calculate the variance of the number of sequences of length three cards, such that every card has value bigger than the value of the card before him.

Question

given a standard deck of cards (ordered randomly), (13 card of every type, numbered 1-13, 1 is Ace). Let X be the number of sequences of length three cards, such that every card has value bigger than the value of the card before him. what is Variance of X?

I am stuck with this problem, I tried to divide the problem into indicators and then try to compute the covariance of two indicators. but I didn't know how to do it. can you please help me? thanks

Answer 1

Place the cards on a row on spots that are numbered with $1,2,\dots,52$ and let $V_{i}$ denote the value of the card on spot $i$.

For $i=1,\dots,50$ let $B_{i}$ take value $1$ if $V_{i}<V_{i+1}<V_{i+2}$ and let $B_{i}$ take value $0$ otherwise.

Then $X=\sum_{i=1}^{50}B_{i}$ so that: $$\mathsf{Var}\left(X\right)=\mathsf{Cov}\left(X,X\right)=\sum_{i=1}^{50}\sum_{j=1}^{50}\mathsf{Cov}\left(B_{i},B_{j}\right)=$$$$50\mathsf{Cov}\left(B_{1},B_{1}\right)+98\mathsf{Cov}\left(B_{1},B_{2}\right)+96\mathsf{Cov}\left(B_{1},B_{3}\right)+2256\mathsf{Cov}\left(B_{1},B_{4}\right)$$

Here: $$\mathsf{Cov}\left(B_{1},B_{j}\right)=\mathbb{E}B_{1}B_{j}-\mathbb{E}B_{1}\mathbb{E}B_{j}=\mathbb{E}B_{1}B_{j}-\left(\mathbb{E}B_{1}\right)^{2}=P\left(B_{1}=1=B_{j}\right)-P\left(B_{1}=1\right)^{2}$$ so we are ready if we can find the following probabilities:

• $P\left(B_{1}=1\right)=P\left(V_{1}<V_{2}<V_{3}\right)$
• $P\left(B_{1}=1=B_{2}\right)=P\left(V_{1}<V_{2}<V_{3}<V_{4}\right)$
• $P\left(B_{1}=1=B_{3}\right)=P\left(V_{1}<V_{2}<V_{3}<V_{4}<V_{5}\right)$
• $P\left(B_{1}=1=B_{4}\right)=P\left(V_{4}<V_{5}<V_{6}\mid V_{1}<V_{2}<V_{3}\right)P\left(V_{1}<V_{2}<V_{3}\right)$

Let $E_{n}$ denote the event that the values $V_{1},\dots,V_{n}$ are distinct and let $E$ denote the event that the values $V_{4}$, $V_{5}$ and $V_{6}$ are distinct. Then applying symmetry we can rewrite the bullets as:

• $P\left(B_{1}=1\right)=\frac{1}{3!}P\left(E_{3}\right)=\frac{1}{6}\frac{48}{51}\frac{44}{50}$
• $P\left(B_{1}=1=B_{2}\right)=\frac{1}{4!}P\left(E_{4}\right)=\frac{1}{24}\frac{48}{51}\frac{44}{50}\frac{40}{49}$
• $P\left(B_{1}=1=B_{3}\right)=\frac{1}{5!}P\left(E_{5}\right)=\frac{1}{120}\frac{48}{51}\frac{44}{50}\frac{40}{49}\frac{36}{48}$
• $P\left(B_{1}=1=B_{4}\right)=\frac{1}{3!}P\left(E\mid V_{1}<V_{2}<V_{3}\right)=\frac{1}{6}\left[\frac{9}{49}\frac{6}{48}\frac{43}{47}+2\times\frac{9}{49}\frac{40}{48}\frac{42}{47}+\frac{40}{49}\frac{36}{48}\frac{41}{47}\right]$