a. See the first paragraph of part d.
b.
The first equality comes from Bayes’ theorem. The
terms
come from the ways to have the first j-1 cards be non-aces. 4*3 refers to
combinations of 2 adjacent aces in the numerator. The ending factorials in the
numerator and denominator come from ordering the rest of the cards.
c. We have
With the LOTP, part b, and the power of R, we can compute
d. Argument by symmetry: Consider the events "the first card after the first
ace is an ace" and "the last card after the first ace is an ace". The second event is
equivalent to the last card in the deck being an ace. In addition, the two events
must have the same probability, as every card drawn after the first ace is
equally likely to be an ace. Therefore, the probability of the first event is
1/13.
For a proof using conditional probability:
Consider the ace of hearts and the ace of spades. The probability that the ace of hearts is the first ace to appear followed immediately by the ace of spades (call this event A) is the probability that they appear adjacent to each other in that order (call this event B) and that those two aces appear before the other two aces (call this event C).
We have
Now, to compute
consider that if the aces of hearts and of spades appear adjacent to each
other in that order, we can consider them as a card "glued together".
There are 3!=6 possible orderings of the glued together card and
the other 2 aces - in 2 of them, the glued together card is first. So
.
We also have ,
as there are 51 sets of two adjacent spaces the two aces could be, and the rest of the cards can
be ordered in
ways.
Then, we now find .
Now, instead of the aces of hearts and spades specifically, consider there are 12 possible pairs of aces that can be adjacent to each other. Then the total probability that the card after the first ace is another ace is