Diamond Variance
In a game of cards, \(10\) cards are dealt from a standard \(52\)-card deck. What is the variance of the number of diamonds among these \(10\) cards?
Answer
Let \(D\) be the number of diamonds in the \(10\)-card hand. For \(i=1,\dots,10\), define the indicator
\[
I_i=\mathbf{1}\{\text{the \(i\)-th dealt card is a diamond}\}.
\]
Then
\[
D=\sum_{i=1}^{10} I_i
\quad\Longrightarrow\quad
\mathrm{Var}(D)=\sum_{i=1}^{10} \mathrm{Var}(I_i)+2\sum_{1\le i<j\le 10}\mathrm{Cov}(I_i,I_j).
\]
First,
\[
\mathbb{P}(I_i=1)=\frac{13}{52}=\frac14
\quad\Rightarrow\quad
\mathrm{Var}(I_i)=\frac14\cdot\frac34=\frac{3}{16}.
\]
For \(i\ne j\),
\[
\mathbb{P}(I_i=1,I_j=1)=\frac{13}{52}\cdot\frac{12}{51}=\frac{1}{17},
\]
so
\[
\mathrm{Cov}(I_i,I_j)=\mathbb{P}(I_i=1,I_j=1)-\mathbb{P}(I_i=1)\mathbb{P}(I_j=1)
=\frac{1}{17}-\left(\frac14\right)^2
=-\frac{1}{272}.
\]
Now plug in:
\[
\mathrm{Var}(D)=10\cdot\frac{3}{16}+2\binom{10}{2}\left(-\frac{1}{272}\right)
=\frac{30}{16}-\frac{90}{272}
=\frac{15}{8}-\frac{45}{136}
=\frac{105}{68}.
\]
\[
\boxed{\mathrm{Var}(D)=\frac{105}{68}.}
\]
Notes and comments
Comment 1: For indicator r.v.'s \(I\), if \(\mathbb{P}(I=1)=p\) then
\[
\mathbb{E}[I]=p,\qquad \mathrm{Var}(I)=p(1-p),
\]
and for two indicators \(I,J\),
\[
\mathrm{Cov}(I,J)=\mathbb{E}[IJ]-\mathbb{E}[I]\mathbb{E}[J]
=\mathbb{P}(I=1,J=1)-p^2.
\]