I'm sure that by now most Team Fortress 2 players know or have heard about unusual hats. They are very rare and can only be found in Mann Co. Supply Crates which adds to their value. The probability of getting an unusual hat from a normal crate(not festive) is 1% or .01 but the probability increases the more crates you open. This is[only] post so far that actually explains the process and shows some math to back it up. I really hope this helps you guys because I put some time into this one.
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Let p = Probability of an unusual hat from a single crate(.01)
Let n = The amount of crates opened
Let k = A number from 1 to n which represents the number of unusual hats you get from n number of crates opened 
Assume that we want AT LEAST ONE unusual hat

1) The probability of getting at least one unusual hat from opening 1 crate is p
2) The probability of getting at least one unusual hat from opening 2 crates is (p^2) + 2p(1-p); The first term is the probability of getting two unusuals in a row, plus the second term, which is the possible combinations of getting at least one unusual from two crates(two possible combinations, therefore the coefficient is 2p)
3) The probability of getting at least one unusual hat from opening 3 crates is (p^3) + 3p^2(1-p) + 3p(1-p)^2; The first term is the probability of getting three unusuals in a row, plus the second term, which is the possible combinations of getting at least two unusuals(three possible combinations) plus the third term, which is the probability of getting at least one unusual(three possible combinations)

Since you guys(or gals) are smart, you have probably started to notice a pattern in these expressions, and that means we can derive a general formula for the probability of getting at least one unusual hat from n number of crates.

n) The probability of getting at least one unusual hat from opening n number of crates is:
(p^n) + {1 C n}[p^(n-1)](1-p) + {2 C n}[p^(n-2)](1-p) + ...
Forgive me for the poor notation, but {1 C n} means the possible combinations for getting at least one unusual from n number of crates.

Cutting to the chase, the final formula is as follows:

{n Σ k=1}({k C n}[p^k](1-p)^(n-k)) which really means:

Find the sum of the total possible combinations of least amount of unusual hats(k) from n number of crates multiplied by the probability of getting one unusual from one crate to the power of our constant(k), multiplied by the chance of NOT getting an unusual hat to the power of the number of crates minus our constant.

What a doozy, but let's see if this checks out:
Plugging in one crate, we get the .01 or 1% chance of getting at least one unusual hat from one crate. If we plug in 69 crates, we get .500 or 50% chance to get at least one unusual hat from 100 crates. Finally, if we open 100 crates, the probability of getting at least one unusual hat is .634 or a 63.4% chance.
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