A player is dealt six cards out of a normal deck of cards. He looks at the first four and notices there is no ace among them. What is the probability that he does not have an ace at all.
This sounds like a conditional probability problem, but I reckon it just as well can be a counting problem: the required probability would be one minus the probability of a hand with four non-aces, one ace, and one arbitrary card.
So by that reasoning, $$p=1-\frac{{48\choose 4}{4\choose 1}{46\choose 1}}{ 52\choose{6}}$$
Is this right?
Best Answer
Rajkumar's direct solution works nicely. You mentioned that the problem could be considered as conditional probability, and I'll show here that that method gives the same answer.
As I mentioned in a comment, the main reason for the inaccuracy of your original answer is overcounting the number of ordered $6$-card hands with the first four being non-aces, and at least one of the last two being aces.
Let $A$ be the event that the player is dealt $6$ non-ace cards that are randomly drawn from a standard deck of $52$ cards.
Let $B$ be the event that the first four of the player's $6$ cards are not aces.
$P(B \mid A$) is clearly $1$, since the first four must be non-aces if the six are all non-aces.
Since each card is chosen independently without replacement, we can find the probability of never getting an ace as
$$P(A) = \frac{48 \cdot 47 \cdot 46 \cdot 45 \cdot 44 \cdot 43}{52 \cdot 51 \cdot 50 \cdot 49 \cdot 48 \cdot 47}$$
Likewise
$$P(B) = \frac{48 \cdot 47 \cdot 46 \cdot 45}{52 \cdot 51 \cdot 50 \cdot 49}$$
Now we can use Bayes' rule:
$$P(A\mid B) = \frac{P(B \mid A) \cdot P(A)}{P(B)}$$
But $P(B \mid A)$ is $1$, so
$$P(A\mid B) = \frac{P(A)}{P(B)}$$
$$P(A\mid B) = \frac{44 \cdot 43}{48 \cdot 47}$$