BMAD

I wrote my text in white as an impromptu spoiler in the above post.

could it be as simple as... (I couldn't find the spoiler button, so highlight below to see the text) about 1 bus every 15 minutes on average x bus : 15 minutes 1 bus: 20 minutes x = .75 bus which is about 1 bus in a 15 minute period and therefore all parties can be right. Highlight above.

The premature retirement of the beloved leader Aldan has left a vacancy on the ruling triumvirate of the Malthusian city of Zocor. There are two political parties, The Meat Lovers Party (MLP)and the Vegetarian Party of the People (VPP). At present each party has one representative in the triumvirate, with the MLP representative Arctan being the most senior.The other ruling member being Arcsin. The arcane voting rules for replacing Aldan are as follows. There is a large Praesidium with n people in it. At the moment all the members are in the VPP. Each member has an integer valued priority. The selection process will now go in rounds: At the beginning of a round Arcsin will propose some subset S of the remaining candidates C of the Praesidium. (Initially C will be the whole Praesidium). Arctan then has two choices. He can devour S, but then he is forced to decrease the priority value of each member of C \ S by one. Alternatively, he can devour C \ S and decrease the priority values of the members of S by one. The process continues until one of two things happen: The whole Praesidium has been eaten and then Arctan can choose any member of his party to fill the vacancy. The other possibility is that someone will reach priority zero and then Arcsin can choose any zero priority person to fill the vacancy. Given that the initial priorities are a1, a2, .... an, which party will get to fill the vacancy?

Turkey Sandwich was worried about an upcoming test in Discrete Mathematics and was finding it hard to get to sleep. Turkey awoke early in the morning, aroused by devilish laughter, only to see an impish looking homunculus sitting at the bottom of the bed next to a seemingly infinite pile of chips. Hello Turkey it said, would you like to play a little game? This pile contains 43546758343209876 chips and the bottom chip represents your immortal soul. The rules are quite simple. The first player takes some chips, but not all of them. After that we take it in turns to take some chips. The only rule now is that a player cannot take more in their turn than the previous player took. The winner is the player who takes the last chip. If I win I get to keep your soul and if you win, you get an A in the test. Would you like to go first or second? This seemed a reasonable bet to Turkey. Can you give Turkey a strategy for playing no matter how many chips there are? Was that too easy? What if a player can take up to twice the number of chips that the previous player took. What is the best strategy now?

The FBI has surrounded the headquarters of the Norne corporation. There are n people in the building. Each person is either an engineer or a manager. All computer files have been deleted, and all documents have been shredded by the managers. The problem confronting the FBI interrogation team is to separate the people into these two classes, so that all the managers can be locked up and all the engineers can be freed. Each of the n people knows the status of all the others. The interrogation consists entirely of asking person iif person j is an engineer or a manager. The engineers always tell the truth. What makes it hard is that the managers may not tell the truth. In fact, the managers are evil geniuses who are conspiring to confuse the interrogators. Under the assumption that more than half of the people are engineers, can you find a strategy for the FBI to find one engineer with at most n-1 questions? Is this possible in any number of questions if half the people are managers? Once an engineer is found, he/she can classify everybody else. Is there a way to classify everybody in fewer questions?

A teacher wrote a large number on the board and asked the students to tell about the divisors of the number one by one. The 1st student said, "The number is divisible by 2." The 2nd student said, "The number is divisible by 3." The 3rd student said, "The number is divisible by 4." . . . (and so on) The 30th student said, "The number is divisible by 31. The teacher then commented that exactly two students, who spoke consecutively, spoke wrongly. Which two students spoke wrongly?

Agree with Ed on first part. Can you clarify the second question? On the first one we had an entire column of false coins. Now in this situation, each column has precisely 8 false coins. What is the fewest amount of weighings needed to find all of the false coins?

Given thirteen gears, each weighing an integral number of grams. It is known that any twelve of them may be placed on a pan balance, six on each pan, in such a way that the scale will be in equilibrium. Prove that all gears must be of equal weight.

Two hundred students are positioned in 10 rows, each containing 20 students. From each of the 20 columns thus formed the shortest student is selected and the tallest of these 20 show students is tagged A. These students now return to their initial places. Next the tallest student in each row is selected, and from these 10 tall students the shortest is tagged B. Which of the two tagged students is the taller if they are different people?

There are twenty-four half-dollar coins stacked in three identical piles. Two piles are real and one is counterfeit. The individual counterfeit coins look exactly like the others but either weigh a gram more or less from the others. The net weight of each pile is exactly the same. How many weighs on a balance will it take to determine which pile is false? How many weighs would it take if there were an equal distribution of false coins in each pile?

After reading your post and reconsidering bonanova's earlier post I think my theoretical maybe shortsighted but what i originally calculated was rather simple:

Interestingly I believe i found the theoretical probability to be 50% but my simulations have it around 60-62%.

There is an old game that I learned back in my home country when I was a child but the name escapes me. When teaching it to my students, I call it 'dash' whose meaning will be obvious. Setup the 'game board' as follows: Row 1: 1 dash, Row 2: 3 dashes, Row 3: 5 dashes, and Row 4: 7 dashes - - - - - - - - - - - - - - - - The requirements of the game are simple: Each player takes turns circling any amount they like in any row. Each player must circle at least 1 not-circled dash on their turn. A player does can circle dashes that are not adjacent. The player may not circle dashes in more than one row in a single turn. The last player to circle a dash is the loser.Is the advantage for the first player or second player or is this a fair game?

This is an extension of my previous question. I honestly am torn between two different solutions (one found in theory and the other in simulation). I am curious what solution we will find here. Recall that we have a unit circle where three random points define three arcs. What is the probability that the longest arc contains the point (1,0)?

Are the digits that represent pi (e.g. 3.1415926....) a uniform distribution? What if we examine pi in base 2, is it a uniform distribution?

Nope: Hidden Content my suspicion is that given that most people would pick a particular round to use their ability that there is in fact a particular advantage even if the expected payouts are the same. Maybe game theory should be utilized here over strict statistics?? This feels very subjective. I simulated the event using the three different options 1M times each. Here's what I get with graphs. Seems like it totally depends on your appetite for risk: Hidden Content One follow-up: Hidden Content would this not imply that the first two rounds use for the magic power are worst. Imagine that you are trying to beat your best score, your opponent is yourself. An opponent with the exact same power. Trying to do the best in this case, according to your previous post would include using this power at the end. Hence there is an advantage.

Nope: Hidden Content my suspicion is that given that most people would pick a particular round to use their ability that there is in fact a particular advantage even if the expected payouts are the same. Maybe game theory should be utilized here over strict statistics??

Where is my flaw... *contains what i believe to be the solution*

i wanted 4 rounds so my apologies. There should be 16 teams. Each game has a 50-50 chance of being picked by you in that you have no knowledge of who you should pick though one team may or may not be better than the other team.

Hmmm i am not sold that the answer Isnt

On April 1st a typist found the hammers of the typewriter re-soldered in an arbitrary order, so typing a text resulted in gibberish. The typist decided to type a document in its entirety using this typewriter. Afterwards, if the result does not represent the original text, he will type the result, and so on. Prove that the clear text will emerge sooner or later. How many iterations are enough to guarantee that the clear text will appear, if the typewriter has, say, 46 keys? N keys

Choose, at random, three points on the circle x2+y2= 1. Interpret them as cuts that divide the circle into three arcs. Compute the expected length of the arc that contains the point (1, 0).

Say you are granted the ability to know the outcome of two consecutive rounds in a 8-team elimination tournament. Each round that you win, the amount you bet is doubled while losing any bets when you pick the wrong team. Every round you take all of the money in your possession and split it evenly over each team that you believe will win. Is it in your best interest to use your ability of foresight in the first two rounds, middle two rounds, last two rounds or does it not matter?

Oh, i think the issue is the assumption (or a poor translation) behind what i meant when i said it is long enough to hold 11 balls. Clearly a chute that can hold 12 can also hold 11. The task was to find the smallest such chute that can hold 11 balls and allow the black balls to exit.

hmm, I see where I remarked that a chute shorter than 15 were possible but I am unsure of a solution of a size less than 12.