Four players, each received $13$ cards, player A saw that their neighbour has ace of spades. What is the probability that player A doesn't have ace.
My textbook says it is $$\dfrac{{39 \choose 13}}{{51 \choose 13}},$$ but I have no idea why.
Can someone explain me why this is answer or provide another if this is not correct.
This is a long-winded comment, that bends if not breaks the rule that an answer should not be provided, for low quality postings.
The reason that I am responding is that the book's answer is wrong. I question how reasonable it is to ask the OP (i.e. original poster) to show work, when the computation that the OP would be trying to establish is wrong.
In a probability problem like this, a Combinatorics answer of $~\dfrac{N}{D}~$ will typically represent (as in this case) that:
There are $~D~$ equally likely possible ways that event $E_1$ can occur.
Event $E_2$, which represents a subset of the event $E_1$ can occur in $N$ ways.
So, you let $E_1$ represent the event that a player is receiving $13$ cards from a $~51~$ card deck. That is, since the Ace of Spades has been seen in someone else's hand, the player is (in effect) being dealt $~13~$ cards from a deck with only $~51~$ cards remaining.
This explains the computation of $\displaystyle ~D~ = \binom{51}{13}.$
I agree with the OP that the computation $\displaystyle ~N~ = \binom{39}{13}~$ is wrong. Instead, the correct computation is $~\displaystyle N = \binom{48}{13}.$
This is because when you remove the three other Aces from the $51$ cards remaining in the deck, you are left with $48$ cards, none of which are an Ace.
So, the probability of not receiving an Ace, from a 51 card deck that contains $3$ Aces is
$$\frac{\binom{48}{13}}{\binom{51}{13}}.$$
Most likely explanation is that either the OP mis-stated the problem or that the problem composer made a mistake. Consider the alternative problem of :
The probability that the player did not receive any spades, given that the Ace of spades is known to be in someone else's hand.
This probability would in fact be
$$\frac{\binom{39}{13}}{\binom{51}{13}},$$
and is explained by reasoning that when you remove the $12$ remaining spades from the $51$ card deck, you have $39$ non-spades. $~\binom{39}{13}~$ represents the number of ways of receiving $~13~$ cards, none of which are spades.