Find the exact formula for P(Sn=k)

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)?

sillybuffalo said:

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)?

Click to expand...

I'll put down some random thoughts, to see if they open any possibilities.

The drawing of any particular ticket has a binary distribution with n trials
p(0) = p(2) = 1/4, ...q(0) = q(2) = 3/4
p(1) = 1/2, ............q(1) = 1/2

The sample space is [0,2n],...E[sum] = n

Suppose you draw a 0s, b 1s, and c 2s
then...................n = a + b + c
.........................k = a*0 + b*1 + c*2 = n + c - a
That looks interesting ..
since n is a constant, you only have to deal with the binary distributions of a and c,
selected such that c - a = k - n