Allright lets play a game. We each sit at a rectangular table with an infinite stack of quarters. We each take turns putting one quarter on the table at a time, untill there is no more room left. Whoever cannot put down a quater looses. So do you want to go first or second, and why?
Just a couple of points I'm not sure about but I'm going to make assumptions so I don't have to wait for your reply before making my first guess. Firstly, we're talking about putting the quarters directly onto the surface, rather than on top of other quarters and by not being able to put a quarter on the table, you mean not being able to put one one other than by sitting it on top of another quarter? Secondly, we're talking about placing the coins onto predefined spaces so that they form a grid of rows and columns. If one could place the coins anywhere, without being allowed to move the coins already on the table, the game could be played out very inefficiently with a lot of space wasted and so the outcome of whether an odd or even number of coins would be on the table at the end would be quite random. So I'm just going to assume that it is one layer of quarters placed to use all available space. So we've got a table that is X coins long and Y coins wide (with both X and Y rounded as far as the last coin that can fit on with its centre of gravity still on the table so that it won't fall off). Let us first represent X and Y as both even numbers, by calling them 2p and 2q where p is X/2 and q is Y/q. Both are integers as X and Y are even. If you multiply X and Y together to get the number of coins that can fit on the table that gives you 4pq which must be an even number as it is divisible by 4. So, if both X and Y are even numbers, then the number of coins that can fit on the table will be even. Let us represent X and Y as both odd numbers. We'll call them X and Y and state that neither has a factor of 2. XY therefore will not have a factor of two either, as all the prime factors are multiplied together but there is no factor 2 to begin with, so there will be no factor of two in the product either. So, if both X and Y are odd numbers, then the number of coins that can fit on the table will be odd. Now then, if one is even and one is odd, then the factor of two in whichever of the dimensions is even will be a factor in the product XY so the product will also be even. The probability that either X or Y is even is 1/2. So the probability that both are odd is 1/2 x 1/2 = 1/4. Which means that the probability that either or both is Even is 3/4. So there is a 75% chance of the number of coins that can fit on the table is even. Therefore if you go second you have a 75% of winning. So I would go second.
yer but youd need too look into it more if your saying that coins can be placed in any fashion. youd need to work with the probability of the amount of rows fitting in if you fit them into the grooves of eachother. its not a neat number, caus its working with circles. caus see, you can move otehr people's coins. depending on the surface size, sliding teh rows into eachotehrs grooves may allow for an extra coin to be placed or not, but this would require too much paperwork to bother. so there is a more accurate answer, which may or may not be the same as yours.
It doesn't matter HOW you fit the rows together. It's still going to be true that there will be a number of rows that all contain the same number of coins. The rows will tesselate rather than be all in line with each other, but each row will still be the same number of coins wide.
Yes you win again, it was my fault for explaining it poorly. The quarters cannot overlap, and there are no predefined places, they can go anywhere. The real solution is as follows. Go first. If you put a quarter in the exact center of the table, you have won right there. Then wherever he puts his quarter, you put a quater on the reflection over on the other side of the table. This way he cannot put a quarter in a spot that you can't reflect it to. Oh yeah, whats with your signature?
no sax, it does matter, for depending on teh tables size in relation to the coins, you could have rows going |o o o o o o| | o o o o o | |o o o o o o | | o o o o o o| but yeah that solution is pretty cool LATERAL THINKNING!!!!!
In that first diagram, you can see that the right hand edges of the table don't line up due to the width of the characters in the line. Now if that right-hand edeg in the second line were moved to be in line with the one in the first line, there would be enough space to fit in one more coin, at least so the centre of gravity is on the table surface, so that it won't topple, espeically if you push the top row over to the far left, so the left most coin JUST has its centre of gravity on the table. Then you'd have the same number of coins in each row, as the pattern would repeat. In the second diagram you DO have the same number of coins in each row. My solution applies as long as their is the same number of coins in each row. As far as edges go, well I'm not sure that a coin would fit in the gap so that it was supported by the table and not by the flat coins. However, if you can do that then that completely changes the rules of the game, and opens up possibilities for far more innovative ways of getting coins on the table.
dude, it wasnt meant to be a scale diagram! its meant to be a demonstration! if you squash all the coins together from that diagram, it goes to show that if you squash em in, you cant fit them on. the table used is not going to be limited by ascii values! and i dont think were tlaking about having them touch teh table, they need to be actualyl on the table completely
They just need to be on the table in a way that means they won't fall off it. Probably best to let the threadstarter clarify that one. Where is he/she anyway?
Nope, I said the answer earlyer, but your logic is sound. I heard this from a buddy who went to a job interview and they asked him this question.
well actually it doesnt matter if thats true of not, my logic still stands for the interlacing coins. jsut adds perhaps more algebra
Infinite amounts of quarters? Here's what I'd do 1. get a wheelbarrow...or a dump truck. 2. head to the bank.
You've stumbled upon something quite profound, there, Blackie. you've got an infinite number of quarters. The bank charge a certain percentage commission for conversion into proper money. So they must therefore take away from you the equivalent value of an infinite number of quarters. So you're left with infinite money minus infinite money ie fuck all. Or to put it another way, what would get you more pissed? Drinking an infinite number of pints of beer? Or an infinite number of half pints of beer? There's a whole branch of mathematics devoted to who there are lots of different kinds of infinity. Thank fuck I don't do maths at university. Incidentally, Two farthings in a ha'penny. Two ha'pennies in a penny. Two pennies is twopence (pronounced tuppence), six pennies is sixpence. Twelve pennies is a shilling. Two shillings and sixpence is a half crown. Twenty shillings is a pound. Twenty-one shillings is a guinea. Couldn't be simpler
Haha. Well the metric system's rather taken over here. We need both, one for physics lessons and the other for travelling and buying food and drink. What is it with Americans and the imperial system. On the one hand you all think you invented it- I had a yank lady come up to me in a supermarket when I worked there asking for 113g of mince or something, and when i gave it to her and said "there you go, a quarter of a pound" she was completely shocked and say "Oh! you're bi-lingual! Fantastic!" In which country did she think she was? From which Empire does the name 'Imperial' come? And on the other hand you can't get it right. There are 20 Fluid Ounces in a pint, you bloody cheap skates! And as for football, well you'd better learn it because it's most popular game in the world. It's even bigger than hide and seek!
Sax Machine, I think I am in love. You are both geeky and hilarious....Come to the "evil empire" and be mine. LMAO Not really, but hey, you rock for being able to do all that algebra like that. I am horrid at it. Wanna help with my homework? Holly
Oooh gosh. Ok, just don't tell Blackie or Shary Bobbins. They might get a bit jealous If you want a bit of help with your maths then just ask. I've got these instant messenger thingies which are quite useful too.
