Help ME with some math!

[Streetwolf]

Alright guys who is up for a challenge??

I know there's a few of you who actually do enjoy math (yes nerds, dorks, geeks, whatever, I don't care what you call us). I have recently been speaking to a couple of the grad students at my school, and they are taking this really cool new course about Game Theory and Word Problems. One of the grad students photocopied this assignment sheet for me. I've removed some of the answers he had listed so that no one bases their reasoning off of a given answer.

It is oftentimes harder to show WHY then to show the answer itself. For instance, I got the right answer to #3 already but am still working on the reasoning.

These are really challenging problems and were given to me so that I could do them FOR FUN. This is not HW. If anyone who enjoys tough problems wants to share their thoughts on any of this, feel free!!

I posted this in the academy because it is certainly academic in nature.

Here is the link: http://people.brandeis.edu/~jshapiro/mathprobs.jpg

[SithDrummer]

Very cool, I'll be working on it. :)

[dook]

For instance, I got the right answer to #3 already but am still working on the reasoning.
For #3 when the game is played with 4^n numbers, A should force B into paying 2^n. Proof by induction on n once you're convinced of what A's and B's strategy should be.

Looks like fun. I'm a busy man* but let me know which ones you're curious about and I'll take a look too.

*that's a lie. i have nothing but my own research until mid-january

[Streetwolf]

For #3 when the game is played with 4^n numbers, A should force B into paying 2^n. Proof by induction on n once you're convinced of what A's and B's strategy should be.

Looks like fun. I'm a busy man* but let me know which ones you're curious about and I'll take a look too.

*that's a lie. i have nothing but my own research until mid-january
The game is being played with 257 numbers, not 256. But given what you said, you are right that B pays 2^n when they both use their best strategies.

[dook]

it'll be a cold day in hell before i recognize 87 as a number

[dook]

bump for the thread. coolest thing to hit the academy in a while and it's getting no love.

i think #7 has a cool solution. gonna have to remember that one

[SithDrummer]

I lied. Work has been brutal to me this week! I'll look at it again and put it away regretfully, I swear! :(

[SithDrummer]

Had some fun solving #2 today! :)

[Mexi]

it'll be a cold day in hell before i recognize 87 as a number

i laughed



bump!

[spin]

ugh im in calc 3 (and im pretty good at it) and just glancing at that gave me a headache :( if im bored over break (after finals) i may return and work on them for fun :)

[dook]

Had some fun solving #2 today! :)

#2 seemed easy

When I first read it I thought it meant without replacement, in which case for p>=1000, A can always force a win by playing the lowest number available. But then the question makes no sense for part b since 9! = 362880 which is less than a million.

With replacement the problem is easier, cause the players have no influence over what the other player has available to choose from... so just work backwards.. for instance, if you win at >= 10^6, then you'd lose if you hit between 111,112 and 999,999... and you'd win if you forced somebody into that band, which means you win if you hit from 55,556 to 111,111. Keep going and you'll get that you win if you hit 10 <= p <= 19 and so you're screwed if you go first. :(

[dook]

Some of these I could see right away, but #4 has been in the back of my head a few days. I finally got it, and just wanted to say that both I and that problem are retarded. ty.