Pot

[Alan]

Transcript of my AIM chat with Vinny:

KazemaF: you should sign up for puzzle pirates
KazemaF: hehe
VLSmooth42: gimme next week
VLSmooth42: unfortunately, I have a really busy schedule this week
VLSmooth42: actually just finished having a poker night at my place with 4 other people
KazemaF: heh
KazemaF you lose all your money eh?
VLSmooth42: have to go to a business breakfast at 9 am, a meeting at noon, and a movie with jason and co tomrrow evening
KazemaF: that's ok, real men don't need money
VLSmooth42: nope
VLSmooth42: started with $18
VLSmooth42: when down to ~90 cents
VLSmooth42: then shot up to $18 at the end
VLSmooth42: it was CRAZY!
VLSmooth42: er
VLSmooth42: started with $10
VLSmooth42: $50 pot, 5 people
KazemaF: how was the $50 pot?
VLSmooth42: $10 of chips per person
KazemaF: how much did you get for $50?
VLSmooth42: 3 colors of chips
VLSmooth42: blue/red/white, 25/10/5 cent chips
VLSmooth42: ante was a nickel
KazemaF: so was it good?
VLSmooth42: yep
VLSmooth42: I got $18
VLSmooth42: an $8 gain
VLSmooth42: ended up playing for a little over 4 hours
VLSmooth42: remarkably amusing since we were also playing dealer's choice
KazemaF: seems like a $32 loss but I guess you must have put in oregano or something
VLSmooth42: ie. the dealer could pick the game
KazemaF: get more out of less
KazemaF: yeah if you're the dealer everyone has to play by your rules
VLSmooth42: we played at least 5 different games from polish poker, 5 card draw to anaconda and man or mouse
VLSmooth42: ah
VLSmooth42: we each put in $10
VLSmooth42: the $50 was the total for 5 people
VLSmooth42: sorry, I guess pot was the wrong therm
VLSmooth42: we each got $10 in chips
KazemaF: seems like you can't get much out of just $50 pot
KazemaF: for 5 people
VLSmooth42: it wasn't to make money -_-
VLSmooth42: my apartment is NOT las vegas ^_^;
KazemaF: then why deal it? just keep it for yourself
VLSmooth42: ?
VLSmooth42: I'm confused with what you're saying
VLSmooth42: we each put $10 into the card box
VLSmooth42: and each got $10 of cihps
VLSmooth42: we played for a little over $4
VLSmooth42: 4 hours*
VLSmooth42: then cashed out
KazemaF: $50 pot isn't much to begin with, if you're not gonna try to make money out of it, why deal it out?
VLSmooth42: the money put some bite in the game
VLSmooth42: a $9 pot is actually amazing
KazemaF: really
KazemaF: $9 pot is pretty crap from what I hear
VLSmooth42: the money was to add bite to the game
VLSmooth42: so people don't just screw around
VLSmooth42: well, we'd rather not smoke plastic chips
KazemaF: all you can do with $9 pot is screw around really
VLSmooth42: 1) they don't smell good
KazemaF: haha was wondering when you'd catch on
VLSmooth42: 2) they're noxious
VLSmooth42: heh

[quantus]

what? vinny's a drug addict? Man, he better hope he doesn't get a random drug test at work. I'd feel bad too because we just didn't pay enough attention and catch him at it before it became an addiction.

[Peijen]

what? vinny's a drug addict? Man, he better hope he doesn't get a random drug test at work. I'd feel bad too because we just didn't pay enough attention and catch him at it before it became an addiction.

the sign was all there too. Paranoia, day sleeper, doesn't eat, lock himself in the bathroom hours at time ... wait a minute he really is doing drug!!

[Dave]

!huggle

[quantus]

doesn't eat

I believe that vinny would get the munchies if he were doing pot... but I guess he could still pass out if he doesn't have anything to drink with all that food...

[bob]

When we had our most recent party, a few weekends ago, I observed some astounding mathematical principles*.

1. pot + alcohol = someone running into the screen door
2. pot + alcohol + time = someone else running into the screen door
3. pot + alcohol + more time = first person running into the screen door and not stopping

Homework: Given a screen door and a sufficient supply of alcohol and pot, there is some probability p that a given person will run into the screen door during any given minute. Assuming n people at such a party, the expected number of times that the screen door will be run into after t minutes is pnt. Now, assume that for each time someone runs into the screen door, there is a probability s that the person will not stop and will inadvertently remove the screen door from its proper place, thus making it impossible for anyone else to run into it. Over the course of a party lasting t minutes, what is the expected number of times that the screen door will be run into?

*remember kids, the person at school is spelled "princiPAL", because he's your friend[/i]

[Alan]

u nerd

[quantus]

When we had our most recent party, a few weekends ago, I observed some astounding mathematical principles*.

1. pot + alcohol = someone running into the screen door
2. pot + alcohol + time = someone else running into the screen door
3. pot + alcohol + more time = first person running into the screen door and not stopping

Homework: Given a screen door and a sufficient supply of alcohol and pot, there is some probability p that a given person will run into the screen door during any given minute. Assuming n people at such a party, the expected number of times that the screen door will be run into after t minutes is pnt. Now, assume that for each time someone runs into the screen door, there is a probability s that the person will not stop and will inadvertently remove the screen door from its proper place, thus making it impossible for anyone else to run into it. Over the course of a party lasting t minutes, what is the expected number of times that the screen door will be run into?

*remember kids, the person at school is spelled "princiPAL", because he's your friend[/i]

We're to assume that there's no replacement of the screen door? Or is there a time penalty to replacing it?

[Jonathan]

I think, for a given party, the probability P of there being one or more replacement screen doors is very very low, such that P =~ 0.

[quantus]

As it turns out, number of times to bump into the screen door until it breaks is a geometric random variable and as such has an expected value of 1/s. So, solve for t in npt = 1/s and get t = 1/(snp) for the expected life of the screen door. So, if T is the length of the party and less than 1/(snp), then the number of times is simply npt, otherwise it's 1/s.

[quantus]

Now, bob's message implies that the screen is simply dislodged and can be replaced. If this is the case, then there is repeatability of this procedure. The more complete answer is then:

Code: Select all

r = replacement time
n = number of people
T = Time of the party
N(T) = number of times the screen is run into
I = int(T/(d+r)) = number of times the screen is dislodged
J = T%(d+r) = the remaining time to bump into the screen but probably not dislodge it
d = 1/(snp) = mean time to dislodge the screen
        
N(T) = /  I/s+npJ   if J < d
       \  (I+1)/s   if d <= J < d+r

This could be done with more statistical accuracy, but this should be good enough.

[quantus]

Can I have a cookie now?

[bob]

cookies.txt has now been created on your harddrive.

[quantus]

yay!

*runs cookie_monster.exe*