You have $10,000 dollars to place a double-or-nothing bet on the Yankees in the World Series (max 7 games, series is over once a team wins 4 games).
Unfortunately, you can only bet on each individual game, not the series as a whole. how much should you bet on each game, so that, if the yanks win the whole series, you expect to get 20k, and if they lose, you expect 0?
Basically, you know that there may be between 4 and 7 games, and you need to decide on a strategy so that whenever the series is over, your final outcome is the same as an overall double-or-nothing bet on the series.
You are an oil mogul considering the purchase of drilling rights to an as yet unexplored tract of land.
The well’s expected value to its current owners is uniformly distributed over [$1..$100]. (i.e., a 1% chance it’s worth each value b/w $1..$100, inclusive).
Because you have greater economies of scale than the current owners, the well will actually be worth 50% more to you than to them (but they don’t know this).
The catch: although you must bid on the well before drilling starts (and hence, before the actual yield of the well is known), the current owner can wait until *after* the well’s actual value is ascertained before accepting your bid or not.
It’s the middle ages, you’re travelling across europe and you want to find the way to vienna. you come to a crossroads, now there are two ways to go. at the crossroads stand a knight and a knave. the knight answers every question truthfully. the knave answers every question falsely. you don’t know which guy is which. how can you figure out which road leads to Vienna by only asking one question?
You find yourself in a duel with two other gunmen. you shoot with 33% accuracy, and the other two shoot with 100% and 50% accuracy, respectively. the rules of the duel are one shot per-person per-round. the shooting order is from worst shooter to best shooter, so you go first, the 50% guy goes second, and the 100% guy goes third.
some of you may have easily solved the pill weighing problem posed here. If so, you are going to love this problem. It is similar but much more difficult.
from my buddy Tom:
Ok, here’s a tough one (i thought). There are no “aha!” tricks - it requires straightforward deductive-reasoning.
You have 12 coins. one of them is counterfeit. All the good coins weigh the same, while the counterfeit one weights either more or less than a good coin. Your task is to find the counterfeit coin using a balance-scale in 3 weighs. Moreover, you want to say whether the coin weighs more or less than is should and, and this is the real kicker, your weighs must be non-adaptive, that is, your choice of what to put on the balance for your second weigh cannot depend on the outcome of the first weigh and your decision about what to weigh for round 3 cannot depend on what happened either your first or second weigh. For example, you can’t say something like “take coin #1 and coin #2 and weigh them. If they balance, then take coins 3,4,5 and weight them against 6,7,8…if 1 and 2 don’t balance, then weigh #1 vs #12…” you have to say something like:
round #1: do this
round #2: do this
round #3: do this
if the results are left tilt, balanced, and left tilt, respectively, then coin #11 is heavier than it should be.
This problem is solvable…it took me about 1-2 hours of working on it to get it. I think even finding the counterfeit using an adaptive solution is tough. Then non-adaptive constraint makes it quite hard and having to find whether it’s heavier and lighter is cruel and unusual riddling ;-)
(you can’t use paper, you have to figure it out in your head)
i have a black triangle, a white triangle, a black circle and a white circle. if i gave you a shape (triangle or circle) and a color (black or white), the “frobby” items would be those that had either the shape or the color, but not both. that is, in order to be frobby, the item must be of the specified color OR the specified shape, but not both the specified shape AND the specified color. i’m thinking of a shape and a color in my head and i tell you that the white triangle is frobby. can you tell me the “frobbiness” of the other items?
there are four cards which have a letter on one side and a number on the other side. i lay them out and the cards appear as 2 5 F E. the rule is that a card with an odd number on one side must have a vowel on the other. what is the minimum number of cards you should turn over to prove the rule is true (and which cards would they be)?
Did you ever wonder how they make those pillsbury cookie dough rolls with the intricate faces inside them? Look here and notice the intricate design they have somehow injected into their cookie rolls? If you examine the roll closely there is no seam between the normal dough and the colored shape, but somehow they get that inside the roll. I emailed them asking them how they do it and they told me it was “doughboy magic”.
I offer to play a card game with you using a normal deck of 52 cards. the rules of the game are that we will turn over two cards at a time. if the cards are both black, they go into my pile. if they are both red, they go into your pile. if there is one red and one black, they go into the discard pile.
We repeat the two card flipping until we’ve gone through all 52 cards. whoever has more cards in their pile at the end wins. i win if there is a tie. if you win, i pay you a dollar. how much would you pay to play this game?
A line of 100 airline passengers is waiting to board a plane. they each hold a ticket to one of the 100 seats on that flight. (for convenience, let’s say that the nth passenger in line has a ticket for the seat number n.)
Unfortunately, the first person in line is crazy, and will ignore the seat number on their ticket, picking a random seat to occupy. all of the other passengers are quite normal, and will go to their proper seat unless it is already occupied. If it is occupied, they will then find a free seat to sit in, at random.
What is the probability that the last (100th) person to board the plane will sit in their proper seat (#100)?
At one point, a remote island’s population of chameleons was divided as follows:
13 red chameleons
15 green chameleons
17 blue chameleons
Each time two different colored chameleons would meet, they would change their color to the third one. (i.e.. If green meets red, they both change their color to blue.) is it ever possible for all chameleons to become the same color? why or why not?”
A man needs to go through a train tunnel. He starts through the tunnel and when he gets 1/4 the way through the tunnel, he hears the train whistle behind him. you don’t know how far away the train is, or how fast it is going, (or how fast he is going). All you know is:
if the man turns around and runs back the way he came, he will just barely make it out of the tunnel alive before the train hits him.
if the man keeps running through the tunnel, he will also just barely make it out of the tunnel alive before the train hits him.
Assume the man runs the same speed whether he goes back to the start or continues on through the tunnel. Also assume that he accelerates to his top speed instantaneously. Assume the train misses him by an infintisimal amount and all those other reasonable assumptions that go along with puzzles like this so that some wanker doesn’t say the problem isn’t well defined.
The warden meets with 23 new prisoners when they arrive. He tells them, “You may meet today and plan a strategy. But after today, you will be in isolated cells and will have no communication with one another.
"In the prison is a switch room, which contains two light switches labeled A and B, each of which can be in either the ‘on’ or the ‘off’ position. I am not telling you their present positions. The switches are not connected to anything.
"After today, from time to time whenever I feel so inclined, I will select one prisoner at random and escort him to the switch room. This prisoner will select one of the two switches and reverse its position. He must move one, but only one of the switches. He can’t move both but he can’t move none either. Then he’ll be led back to his cell.
"No one else will enter the switch room until I lead the next prisoner there, and he’ll be instructed to do the same thing. I’m going to choose prisoners at random. I may choose the same guy three times in a row, or I may jump around and come back.
"But, given enough time, everyone will eventually visit the switch room as many times as everyone else. At any time anyone of you may declare to me, ‘We have all visited the switch room.’ and be 100% sure.
"If it is true, then you will all be set free. If it is false, and somebody has not yet visited the switch room, you will be fed to the alligators."
What is the strategy they come up with so that they can be free?
Five pirates discover a chest full of 100 gold coins. The pirates are ranked by their years of service, Pirate 5 having five years of service, Pirate 4 four years, and so on down to Pirate 1 with only one year of deck scrubbing under his belt. To divide up the loot, they agree on the following:
The most senior pirate will propose a distribution of the booty. All pirates will then vote, including the most senior pirate, and if at least 50% of the pirates on board accept the proposal, the gold is divided as proposed. If not, the most senior pirate is forced to walk the plank and sink to Davy Jones’ locker. Then the process starts over with the next most senior pirate until a plan is approved.
These pirates are not your ordinary swashbucklers. Besides their democratic leanings, they are also perfectly rational and know exactly how the others will vote in every situation. Emotions play no part in their decisions. Their preference is first to remain alive, and next to get as much gold as possible and finally, if given a choice between otherwise equal outcomes, to have fewer pirates on the boat.
The most senior pirate thinks for a moment and then proposes a plan that maximizes his gold, and which he knows the others will accept. How does he divide up the coins? What plan would the most senior pirate propose on a boat full of 15 pirates?
Two MIT math grads bump into each other while shopping at Fry’s. They haven’t seen each other in over 20 years.
First grad to the second: “How have you been?” Second: “Great! I got married and I have three daughters now.” First: “Really? How old are they?” Second: “Well, the product of their ages is 72, and the sum of their ages is the same as the number on that building over there…” First: “Right, ok… Oh wait… Hmm, I still don’t know.” Second: “Oh sorry, the oldest one just started to play the piano.” First: “Wonderful! My oldest is the same age!”
Four people need to cross a rickety rope bridge to get back to their camp at night. Unfortunately, they only have one flashlight and it only has enough light left for seventeen minutes. The bridge is too dangerous to cross without a flashlight, and it’s only strong enough to support two people at any given time.
Each of the campers walks at a different speed. One can cross the bridge in 1 minute, another in 2 minutes, the third in 5 minutes, and the slow poke takes 10 minutes to cross. How do the campers make it across in 17 minutes?