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

  • $\begingroup$ 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? $\endgroup$ – user603569 Dec 1 '18 at 20:26
  • 1
    $\begingroup$ 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. $\endgroup$ – platty Dec 1 '18 at 20:27

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