Find the exact formula for P(Sn=k)

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The discussion focuses on deriving the exact formula for P(Sn=k) where Sn represents the sum of numbers drawn from a box containing tickets numbered 0, 1, 1, and 2. The initial probabilities calculated are P(Sn=0) = (1/4)^n, P(Sn=1) = n * (1/4)^(n-1) * (1/2), and P(Sn=2) = n * (1/4)^(n-1) * (1/4) + (n choose 2) * (1/4)^(n-2) * (1/2)^2. The discussion suggests utilizing the trinomial theorem to simplify the derivation of P(Sn=k) for larger values of k, as the combinations become increasingly complex.

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A box contains four tickets, numbered 0,1,1, and 2. Let Sn be the sum of the numbers obtained from n draws at random with replacement from the box.
Find the exact formula for P(Sn=k) (k=0,1,2...)

I started by finding the probability of the first few sums:
P(Sn=0) = (1/4)^n
P(Sn=1) = n * (1/4)^(n-1) * (1/2)
P(Sn=2) = n * (1/4)^(n-1)*(1/4) + (n choose 2) * (1/4)^(n-2)(1/2)^2

Then I get stuck, since the combinations of draws that lead to the subsequent draws get much more complicated.
Is there a more simplified way to find the formula for P(Sn=k)?
 
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