Probability that A wins The Game? Two Players A and B participate in a game of drawing a card from an ordinary deck alternately until one of them gets a club and wins the game. If A starts the game, the chance that A wins the game is?
a)1/7
b)2/7
c)3/7
d)4/7 
 A: A little cheating would be to realize that the person who starts clearly has a higher chance of winning and thus the probability must be $4/7$ given that it is indeed one of the four options. To specifically calculate the probability we need more information. 
A: Denote with $W$ the event that player A wins the game and with $p:=P(W)$ it's probability. Event $W$ can occur in following ways, conditioning on the result of the first draws of the two players:


*

*Player A wins in the first draw. This occurs with probability $$\frac{13}{52}$$

*Player A does not win in the first draw, but player B does not win neither. Then the game starts over again. So, due to the multiplication rule the probability in this case (assuming that they draw with replacement) is equal to: $$\left(\frac{39}{52}\right)^2\cdot p$$


Adding up the two probabilities we have that $$p=\frac{13}{52}+\left(\frac{39}{52}\right)^2p=0.25+0.5625p$$ which gives $$p=\frac{4}{7}$$

Assuming that they draw without replacement the problem cannot be solved recursively, since now if the turn of Player A comes again, then the probability that he wins is equal to $\frac{13}{50}$. Instead in this case we have that $$p=\frac{13}{52}+\frac{39}{52}\cdot\frac{38}{51}\cdot\frac{13}{50}+\frac{39}{52}\cdot\frac{38}{51}\cdot\frac{37}{50}\cdot\frac{36}{49}\cdot\frac{13}{48}+\ldots$$
