# 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

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Are the cards drawn with or without replacement? – Sabyasachi Apr 1 '14 at 15:55
How many cards do you have in your deck? $32$? $36$? $52$? – Jérémy Blanc Apr 1 '14 at 15:56

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.

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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:

1. Player A wins in the first draw. This occurs with probability $$\frac{13}{52}$$
2. 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 $$p=\frac{13}{52}+\frac{39}{52}\cdot\frac{38}{51}\cdot\frac{13}{50}+\frac{39}{52}\cdot\frac{38}{51}\cdot\frac{37}{50}\cdot\frac{36}{49}\cdot\frac{13}{48}+\ldots$$

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Don't you mean $p = 13/52 + 39/52\cdot38/51\cdot13/50 + 39/52\cdot38/51\cdot37/50\cdot36/49\cdot13/48 + \dotsb$? – derpy Apr 1 '14 at 16:15
By the way, in case anyone were curious, here is the result in this case; about 57%, and interestingly enough, differing from 4/7 by only about one part per thousand. – derpy Apr 1 '14 at 16:29
@derpy Yes, sorry, I saw my mistake. This is really interesting!!!! – Jimmy R. Apr 1 '14 at 16:33