Guess which card will be the Ace of spades. Consider the following game played with an ordinary deck of 52 playing cards. The cards are shuffled and then turned over one at a time. At any time the player can guess that the next card to be turned over will be the Ace of spades; if it is, then the player wins. In addition, the player is said to win if the ace of spades has not yet appeared when only 1 card remains and no guess has yet been made. What is a good strategy; what is a bad strategy?
Solution
Every strategy has probability 1/52 of winning. To show this, we will use induction to prove the stronger result that for an n card deck, one of whose cards is the ace of spades, the probability of winning is 1/n, no matter what strategy is employed. Since this is clearly true for n = 1, assume it to be true for an n-1 card deck, and now consider an n card deck. Fix any strategy and let p denote the probability that this strategy guesses that the first card is ace of spades. Given that it does then the players probability of winning is 1/n. 
** On the other hand if the strategy does not guess that first card is ace of spades, then the probability that the player wins is the probability that the first card is not the ace of spades, namely (n-1)/n, multiplied by the conditional probability of winning given that the first card is not the ace of spades. But this latter conditional probability is equal to the probability of winning when using an n-1 card deck containing a single ace of spades; it is thus, by the induction hypothesis, 1/(n-1). Hence, given that the strategy does not guess the first card, the probability of winning is
$ {{n-1}\over {n}} {{1} \over {n-1}} = {1 \over n }$
Thus, letting G be the event that the first card is guessed, we obtain that
$P(win) = P(win|G)P(G) + P(win|G^c)(1 - P(G))$
$P(win) = {1 \over n}p + {1 \over n}(1-p)$
$={1 \over n}$
My problem
I fail to understand what they are doing in the second paragraph of the solution (**). 
They are essentially saying that $ P(win|G^c) = {{n-1}\over {n}} {{1} \over {n-1}} ... (1)$
But $ P(win|G^c) = {P(win \bigcap G^c) \over P(G^c)} ... (2)$
I fail to see how (2) -> (1)
Would appreciate it if someone can explain by expanding the equation correctly.
 A: Here is how you can write out the logic of the second paragraph using the conditional probability identities. Let us call $A$ the event that the first card is the ace of spades, and accordingly, $A^c$ that it is not (I'll also write $W$ instead of "win").
$$ P(W|G^c) = \frac{P(W \cap G^c)}{P(G^c)} = \frac{P(W \cap G^c \cap A) + P(W \cap G^c \cap A^c)}{P(G^c)}, $$
where I have used that $A$ either happens or does not. Note that the first term in the numerator is zero, because you cannot win if you didn't guess, but the ace came up ($G^c \cap A = \rm loss$).
Also note that the other term is
$$ P(W \cap G^c \cap A^c) = P(W \cap (G^c \cap A^c)) = P(G^c \cap A^c) \cdot P(W|(G^c \cap A^c)) .$$
[The first step just groups events into a combined event, the second step uses the conditional probability formula in the form $P(A\cap B) = P(B)P(A|B)$.]
So we are left with
$$ P(W|G^c) = \frac{P(G^c \cap A^c) \cdot P(W|(G^c \cap A^c))}{P(G^c)} =
P(A^c) \cdot P(W|(G^c \cap A^c)), $$
because $A^c$ and $G^c$ are independent, so $P(G^c \cap A^c) = P(G^c)P(A^c)$ and the denominator gets cancelled.
Finally, $P(A^c) = {n-1\over n}$, while $P(W|(G^c \cap A^c)) = {1\over n-1}$ by the inductive assumption (given that you didn't guess, and the ace didn't come up, you just play out the rest of the deck, which is the same as an $(n-1)$-card game with the same rules). Thus you get $(1)$.
A: No real answer to your question, but an alternative way that uses induction.

Assumption: if the game is played with $k<n$ cards then every strategy gives probability $\frac1{k}$ to win.
Now check the strategies when $n$ cards are used.
One of them is guessing that the first card is the ace of spades and evidently it gives probability $\frac1n$ to win.
Now take another strategy and let $A$ denote the event that the first card is ace of space. 
Then $$P(\text{win})=P(\text{win}\mid A)P(A)+P(\text{win}\mid A^{\complement})P(A^{\complement})=0\cdot\frac1{n}+\frac1{n-1}\frac{n-1}{n}=\frac1n$$
Here $P(\text{win}\mid A^{\complement})=\frac1{n-1}$ on base of the assumption.
A: $G$ the event the first card is guessed
$A$ the event the first card is the ace of spades
$W$ the player wins
Notice the paragraph in question begins with "On the other hand, if the strategy does not guess that the first card is the ace of spades".  This tells us the context of the paragraph is the event $G^C$.  The result of the paragraph, $1/n$, is plugged into $P\{ \text{win} | G^C \}$.
Instead of stealth conditioning on $G^C$,
"the probability that the player wins" $P( W | G^C )$
"is the probability that the first card is not the ace of spades" $P( A^C | G^C )$
"multiplied by the conditional probability of winning given that the first card is not the ace of spades"
$\times \, P ( W | A^C \cap G^C )$,
let's work through the paragraph in question with conditioning on $G^C$ exposed.  We will show
\begin{equation*}
P( W | G^C ) = P( A^C | G^C ) \times P ( W | A^C \cap G^C )
\end{equation*}

By the definition of conditional probability, $P( W | G^C ) = \dfrac{ P( W \cap G^C ) }{ P( G^C ) }$.
Partition the set,  $W \cap G^C = ( W \cap G^C \cap A ) \cup ( W \cap G^C \cap A^C )$.
Since
$W \cap G^C \cap A = \emptyset$, $W \cap G^C = W \cap G^C \cap A^C$.
Therefore, $\dfrac{ P( W \cap G^C ) }{ P( G^C ) } = \dfrac{ P( W \cap G^C \cap A^C ) }{ P( G^C ) }$.
By the definition of conditional probability,
$\dfrac{ P( W \cap G^C \cap A^C ) }{ P( G^C ) } = P( W \cap A^C | G^C )$.
Connecting the string of equations,
\begin{equation}
P( W | G^C ) = P( W \cap A^C | G^C )
\end{equation}

In general for any condition $C$, let $P_{ C } ( \cdot ) = P( \cdot | C )$.
The symbols $P( W | C )$ and $P_{ C } ( W )$ are alternate expressions for the same quantity. In particular, $P_{G^C} ( W ) = P( W | G^C )$.
By the definition of conditional probability, $P_{ G^C }( W | A^C ) = \dfrac{ P_{ G^C } ( W \cap A^C ) }{ P_{ G^C } ( A^C ) }$.
$P_{ G^C } ( W \cap A^C ) = P_{ G^C } ( A^C ) \cdot P_{ G^C }( W | A^C )$
See (***) below for a discussion of the important identity that converts a conditional probability from the alternate notation to the standard notation.
$P_{ G^C } ( W | A^C ) = P( W | A^C \cap G^C )$.
Thus, $P_{ G^C } ( W \cap A^C ) = P_{ G^C } ( A^C ) \cdot P( W | A^C \cap G^C )$.
Return to the standard notation for conditional probability.
\begin{equation}
P( W \cap A^C | G^C ) = P( A^C | G^C ) \cdot P( W | A^C \cap G^C )
\end{equation}
We assume $A$ and $G$ are independent events, they do not affect each others probabilities.  By independence, $P( A^C | G^C ) = P( A^C )$.  $P( A^C ) = 1 - P( A ) = 1 - 1/n = (n-1)/n$.  Thus, $P( A^C | G^C ) = (n-1)/n$.
$P( W | A^C \cap G^C )$ "is equal to the probability of winning when using an $n - 1$ card deck containing a single ace of spades; it is thus, by the induction hypothesis, $1/ ( n - 1 )."$
$P( W | A^C \cap G^C ) = 1/(n - 1 )$.
\begin{equation}
P( A^C | G^C ) \cdot P( W | A^C \cap G^C ) = \dfrac{ n - 1 }{ n } \cdot \dfrac{ 1 }{ n - 1 } = 1/n
\end{equation}
Thus,
$P( W | G^C ) = P( W \cap A^C | G^C ) = P( A^C | G^C ) \cdot P( W | A^C \cap G^C ) = 1/n$.

(***) Repeated/Sequential conditioning. An important result.
$P( A | C ) = \dfrac{ P( A \cap C ) }{ P( C ) }$. Let $P_C( A ) = P( A | C )$.  The symbols $P( A | C )$ and $P_C( A )$ are alternate expressions for the same quantity.
\begin{equation*}
P_C( A | D ) = P ( A | C \cap D )
\end{equation*}
Proof. $P_C( A | D ) = \dfrac{ P_C( A \cap D ) }{ P_C( D ) } =
 \dfrac{ P( A \cap D \cap C ) }{ P( C ) } \div \dfrac{ P( D \cap C ) }{ P( C ) } =
 \dfrac{ P( A \cap D \cap C ) }{ P( D \cap C ) } =
P ( A | C \cap D )$
(2 nice explanations: Paul E. Pfeifer, Probability for Applications. 3.3 Repeated Conditioning, and Marcel F. Neuts, Probability. Iterated conditional probabilities. Theorem 3.1.2)

(The example as posted is Example 3c in A First Course in Probability (7th & 8th editions) by Sheldon Ross.  The reader has not yet learned about independence, repeated/sequential conditioning $P( A | C \cap D )$, or conditional probability $P(\cdot | C )$ as a probability measure.)
