Let $n ≥ 3$ be an integer and consider a group $P_1, P_2, . . . , P_n$ of $n$ people. Each of these people has a uniformly random birthday, which is independent of the birthdays of the other people. We ignore leap years; thus, the year has $365$ days.
Define the random variable $X$ to be the number of unordered triples $\{P_i, P_j, P_k\}$ of people (i.e., subsets consisting of three people) that have the same birthday.
What is the expected value $E(X)$ of $X$?
Hint: Use indicator random variablesAnswer: $\frac{1}{365^2}$$ n \choose3$
The answer is intuitive without using indicator random variables (Expected value of binomial distribution), but I'm not quite sure what the indicator random variable would be here if I took that route.
This is my wild guess, but it's clearly wrong:
Random variable:
$X =$ # of unordered triplets with the same birthday
Indicator Random Variable
$X_i = 1$, if $X_i = X_{i+1} = X_{i+2}$
$X_i = 0$, otherwise
$Pr(X_i) = 1/4$ because $X_i = 1$ if you have the substring $111$ or $000$.
Then, to get the expected value, it has to be the sum of $X_i$ up to $n-3$.
I know this is wrong, so I will not continue, but I have tried to show some thought process behind this.
Thanks
I agree that indicator variables aren't really necessary. If I wanted to be explicit about using them, it would be something like this:
For all possible triples $T_i$, let $X_i = 1$ if the triple all have the same birthday, $0$ otherwise.
$\operatorname{E}[X_i] = \frac{1}{365^2}$ and there are $k=\binom{n}{3}$ such $X_i$, so $$\operatorname{E}[X] = \operatorname{E}[X_1 + \dots X_k] = \binom{n}{3} \operatorname{E}[X_i] = \frac{1}{365^2}\binom{n}{3}$$