# Probability problem of 220 people randomly selecting only 12 of 35 exclusive options.

There are 220 people and 35 boxes filled with trinkets.

Each person takes one trinket out of a random box.

What is the probability that the 220 people will have grabbed a trinket from exactly 12 different boxes?

I'm trying to calculate the probability of grabbing a trinket from at most 12 boxes, $P(12)$. Then calculate $P(11)$ with the answer being $P(12)-P(11)$ but I'm drawing blank.
$P(12) = 1-(23/35)^{220}$ doesn't look right to me.

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Very few discrete probability calculations result in a cumulative probability, like $P(X\leq C)$. – rschwieb Jun 1 '12 at 11:56
@rschwieb How does this affect my problem? – Captain Giraffe Jun 1 '12 at 12:08
I'm still working on an explanation. I just mean that it looks like you're treating all probability computations as if they were cumulative. It also looks like you (arbitrarily) attempted a binomial probability computation, when the givens don't resemble a simple binomial experiment. It's a bit more complex than that, I think :) I could be (and probably am) wrong about your intentions, but I just wanted to alert you to habits you might not have realized you were exercising. – rschwieb Jun 1 '12 at 12:20
@rschwieb I see, and you are correct in your assumption of my previous attempt. – Captain Giraffe Jun 1 '12 at 12:21

If the probability that the first $n$ people have chosen from exactly $c$ boxes out of a possible $t$ total boxes [$t=35$ in this case] is $p(n,c,t)$ then $$p(n,c,t)=\frac{c \times p(n-1,c,t)+(t-c+1)\times p(n-1,c-1,t)}{t}$$ starting with $p(0,c,t)=0$ and $p(n,0,t)=0$ except $p(0,0,t)=1$.
Using this gives $p(220,12,35) \approx 4.42899922 \times 10^{-94}$. This is close to but not exactly the naive ${35 \choose 12}\times \left(\frac{12}{35}\right)^{220} \approx 4.42899948\times 10^{-94}$.
The probability the first $n$ have chosen from $c$ boxes is the probability the first $n-1$ have chosen from $c$ boxes times the probability the $n$th chooses from one of the $c$ plus the probability the first $n-1$ have chosen from $c-1$ boxes times the probability the $n$th chooses from one of the rest. – Henry Jun 2 '12 at 6:42