bonanova

Should we understand that, in a single 90-second stop, people could both load [enter] and unload [leave]?

The sum he creates. Chuck's sum e. g. Is Al's plus Bob's integers.

With a tip of the cap to Anza Power, here's a Fibonacci puzzle with smaller numbers. In today's class, Mr. Bigollo taught his students the Fibonacci series and Golden Ratio. He noted the series traditionally begins 0, 1, 1, 2, 3, 5, 8, ... , but actually any two numbers, not necessarily in ascending order, can seed the process. To make the subject compelling, Mr. Bigollo announces to the class that you have proven yourself among the Brain Denizens as particularly adept at solving seemingly impossible mathematical puzzles. Furthermore, you have accepted his invitation to their session today to participate in a demonstration of the amazing properties of the Fibonacci series. You stand and smile, graciously, at the adoring stares of the class. To begin, Mr. B selects ten students to line up in the front of the classroom. Amazingly, their names are Al, Bob, Chuck, Dave, Ed, Frank, Geoff, Hal, Irv and Jack. Here's what he asks them to do. Al is to select an integer at random and whisper it to Chuck. Bob also selects an integer at random and whispers it to Chuck. Chuck sums the numbers he hears and whispers his sum and Bob's number to Dave. Dave sums the numbers he hears and whispers his and Chuck's sums to Ed. Ed sums the numbers he hears and whispers his and Dave's sums to Frank. Frank sums the numbers he hears and whispers his and Ed's sums to Geoff. Geoff sums the numbers he hears and whispers his and Frank's sums to Hal. Hal sums the numbers he hears and whispers his and Geoff's sums to Irv. Irv sums the numbers he hears and whispers his and Hal's sums to Jack. Jack sums the numbers he hears. Finally, each student whispers his own sum [Al and Bob whisper their chosen integers] to Mr. Bigollo, who writes the sum of the ten numbers on a sheet of paper, which he then places in an envelope. To clarify: the envelope contains the sum of the first ten Fibonacci numbers generated by Al and Bob's initial integer choices. No one but Mr. Bigollo knows this sum. The students take their seats and Mr. Bigollo invites you to the front of the classroom and announces to the class that you are now prepared to announce the number contained in the envelope. Noting your shock, and before you can protest, Mr, B. smiles and says: Well that would be impossible of course. So, you may have one piece of information. You may ask one of the students what his sum was. [That eliminates Al and Bob, they didn't sum anything, but neither of them alone could supply useful information.] Now your reputation is on the line. Which student do you ask, and what is the sum?

Bingo. And the coveted bonanova gold star to boot:

Well it's not 1 11 21 1211 111221 312211 .... Which it resembles at first glance.

Some responders have assumed a limitation on choice of game. No limitation is intended. Bushindo has given one of the two answers I was looking for. It's certainly what the Cardinals would now wish for, down 2-3. But there is in fact one game the series winner always wins.

Studies have been made of pivotal games in a series of contests. Third and fifth games have been given importance in that regard. If your team is evenly matched in a best 4 of 7 series of games, and you could be assured of winning one of the games, which game would you choose? See if you can solve this without enumerating all the outcome permutations.

Concepts like "order" and "previous" and "dependence" don't apply here, even tho the experiment is stated as a sequence of steps. The chance that card x modulo 13 has a value of x does not depend on x. I.e. there is no way to say the third card's chance of being 3 differs from the fifth card's chance of being 5. Corollary 1: It does not matter what order you use to inspect the cards. I.e. inspecting card number 3 before you inspect card number 5 does not alter card 5's chance of being a 5. Corollary 2: Since the deck is well shuffled, you can call out the cards in ANY legal sequence, e..g., AAAA222233334444 .... and get the same average number of discards. A different number for each shuffle, of course, but the same expectation. Sorry I can't spoiler from mt iPad, but I'm not giving a solution, just discussing the process.

If you don't mind the word "degenerate" you get a larger result.

Bravo EH.

I think so.

Some of those have more than 2 non-zero elements per row and column. But I think there is no limit, based on the model of those that don't.

Yeh Ed, good one. I forgot you could ignore cases, I was trapped in powers of two.

Next try.

Here's a start.

Here's a quickie you should be able to get in a few moments of thought. Name words or a word that behave as follows: A verb that becomes past tense by moving the initial letter to the last position A noun that becomes plural by adding a "c". A plural noun that becomes singular by appending an "s" at the end. Enjoy!

Every school boy or girl knows that with best play tic tac toe is a drawn game. That is, getting three markers X or O in a row playing alternately on a 3x3 grid is impossible if your opponent plays correctly. But suppose we extend the board to the plane - there is an unlimited number of squares on which to play. Certainly then the first player can always achieve a row of 3 markers no matter how her opponent plays. What is the largest number of markers in a row that first player can achieve in this case?

I love this problem. The closest approach to a procedure I can imagine is some variation of the procedure for the binary [black hat, white hat] case where exactly half must be right. Assuming an even number of prisoners.

The wiki discussion of this paradox contains some loose arguments:

I'll describe two other twists: but first let's establish a test for emptiness of the box. To claim that a ball or balls remain in the box, their number must be specified.

Another consideration is that the probability of finding subgroups is zero. Well not exactly zero, since "exactly 2 liters" is an integer number of HOH molecules. If we discuss balls in urns, subgroups would occur with more useful probability.