Your friend starts first. What is the probability that you win? Imagine you’re playing a game of cards with a friend.  
You decide to keep it a simple game: taking turns, you both take a card from a standard $52$-card deck (first your friend takes a card, then you, then your friend, then you, etc...). 
The first one to pull a card of hearts wins the game. 
If you don’t pull a game of hearts, you put the card back in the deck (after which the deck is shuffled). 
The game continues until someone wins. 
Your friend starts first. What is the probability that you win? 
I think that the probability that I win is $(1-13/52)*(13/52)=10/53$ right?  Knowing that in a $52$-card deck we have $13$ hearts and the card is put back in the deck after each turn?
 A: Denote by $A$ the event "the first player wins", and let $P(A)=p$; thus, $\overline A$ is the event "the second player wins", and $P(\overline A)=1-p$. Also, let $B$ be the event "the first card drawn is a heart". By the law of total probability,
  $$ p = P(A) = P(A|B)P(B) + P(A|\overline B) P(\overline B)= 1\cdot\frac14 + (1-p)\cdot\frac34 = \frac14+\frac34(1-p), $$
implying $p=\frac47$. 
(The explanation for $P(A|\overline B)=1-p$ is that if the first card is not a heart, then the roles of the players switch.)
A: If $q$ is the probability that your friend wins then:$$q=\frac14+\frac34(1-q)$$leading to $q=\frac47$.
So your probability to win is $1-\frac47=\frac37$.
A: The probability of picking a hearts is $\frac{1}{4}$ and that you don't is $\frac{3}{4}$
The probability that you win is pick a heart in your first try after your friend fails in his first try, then you and your opponent did not pick it in the first try and you pick it in your second try after your friend fails in the second try and keeps going and thus the probability that you win is
$$=\frac{3}{4}\frac{1}{4}+(\frac{3}{4})^3.\frac{1}{4}+(\frac{3}{4})^5.\frac{1}{4}+\cdots$$
$$ = \frac{3}{16}(1+\frac{9}{16}+(\frac{9}{16})^2+\cdots$$
$$ = \frac{3}{16}\dfrac{1}{(1-\frac{9}{16})}$$
$$ = \frac{3}{7}$$
