How to determine if two events are independent when dealing a 5-card hand?

Question

Let's assume we have a standard deck of 52 cards. A dealt hand will be 5 cards. Consider the following probabilities:

P(dealt at least 1 ace)
P(dealt no 2s)
P(dealt at least 1 ace AND dealt no 2s) = P(dealt at least 1 ace) * P(dealt no 2s)???

Are these independent events? My hunch is that these are dependent events because if I get dealt at least 1 ace, then there are only 4 cards left, which affects the probability of P(dealt no 2s). Normally P(dealt no 2s) would be calculated like so:

$${48 \choose 5} \div {52 \choose 5}$$

But if we open at least 1 ace, then we would have to choose 4, no?

Answer 1

Let's consider the set $A$ of all possible hands that do not contain any twos.

Now consider all possible hands that do not contain any twos and also do not contain any aces. Let $B$ be the set of all such hands.

Then $B$ is a subset of $A.$ And if you remove all hands in $B$ from the set $A,$ what is left is a set of hands that do not contain any twos, but every hand contains at least one ace. This remaining set is denoted $A \setminus B.$

We can count the number of hands in the set $A$: $\lvert A\rvert = \binom{48}{5}.$ The number of hands in $B$ is $\lvert B\rvert = \binom{44}{5}.$ Since $B$ is a subset of $A,$ we can find the number of hands in $A \setminus B$ by setting $\lvert A \setminus B\rvert = \lvert A\rvert - \lvert B\rvert = \binom{48}{5} - \binom{44}{5}.$

So the probability to have at least one ace, but no twos, is $$ P_1 = \frac{\binom{48}{5} - \binom{44}{5}}{\binom{52}{5}} $$

whereas the probability of at least one ace is $\binom{48}{5}/\binom{52}{5},$ the probability of no twos is $1 - \binom{48}{5}/\binom{52}{5},$ and the product of the two probabilities is $$ P_2 = \frac{\binom{48}{5}}{\binom{52}{5}} \left(1 - \frac{\binom{48}{5}}{\binom{52}{5}}\right). $$

If you correctly compare these quantities, then you will find that $P_1$ is slightly greater that $P_2.$ It may help with the comparison if you rewrite each of the quantities as follows: \begin{align} P_1 &= \frac{\binom{48}{5}}{\binom{52}{5}} - \frac{1}{\binom{52}{5}} \binom{44}{5}, \\ P_2 &= \frac{\binom{48}{5}}{\binom{52}{5}} - \frac{1}{\binom{52}{5}} \left(\frac{\binom{48}{5}^2}{\binom{52}{5}}\right). \end{align} Then you just need to show that $\frac{\binom{48}{5}^2}{\binom{52}{5}} > \binom{44}{5}.$ After a few more cancellations of common factors, this comes down to $$ \frac{(48\cdot47\cdot46\cdot45\cdot44)^2}{52\cdot51\cdot50\cdot49\cdot48} > 44\cdot43\cdot42\cdot41\cdot40, $$ which is true because in general, $(n+4)^2 > n(n+8).$

Answer 2

You are right, these are dependent events. Let $A$ be the event of getting at least one ace, $B$ - getting no $2$'s. Then, $$ \mathbb{P}[A] = 1 - \mathbb{P}[\text{not } A] = 1 - \binom{48}{5} \div \binom{52}{5} $$ and $$ \mathbb{P}[B] = \binom{48}{5} \div \binom{52}{5}, $$ but can you compute $\mathbb{P}[A \cap B]$?