A few questions regarding TF2 math...

I would say "puzzle's too hard", but you have no pending giveaways, so I'll just say free bump.

For A or indeed C (ASSUMING p CONSTANT!), you have a Binomial distribution. Written as X~Bin(n, p).

Probability of getting more than k successes:
P(X>= k) = 1 - P(X<= k) = 1 - SUM[from i = 0 to k]((n choose i)*(p^i)*(1-p)^(n-i))

Which looks horrific. For clarity, n choose i is n! / (i!*(n-i)!)

EDIT: It's been a while since I've done stats. Kind of glad no one posted this whilst I was thinking it over...

answer: be prepared to spend a alot of keys in hopes to make profit

I'll nail this, when I get home.

42

by watching Mike's uncrates, 1 unusual ~150 crates is a good measure.

Been looking at (F).

I reckon (and this is going to be messy as fuck, I apologise). I'm also not clear on whether unusuals can have the same name and different effects? Lets assume so.

Lets define some Binomial distributions, since that is what I think they are (and define a few other things while we are at it).
Let h = number of hats. Let I = number of items in TF2 that can be unusual. Let m = number of different effects
So we have the chance of getting an unusual from opening crates: U~Bin(n, p).
Chance of that unusual being a hat: h/I
Chance of it being our fav. hat: 1/h --> chance of it being our fav hat & unusual: 1/h h/I = 1/I.
Chance of it being our fav. unusual hat with our fav. effect: 1/I 1/m = 1/Im

Sooo, I reckon we can now say that the probability that the crate opens and gives our fav. hat, unusual and effect is p/Im. Because p is already a probability so we don't need to take it's reciprocal...

Now we can define a new Binomial, since U was Binomial in the first place. Lets use Z, because you know, why the fuck not? Z~Bin(n, p/Im)

P(Z>= k ) = 1 - P(Z<= k) = 1 - SUM[from i = 0 to k](n choose i)*(p/Im^i)*(1-p/Im)^(n-i))

EDIT: So if unusuals can only have one effect, the formula becomes:
P(Z>= k ) = 1 - P(Z<= k) = 1 - SUM[from i = 0 to k](n choose i)*(p/I^i)*(1-p/I)^(n-i))

simple language please i just woke up understood nothing -,-

Fuck math.

Bump for solved.

the simple answer is hats

Buy keys
.........
Profit

Probability -> 0 as fucks given -> 0.

Thanks for skyrim

Probability can't be too much since i never play it.

I thought TF2 math was something like "I had 5 hats and gave 2 to a friend, how many hats do I have now?"

Didn't do anything like that since... 4+ years ago I guess? ;_;
The only thing I know is, I know a guy who had opened 600+ crates and got shit, and I know a guy who got 2 unusuals in 9 crates.