We shuffle up a standard deck of cards. What is the expected number of adjacent pairs of Aces in the deck? I was thinking about using the indicator method, and having each indicator variable be 1 (success) if the first two cards are adjacent pair of aces or 0 otherwise, and do that for all the pairs of cards in the shuffled deck. However this seems long and tedious is there a better method of using indicator variables?
[Math] Finding adjacent cards using indicator method
discrete mathematicsprobability distributionsrandom variables
Best Answer
I don't think your method is tedious. One can use the linearity of expectation to make the calculation quite short.
For $i = 1, \ldots, 51$, let $E_i$ be the event the $i^\text{th}$ and $(i+1)^\text{st}$ cards are both aces. One can compute this probability by counting unordered pairs, and we see that $Pr(E_i)$ is the same for all $i$. Denote this probability by $p$.
Letting $X_i = \mathbb{1}_{E_i}$ be the indicator variable for $E_i$, then $\mathbb{E}(X_i) = Pr(E_i)$. By the linearity of expectation, then \begin{align*} \mathbb{E}\left(\sum_{i=1}^{51} X_i\right) &= \sum_{i=1}^{51}\mathbb{E}\left(X_i\right) = \sum_{i=1}^{51}Pr(E_i) = 51 p \, . \end{align*} I'll leave it to you to compute $Pr(E_i)$. In the end, I find that the expectation for the number of consecutive aces is $\approx 0.230769230769231$.