Understanding example 3j in Ross Probability For a randomly shuffled deck, Let $E$ be the event that the card following the firs ace is some specified card, say card $x$. To compute $P(E)$, we ignore card $x$ and condition on the relative ordering of the other 51 cards in the deck. Letting $\bf O$ be the ordering gives
$$ P( E ) = \sum_{\bf O} P(E | {\bf O} ) P(\bf O) $$
Now, given $\bf O$ there are 52 possible orderings of the cards, corresponding to having card $x$ being the ith card in the deck ($i=1,2,...,52$). But because all $52!$ possible orderings were initially equally likely, it follows that conditional on $\bf O$, each of the remaining $52$ remaining possible orderings is equally likely. Because card $x$ will follow the first ace for only one of these orderings, $P(E | {\bf O} ) = 1/52 $ and so $P(E) = 1/52$
my question:
I am having difficulties understanding what $\bf O$ really means. Also, how do we get $P(E) = 1/52$ ? I find this example extremely hard to grasp.
 A: Solution:
$$ P( E ) = \sum_{\bf O} P(E | {\bf O} ) P(\bf O) $$ 
$$= \sum_{r=1}^{51!} P(E | {O_r} ) P(O_r) \tag{1}$$ 
$$= \sum_{r=1}^{51!} \frac{1}{52} P(O_r) \tag{2}$$ 
$$= \sum_{r=1}^{51!} \frac{1}{52} \frac{1}{51!} \tag{3}$$ 
$$= 51! \frac{1}{52} \frac{1}{51!} $$ 
$$= \frac{1}{52} $$ 

Discussion:
$(1)$ If you take out a card, let's say jack of diamonds ($x$ in the problem) from a standard deck of cards, there are 51 cards left and then there are 51! orderings of 51 cards. Name ordering with $O_1, ..., O_{51!}$. These orderings actually form a partition of $\Omega$, i.e. if you take out jack of diamonds, then the remaining cards follow exactly 1 of those orders $O_r$.
$(2)$ Now let's pick an order $O_r$ and then put back the jack of diamonds there somewhere there. What's the probability that the first ace is after the jack of diamonds? There's exactly 1 one way to put the jack of diamonds in $O_r$ s.t. it is after the first ace. Assuming all the 52 ways have equal probability. The probability is $1/52$.
$(3)$ Assuming the $O_r$'s have equal probability.
A: $\bf O$ is one of the $51!$ possible permutations of the other $51$ cards.
Given a particular $\bf O$, there are 52 places to insert $x$, each of the 52 places being equally likely.  So $P(E|\mathbf{O})= 1/52$.  Now normally $P(E|\mathbf{O})$ should be a function of the particular $\bf O$, but you have just proven that it is in fact the same value ($1/52$) for any $\bf O$, so can you see how to conclude $P(E) = 1/52$?
