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Player A throws a fair coin n+1 times. Player B throws the same coin n times. Player A wins if she throws more heads than Player B. What is the probability of player A winning the game?

My approach has been that since the answer is independent of n, so I chose n=1. So, A can win 3 out 8 times (assuming that Head-Tail is same as Tail-Head) hence the probability is 3/8. Is my assumption correct? What about the approach? How can the solution be generalized?


marked as duplicate by lulu, Community Oct 2 '18 at 0:33

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  • $\begingroup$ "Assuming that Head-Tail is same as Tail-Head" Why do you say that? If you wish to find probabilities by counting, then you must recognize what you are counting. We took a ratio over $8$. Those $8$ things that we took a ratio of are the ordered ways in which three coin flips can occur. The order matters in such a scenario so we may not just say "head-tail is same as tail-head." There are four ways out of the eight equally likely ways in which $A$ wins, not three. $\endgroup$ – JMoravitz Oct 2 '18 at 0:34
  • $\begingroup$ @JMoravitz Oh yes. Thank you! $\endgroup$ – prashant Oct 2 '18 at 0:39