Probability that a player wins a card guessing game $ \textbf{Question:} $ A card game consists of $ n $ cards $ (n \ge 1), $ one of which is a special card. The cards are shuffled randomly and then turned over one at a time. At any time, a player must guess whether the current card is the special card before it is revealed. The player wins when he correctly guesses the special card. What is the probability that the player wins the game? 
I approach this problem by letting $ E $ be the event that the current card is a special card and $ F $ be the event that the player will guess the current card is the special card, so finding the probability that the player wins the game means finding the probability that both $ E $ and $ F $ occur. Now I have $ \displaystyle P(EF) = P(E)P(F|E) $ with $ \displaystyle P(E) = \frac{1}{n}, $ but I don't know how to compute $ P(F|E). $ It seems reasonable (to me) that $ P(F|E) = 1 $ since once the player knows apriori that the current card is the special card, he will just make the guess. 
Suppose that the problem provides an extra information that at any time, the probability that the player guesses the current card is the special card is $ 30\% $ (meaning $ P(F) = 0.3), $ will that change the value of $ P(F|E) $ from $ 1 $ to $ 0.3? $    
 A: The setup as I understand: Each turn the player may guess or hold, then one card is turned over. 


*

*Guess and turn special: win

*Guess and turn ordinary: loose

*Hold and turn special: loose

*Hold and turn ordinary: continue the game with one less card.


Assuming that the probability that the player guesses on any given draw is inversely related to the number of cards unturned; you can recursively define the probability of winning when $k$ cards are in the deck as:
$$P(k) = \frac{1+(k-1)P(k-1)}{k^2}; P(1)=1$$
And so find $P(n)=\tfrac 1n$, by observing a pattern in evaluating the first few of the sequence, $P(2), P(3), ...$ and then using an inductive proof to confirm.

Alternatively:
Let $X$ be the count draws until the special card is drawn, and $Y$ be the count of draws until the player makes a guess; with the understanding that $Y=X+1$ is the event of a loss through not guessing before the special card is drawn. 
Now clearly: $\mathsf P(X=k) = \tfrac 1 n \quad[1\leq k\leq n]$
For any given $X$, the probability that the player guesses then, is $$\mathsf P(Y=k\mid X=k) = \frac{1}{n}~[k\in\{1,..,n\}]$$
Because it is the probability that the player has not guessed in the first $k-1$ draws times the probability that the players guesses on right draw.
Thus the event of a win is:
$$\mathsf P(X = Y) = \sum_{k=1}^n \mathsf P(Y=k\mid X=k)\mathsf P(X=k) = \frac 1 n$$

Remark, this is assuming that the player chooses to guess or not based exclusively on how many cards are left.   Naturally a different result will occur if a different strategy is employed.
A: I assume that the player keeps  the same decision pattern throughout. This can be generalised easily to more realistic patterns.
Assume  at every  and until the  $(n-1)$st trial  : $S$=decides the card is special, $P(S)=p$ and $NS$=decides the card is not  special  $P(NS)=q=1-p$
Here we have a sequence of pairs of events.  In each pair, the first component is the decision of the player and the second is the card that appears. These should be independent.
The player wins(game stops) at the $k$-th trial, $k=1,2,\ldots,n-1$,  when these events happen concurrently: ($NS$ and the card is not special) in $k-1$ trials and ($S$ and the card is special) in the $k$-th trial.
For example when $k=3$ the probability  of these events is:  $$[q(n-1/n)][q(n-2/n-1)][p(1/n-2)]=q^2p\cdot 1/n$$ 
In general, 
$$P(\text{wins at k-th trial})=q^{k-1}p\cdot 1/n $$ 
When $k=n$  $$P(\text{wins at n-th trial})=\{1-\sum_{k=1}^{n-1}q^{k-1}p\}\cdot 1/n=q^{n-1}/n$$ 
A: The answer is 1/n.  There is no strategy to vary the odds of winning.
with a 2 card game, his probability of winning is 1/2 with any strategy. If he goes for the first card, then he has a 1/2 chance of winning.
if he waits for the second card, his win chance is the 1/2 chance that the first card does not win, so however he decides how to play, his win chance is 1/2
so
P(2) = 1/2
what is P(3)?  The player can take an immediate 1/3 shot at winning on the first card, or risk a 1/3 chance of losing by waiting, and go into a 2 card game, with prob of winning P(2) = 1/2, so that will give him a 1/3 chance of winning,
whatever his strategy, he will have a 1/3 chance of winning
P(4) = 1/4 on first go, or take a 3/4 chance of going to P(3) - so he has 1/4 chance of winning.
Inductively, the addition of a card to the game gives the player a 1/n chance of winning on the first go, or taking an (n -1) / n chance of going into a game with a 1/(n-1) chance of winning, meaning that his chance is also 1/n
I've also shown inductively that with n cards remaining, the choice of whether or not to take the guess gives a 1/n chance of winning in both cases.  There is no justification that the first card is better than a 1/n shot, and if he reaches the n-1 game, then no strategy is available to increase his odds.
It is true that a player taking no guesses in an n game seems to be getting a higher chance of winning as the game goes on, but only if he fails to lose.  By the time he has 1 card left, he had avoided an (n-1) / n chance of defeat
