Here is my attempt to explain the solution in the best possible manner.
Hope this will help you and others as well 
First : what really matters is the number of dishes with 0, 1, 2 and 3 sushis and not the order of the dishes.
So answer for 2,1,0,2,1 is same as answer for 0,1,1,2,2.
Number of dishes with 0 sushis is easily determined by N - one - two - three, where one is the number of dishes with 1 sushi and N is the total number of dishes in the input.
Let F(x, y, z) be the expected moves needed for x dishes with 1 sushi, y with 2 and z with 3.
Now in the next move we can pick a dish with 1 sushi with a probability of x/N or p1. we can pick a dish with 2 sushi with a probability of y/N or p2. we can pick a dish with 3 sushi with a probability of z/N or p3. we can pick a dish with 0 sushi with a probability of (N - (x + y + z))/N or p0.
Now try to understand this :
F(x,y,z) = 1 + p0F(x,y,z) + p1F(x-1,y,z) + p2F(x+1,y-1,z) + p3F(x,y+1,z-1)
Here we add a 1 for the current move that we are making.
(Note : if you pick a dish with 3 sushi z decreases but y increases)
This equation now becomes :
(1 - p0) F(x,y,z) = 1 + p1F(x-1,y,z) + p2F(x+1,y-1,z) + p3*F(x,y+1,z-1)
simplifies to:
F(x,y,z) = (p1F(x-1,y,z) + p2F(x+1,y-1,z) + p3*F(x,y+1,z-1))/(1-p0)
This equation can be easily evaluated using dynamic programming
However try and simplify the equation to minimize the number of divisions for higher precision.
Implementation : Submission #9551520 - Educational DP Contest
Let me know if you find something is not clear.