To find the probability that exactly n cards are dealt before the first ace appears, we can use the concept of a geometric distribution. In a geometric distribution, we're interested in the number of trials (in this case, card draws) required for a success to occur (in this case, drawing an ace) for the first time.
The probability of drawing an ace in a single draw from a well-shuffled pack of 52 cards is 4/52 because there are 4 aces out of 52 cards.
So, the probability of drawing a non-ace in a single draw is (52 - 4)/52 = 48/52.
Now, let X be the random variable representing the number of cards drawn before the first ace appears. X follows a geometric distribution with parameter p, where p is the probability of success on a single trial.
P(X = n) = (1 - p)^(n - 1) * p
In this case, p is the probability of drawing an ace on a single trial, which is 4/52, and n is the number of cards drawn before the first ace.
So, the probability that exactly n cards are dealt before the first ace appears is:
P(X = n) = (1 - 4/52)^(n - 1) * (4/52)
Now, to find the probability that exactly k cards are dealt in all before the second ace appears, we need to consider two scenarios:
1. The first ace appears on the nth card, and the second ace appears on the kth card after that. This is represented by P(X = n) * P(X = k).
2. The first ace appears on the kth card, and the second ace appears on the nth card after that. This is represented by P(X = k) * P(X = n).
So, the total probability that exactly k cards are dealt before the second ace appears is:
P(X = n) * P(X = k) + P(X = k) * P(X = n)
You can calculate this probability using the formula for the geometric distribution with p = 4/52 as mentioned earlier for both P(X = n) and P(X = k).