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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?

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    $\begingroup$ Hint: geometric series. $\endgroup$ – Don Thousand Jan 19 at 18:03
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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.)

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  • $\begingroup$ Why is $3/4(1-p)$ in the right hand side? $\endgroup$ – Luke Marci Jan 19 at 18:14
  • $\begingroup$ I understand that the probability that my friend wins is 1/4 (the probability that he pulls a heart the first time) plus $3/4$ times the probability that I win, but why $3/4$? $\endgroup$ – Luke Marci Jan 19 at 18:18
  • $\begingroup$ @LukeMarci: I added an explanation, hope it explains the things. $\endgroup$ – W-t-P Jan 19 at 18:30
  • $\begingroup$ so $P(A|\overline{B})$ is the probability that the first player wins given that the first card is not a heart, so why is $1-p$? $1-p$ is not the probability that the second player wins? $\endgroup$ – Luke Marci Jan 19 at 18:51
  • $\begingroup$ @LukeMarci: If the first card is not a heart, then the second player takes his turn becoming the first, so the probability that he wins is $p$, and the probability that the first players (who became now the second) wins is $1-p$. $\endgroup$ – W-t-P Jan 19 at 18:56
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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 $1-\frac47=\frac37$.

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  • $\begingroup$ Why is $3/4(1-p)$ in the right hand side? $\endgroup$ – Luke Marci Jan 19 at 18:14
  • $\begingroup$ There is a probability of $\frac14$ that the friend gets a heats. There is a probability of $\frac34$ that this does not happen. In that situation (you are on move now) the probability that your friend wins is $1-q$. $\endgroup$ – drhab Jan 19 at 18:20
  • $\begingroup$ Ohh it is true, I'm so stupid ahahah Thanks a lot! $\endgroup$ – Luke Marci Jan 19 at 18:25
  • $\begingroup$ You are welcome. $\endgroup$ – drhab Jan 19 at 18:37
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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

$$=\frac{3}{4}\frac{1}{4}+(\frac{3}{4})^3.\frac{1}{4}+(\frac{3}{4})^5.\frac{1}{4}+\cdots$$

$$ = \frac{3}{16}(1+\frac{9}{16}+(\frac{9}{16})^2+\cdots$$

$$ = \frac{3}{16}\dfrac{1}{(1-\frac{9}{16})}$$

$$ = \frac{3}{7}$$

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  • $\begingroup$ thanks for the answer $\endgroup$ – Luke Marci Jan 19 at 18:37
  • $\begingroup$ You are welcome!!. This is the geometric series what Don was referring to!! $\endgroup$ – Satish Ramanathan Jan 19 at 18:38

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