Expected number of players to have pocket aces

Question

There are 6 people, each dealt 2 cards from a standard deck. What is the expected number of players that have two aces?

My approach to this was multiplying probability of 1 player gets 2 aces with 1 and added that with probability that 2 players gets two aces and multiplying that by two. Is this the best way to solve or is there a more clever approach?

Answer 1

Let $X_i$ be an indicator variable that is equal to $1$ if person $i$ gets $2$ aces, and $0$ otherwise, then

$P(X_i) = \Large\frac{\binom42}{\binom{52}2}=\frac1{221}$

The expectation of an indicator variable is just the probability of the event it indicates, thus

$\Bbb E[X_i] = \Large\frac1{221},$

and by linearity of expectation, which applies even when the variables ar not independent,

$\Bbb E[X] = \Sigma \Bbb E[Xi] = \Large\frac{6}{221}$