# Indicator Random Variables for Card Game (2 Red in a row)

Working through some practice problems in Introduction to Probability, Blitzstein. (I found similar problems to this one but want to make sure I understand this concept.)

You have a well-shuffed 52-card deck. On average, how many pairs of adjacent cards are there such that both cards are red?

Create indicator random variable I, where I=1 if both red, else 0

Cards could be RR, RB, BB, BR, so probability of indicator variable success is:

• P(I=0)=$$\frac{3}{4}$$
• P(I=1)=$$\frac{1}{4}$$

You don't have to check the 1st card alone, so there are 52-1 cards to check. Expectation is therefore:

$$E(I) = \sum_{i=1}^{51} P(I_i=1) = 51(0*\frac{3}{4}+1\frac{1}{4}) = \frac{51}{4}$$

Questions: 1. Is the logic correct? 2. Can someone verify and further explain why sum from 1 to 51 is correct (or incorrect if I'm wrong) 3. Is the nomenclature for what to sum correct?

(Edit: edited P(I=1) and P(I=0) per comment below)

Apart from what looks likes a typo flipping $$P(I_i = 1)$$ and $$P(I_i=0)$$, this looks okay. You should also be careful with being explicit about your random variables; as you define $$I$$, you should have $$E(I) = \frac{1}{4}$$ (what you really want is to define a separate $$I_i$$ for each $$1 \leq i \leq 51$$).
The justification for the sum comes from linearity of expectation. If you define the $$I_i$$'s as described, and let $$I$$ be the random variable denoting the total number of positions where this occurs, we have: $$E\left(I\right) = E\left(\sum_{i=1}^{51} I_i \right) = \sum_{i=1}^{51} E\left(I_i \right) = \sum_{i=1}^{51} \Pr(I_i = 1) = \frac{51}{4}$$
• Good catch on P(I=1) and P(I=0); I'll edit in the question in case anyone gets confused by that. On semantics: is there any issue with writing $\sum_{i=2}^{52}$ instead? – user603569 Dec 1 '18 at 20:26
• Not really; this depends on how you set up your indicators (does $I_i$ represent $i$ and $i+1$ both being red, or $i$ and $i-1$ both being red?) You should get the same answer either way. – platty Dec 1 '18 at 20:27