TITLE: finding the number of circles we get when randomly placing given patterns into a grid of squares
QUESTION [5 upvotes]: We have an 11$\times$11 table of squares (consist of 121 squares of dimension 1$\times$1). we have 3 tiles shown in the picture. Each tile has dimension 1$\times$1. we now randomly pick 3 tiles into the table. Let $N$ denote the total of number of circles we get in a random drawing. Compute expected value of $N$, mean of $N$ and standard deviation of $N$.
I think of an idea that we construct a $X$-$Y$ coordinate in accordance with the 11$\times$11 table, the only way to get a circle is to put the first tile in square $(x,y)$ and $(x+1,y-1)$ and put the third tile in square $(x+1,y+1)$ and $(x,y-1)$ for $0 \le x,y \le 11$. I want to construct a variable that takes on value 1 if this arrangement occurs and 0 if not, then construct what I want. But now I get stuck. Any help would be really appreciated. thanks
The image of the three tile patterns is below

REPLY [2 votes]: Let $n_k$ be an indicator rv (Bernoulli, $p=1/81$) which takes the value 1 if the $2\times 2 $ block $k$ forms a circle, 0 otherwise. The index $k$ runs over the $(L-1)^2=100$ blocks, where $L=11$ is the number of squares per side.
Then, if $N$ is the number of circles, we have $N =\sum_k n_k$ and $E(n_k)=p$, hence
$$E(N) = (L-1)^2 \, p = \frac{100}{81} = 1.2345679
 $$
Further, 
$$E(N^2) = E\left[\sum_k n_k\right]^2 = (L-1)^2 E(n_k^2) + \sum_{k \ne j} E(n_j n_k)$$
Now, $E(n_j n_k)=0$ if the blocks overlap at two squares; there are $2(L-2)(L-1)=180$ such pairs. 
If the block pair operlaps diagonally, at one square,  $E(n_j n_k) =1/3^7$; there are $(L-2)^2=81$ such pairs. 
And $E(n_j n_k)= E(n_j)E(n_k)= p^2$ if the pair doesn't overlap: $(L-2)^{2}\,( {L}^{2}-2)= 9639$ pairs.
Then $E(N^2)  = 74/27$ and $Var(N) = E(N^2) - E(N)^2=7982/6561=1.2165828$ 
Notice that is a little less that the variance obtained by the Binomial approximation suggested in Ross's answer ($1.2193263$), which was to be expected (the negative correlation tends to lower the variance of the sum).