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)
Best Answer
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: $$E\left(I\right) = E\left(\sum_{i=1}^{51} I_i \right) = \sum_{i=1}^{51} E\left(I_i \right) = \sum_{i=1}^{51} \Pr(I_i = 1) = \frac{51}{4}$$