 mph, how fast must you go (again returning by the same route) for your average speed to be 40 mph?