The cards of a standard 52-card deck are dealt out in a circle. What is the expected number of pairs of adjacent cards which are both black? Express your answer as a common fraction.
I really have no idea how to start. Any help would be great.
[Math] Expected Value with Cards
probability
Related Solutions
The indicator variable is, as you suspected, for the event that a given card is guessed correctly. I'll call it $X_i$ for convenience:
For $i=1,\ldots,52$, let
$$X_i = \begin{cases} 1 & \qquad\text{if card $i$ is guessed correctly} \\ 0 & \qquad\text{otherwise} \end{cases}$$
Then, using linearity of expectation,
$$E(X) = E\left(\sum_{i=1}^{52}{X_i}\right) = \sum_{i=1}^{52}{E\left(X_i\right)} = \sum_{i=1}^{52}{P\left(X_i=1\right)}.$$
We seek $P(X_i = 1)$ now. If we denote by Y (N) a correct (incorrect) guess for any particular card then the event "$X_i=1$" can be represented by the set of all possible strings of length $i$ of Ys and Ns ending in Y.
Take such a string and consider its substrings of maximally consecutive Ns and the following Y. For example, YNNYNNNNYYNY has $5$ such substrings (because it has $5$ Ys) of lengths $1,3,5,1,2$. We can calculate the probability of this outcome, given the rules of the game, to be:
$$\frac{1}{52}\;\cdot\;\frac{50}{51}\cdot\frac{49}{50}\cdot\frac{1}{49}\;\cdot\;\frac{49}{50}\cdot\frac{48}{49}\cdot\frac{47}{48}\cdot\frac{46}{47}\cdot\frac{1}{46}\;\cdot\;\frac{1}{49}\;\cdot\;\frac{47}{48}\cdot\frac{1}{47}\;\; = \;\;\frac{1}{52}\cdot\frac{1}{51}\cdot\frac{1}{50}\cdot\frac{1}{49}\cdot\frac{1}{48}$$
$$\\$$ Hopefully, it's not hard to convince yourself of the pattern here that the probability of any given Y-N string is $\dfrac{(52-k)!}{52!}$ where $k$ is the number of Ys in the string.
Now, of all Y-N strings of length $i$ ending in Y, there are $\binom{i-1}{k-1}$ strings with exactly $k$ Ys since we need to choose $k-1$ positions from $i-1$ possibilities for all Ys but the last. And since this $k$ can range from $1$ to $i$, we have,
\begin{eqnarray*} P(X_i = 1) &=& \sum_{k=1}^{i}{\binom{i-1}{k-1} \dfrac{(52-k)!}{52!}}. \end{eqnarray*}
So, \begin{eqnarray*} E(X) &=& \sum_{i=1}^{52}{\left[ \sum_{k=1}^{i}{\binom{i-1}{k-1} \dfrac{(52-k)!}{52!}} \right]} \\ && \\ &=& \sum_{k=1}^{52}{\left[ \dfrac{(52-k)!}{52!} \sum_{i=k}^{52}{\binom{i-1}{k-1}} \right]} \qquad\text{(swapping the order of summation)} \\ && \\ &=& \sum_{k=1}^{52}{\left[ \dfrac{(52-k)!}{52!} \binom{52}{k} \right]} \qquad\text{(using identity $\sum\limits_{r=p}^{q}{\binom{r}{p}} = \binom{q+1}{p+1}$)} \\ && \\ &=& \sum_{k=1}^{52}{\dfrac{1}{k!}} \\ && \\ \end{eqnarray*}
Note now the Taylor expansion: $\;\;e^x = 1 + \frac{x}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots$.
So, $\;e-1 \;=\; \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \cdots\; = \;\sum\limits_{k=1}^{\infty}{\frac{1}{k!}}$.
We can see $E(X)$ is a partial sum of this series and is extremely close to $e-1$.
The denominator ${52 \choose 5}$ is correct: the number of ways of choosing 5 cards from 52.
If we are going to get no pairs, we can choose any of the 52 for the first card.
But we can only choose 48 for the next card, since 3 of the remaining 51 would give a duplicate. We can only choose 44 for the card after that, 40 after that, and 36 for the last one.
Hence, the probability is $$ \frac{\frac{52*48*44*40*36}{5!}}{52 \choose 5} $$
Which is $$ \frac{4^{5}*{13 \choose 5}}{52 \choose 5} $$
In other words, you're missing a $4^{4}$ in your computation because each of the 5 cards can be any of the 4 suits.
Best Answer
Hint to get you started:
Use indicator variable.
Focus on a pair of card, what is the probability that both are black.
Edit:
Let's introduce the indicator variable $X_i$ such that it is equal to $1$ when it is black and the very next card (counter clockwise) is black as well and it is equal to $0$ otherwise.
Then the number of pairs of adjacent cards which are both black is $\sum_{i=1}^{52} X_i$ and its expected value is $$\mathbb{E} \left[\sum_{i=1}^{52} X_i\right] =\sum_{i=1}^{52} \mathbb{E} \left[X_i\right] =\sum_{i=1}^{52} \frac12 \cdot \frac{25}{51} =\frac{26\cdot 25}{51}=\frac{650}{51}$$