superprismatic

CORRECTION ON THE PROBLEM STATEMENT: "positive integers" in the problem should be "non-negative integers". Sorry, I copied this verbatim from my source (privately published material) and I didn't notice that the problem can't be solved without 0 as one of the coordinates.

What I mean by "coordinates" are the numbers either written down the side of the grid or across the top of it.

Yes.

Sorry! You don't have 23 as one of the coordinates.

One set of eight positive integers is written down the side of an 8x8 square grid, another set of eight across the top. The entry in any cell of the grid is the sum of the corresponding side and top coordinates. These entries are the numbers from 1 to 64 inclusive. If one of the coordinates is 23, what are the others?

anyway I tried to look up this problem you were talking about. can you tell me what this problem is or what website i could find it on. Spot on! Part 1 is indeed that! I can't give you a website because I got it from some privately published papers containing original Walter Penney problems. I'm glad you liked this one -- I'll put out more from time to time. I won't do it too quickly, though. Walter Penney was a master puzzle maker and I don't want to monopolize things. Walter died in 2000 at the age of 87.

Here is my modification of a puzzle by the late great puzzle maker, Walter Penney: There are 309 white balls and 191 black balls in a bag. A man makes a deal in which he draws balls one at a time from the bag, replacing and mixing after each draw, and if the first black ball is drawn on the Nth trial, he is to receive a number of dollars equal to the Nth Fibonacci number (0,1,1,2,3,5,8,13,... for N=1,2,3,4,5,6,7,8,...). Two questions: 1. What is the expected payoff for the man? 2. If he were to surreptitiously place one more white ball in the bag, by how much does it improve his expectation from that in question 1?

Suppose we have a group of N men and another group of N women. Each man makes a preferential list of women he could marry (ties are not allowed). Similarly, each woman makes a preferential list of men she could marry. It is always possible to pair up all of these people into N man-woman pairs so that the pairing is stable. By stable I mean that, if we take any two pairs in this pairing, (m1,w1) and (m2,w2) say, then it is never the case that [m1 would prefer w2 over w1] AND [w2 would prefer m1 over m2]. That is, it is never the case that m1 and w2 would both prefer each other to their paired partners. An easy proof of this can be found at: http://tinyurl.com/nze7cr If, however, we have to pair up 2N people to be roommates, that's a different matter! If each of these people make a preferential list of possible roommates (of all 2N-1 other people -- with no ties), then it is not always possible to make a stable pairing according to the definition above. What is the smallest N(>2) such that a stable pairing is not always possible? For your value of N, give a preferential order (1 being most preferable to 2N-1 being least preferable) for each of the 2N people, for which a stable pairing cannot be made.

What's wrong with the spoiler? I get my last 2 posts after each other. The second spoiler is what I mean to say.

Consider a horse race at a racetrack. Further consider what happens when people bet that a horse will win that race. A percentage, X (<100%), of the total money bet on these horses to win is returned to the people who have bet on the winning horse. A person's share of the winnings is proportional to how much he bet. The rest, (1-X) of the money bet, is used to pay for expenses and prizes to the horse owners. This is, basically, the definition of parimutuel betting. Now, the racetrack continuously updates odds for each horse on a large display for all to see. These updates stop at the start of the race when no further betting is allowed. These odds tell the bettors what the return on their bet would be should their horse win that race. The odds are posted as a fraction A/B which means that for each dollar the bettor bet, he will receive $((A/B)+1) should his horse win. So, for example, if the bettor bet $1 and the odds posted as 15/1 just as the race began, the track would pay the bettor $16 ($15 plus a return of his $1 bet). Odds of 8/5 would return $2.60 ($1.60 plus his $1 bet). Suppose the odds posted at the start of a 7 horse race were 15/1, 7/2, 8/5, 12/1, 5/2, 6/1, and 10/1. You can derive a reasonable estimate for X (defined in the first paragraph) from this information. What value would you give X and why?

EventHorizon has it! Just to check things, I wrote a program to simulate random games. After 1,000,000,000 random games, the average score from the simulator was 91.4135. Congrats!

I used to like to go ten pin bowling quite a bit. I remember that I scored 83 in my first ever game. It occurred to me that I should compare how I did in that game to the expected score in a type of random bowling. In random bowling, if there are N>0 pins standing, there is an equal probability for each of the possible outcomes of your bowl: knocking down 0 pins, 1 pin, ..., N pins. Each of these outcomes has probability 1/(N+1). The game is scored the way standard 10-pin bowling is scored. So, for example, at the beginning of a frame, your first bowl has 1/11 probability of getting a strike, a 1/11 probability of knocking nine pins down, and so on down to a 1/11 probability of knocking no pins down. If you didn't bowl a strike, there are some number of pins, say X, still standing. On your next throw to complete the frame, you have a 1/(X+1) chance of knocking all X down for a spare, a 1/(X+1) chance of knocking down X-1 pins, and so on down to a 1/(X+1) chance of knocking none of them down. With this scenario, what is your expected score?

Would you post precisely which tickets they would be? Just assume that the mathematicians want to cover all subsets of 6 from the numbers {1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16}. I can't seem to find a set of tickets as you describe. (Webmaster, There seems to be a spoiler problem)

No, you're not missing something. You have a convincing argument that you have found an upper bound. Can you prove that it can't be done with fewer tickets?

I would like to restate the problem: The country of Ozendia has a weekly lottery drawing. A set of 6 numbers are drawn without replacement from the set of numbers {1,2,3,...,40} randomly. A play of this lottery consists of buying a ticket with a buyer-chosen guess as to what these 6 numbers will be. The drawing order of the numbers does not matter. Each play costs $1. The payoff is usually in the multi-million dollar range and it is a percentage of the ticket sales since the last winner. Ozendians love to gamble, so there is usually a small number of winning tickets each week. If this number is greater than one, the jackpot is shared equally among the winning tickets. So far, this is much like many lotteries familiar to most people. Ozendians are social gamblers. That is, very often a group of people will buy a bunch of tickets with any winnigs divided up equitably to the group members. Ozendian lottery officials want to encourage this behavior as they believe it increases ticket sales. So they developed a scheme to allow groups to buy lots of sets of numbers on a single ticket. They have what they call a 7-play ticket. You get to choose 7 numbers on a 7-play and the ticket wins if any subset of 6 of these 7 numbers is picked. Since this is equivalent to buying 7 different normal tickets, the cost is $7 for this kind of ticket. Similarly, they have an 8-play, which is equivalent to 28 standard tickets and so, costs $28. In fact, they have N-play tickets right up to N=15 which costs a whopping $5005! They, in fact have all N-plays for N=6,7,8,9,10,11,12,13,14,15 (I included the standard game as a 6-play). Well (wouldn't you know it?), there's a group of 143 mathematicians in the Ozendian Defence Department who wish to buy the equivalent of a 16-play costing $8008 ($56 for each mathematician). They have decided on their 16 favorite numbers. They wish to cover all subsets of 6 out of their 16 numbers using standard N-play tickets (N=6 to 15) with as few N-play tickets as possible (including the standard game as 6-play). After all, they are mathematicians, so they can afford to pay a bit more than $56 each! What is the minimum number of N-play tickets which would achieve the cover of all subsets of 6 out of the 16 numbers? What would they cost?

Sorry, I meant to ask: What is the minimum number of N-play tickets which would achieve the cover of all subsets of 6 out of the 16 numbers?

The country of Ozendia has a weekly lottery drawing. A set of 6 numbers are drawn without replacement from the set of numbers {1,2,3,...,40} randomly. A play of this lottery consists of buying a ticket with a buyer-chosen guess as to what these 6 numbers will be. The drawing order of the numbers does not matter. Each play costs $1. The payoff is usually in the multi-million dollar range and it is a percentage of the ticket sales since the last winner. Ozendians love to gamble, so there is usually a small number of winning tickets each week. If this number is greater than one, the jackpot is shared equally among the winning tickets. So far, this is much like many lotteries familiar to most people. Ozendians are social gamblers. That is, very often a group of people will buy a bunch of tickets with any winnigs divided up equitably to the group members. Ozendian lottery officials want to encourage this behavior as they believe it increases ticket sales. So they developed a scheme to allow groups to buy lots of sets of numbers on a single ticket. They have what they call a 7-play ticket. You get to choose 7 numbers on a 7-play and the ticket wins if any subset of 6 of these 7 numbers is picked. Since this is equivalent to buying 7 different normal tickets, the cost is $7 for this kind of ticket. Similarly, they have an 8-play, which is equivalent to 28 standard tickets and so, costs $28. In fact, they have N-play tickets right up to N=15 which costs a whopping $5005! They, in fact have all N-plays for N=6,7,8,9,10,11,12,13,14,15 (I included the standard game as a 6-play). Well (wouldn't you know it?), there's a group of 143 mathematicians in the Ozendian Defence Department who wish to buy the equivalent of a 16-play costing $8008 ($56 for each mathematician). They have decided on their 16 favorite numbers. They wish to cover all subsets of 6 out of their 16 numbers using standard N-play tickets (N=6 to 15) at minimal cost. After all, they are mathematicians, so they can afford to pay a bit more than $56 each! How many of each type of N-play tickets (include the standatd game as 6-play) are necessary to achieve minimal cost? What is the cost per play?