Suppose that there are ten cards numbered from 1 to 10. The cards are shuffled and the game is played between two players: player 1 and player 2. Player 1 starts the game. Player 1 wins if he selects any card numbered from 1 to 4 both inclusive and the game ends. If player 1 fails to draw the card with desired number, then player 2 draws a card and he wins if any card numbered from 5 to 10 is drawn. If player 2 fails to draw the desired card the game goes back to player 1 and he draws the card again. The players keep on playing the game until one of the player wins. Find the probability that player 1 wins the game.
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$\begingroup$ are cards replaced once they have been choosen $\endgroup$– Manish Kumar SinghCommented Oct 6, 2015 at 16:48
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$\begingroup$ Assuming no replacement, brute force works pretty well, here. If Player 1 wins, it will be on one of his first four draws. There aren't very many terms to calculate. $\endgroup$– user3294068Commented Oct 6, 2015 at 16:56
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$\begingroup$ @ManishKumarSingh with no replacement $\endgroup$– chauka khanCommented Oct 6, 2015 at 17:07
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2 Answers
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Suppose cards are replaced after each draw,$$\begin{align}P(\text{player 1 wins})&=\frac4{10}+(\frac6{10}\frac4{10})\frac4{10}+(\frac6{10}\frac4{10})(\frac6{10}\frac4{10})\frac4{10}+...\\&=\frac4{10}\left(1+0.24+0.24^2+...\right)\\&=\frac4{10}\frac1{1-0.24}\\&=10/19\end{align}$$
For no replacement,$$P(\text{player 1 wins})=\frac4{10}+\frac6{10}\frac49\frac38+\frac6{10}\frac49\frac58\frac37\frac26+\frac6{10}\frac49\frac58\frac37\frac46\frac25\frac14=37/70$$
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I don't have enough rep to comment, so please treat this as a comment.
I think about using the negative binomial distribution for this kind of problem, but I'm not sure if it's the right way to think about this.
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$\begingroup$ Welcome to Math.SE! This does not provide an answer to the question, and comments are not entertained among answers. To critique or request clarification from an author, leave a comment below their post - you can always comment on your own posts, and once you have sufficient reputation, you can comment on any post. $\endgroup$ Commented Oct 15, 2015 at 3:07