# Problem about a coin tossing game

I have a statistics question about coin toss. Say we have $2$ players $A$ and $B$ and the first one to get heads will win the game. So assuming $A$ will start the game what is the probability of winning the game for $A$?

Edit: I assumed to use this formula to find the answer with adding $50\%$ probability of $A'$ winning chance. Since I just found this website I didn't know I had to add this and I can not tag my question as probability so I tagged it as experimental mathematics.

$$P(X=n)=P(r-1 \text{ heads in } n-1 \text{ flips})*p$$

$$P(X=n)=\binom{n-1}{r-1}p^{r-1}{(1-p)}^{n-r}*p$$

This is the formula I am trying to use to get number of flips to get heads as $r$ is head that I need in $n$ flips. Or should I use something else?

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Wrong tag, missing hypotheses, nothing about what the OP tried... somebody might want to read the howtoask page. – Did Mar 20 '13 at 18:38
I do not understand the identity at the end of (the revised version of) your post. Can you explain? – Did Mar 20 '13 at 18:47

HINT: Let $p$ be the probability that A wins. If A tosses heads on his first toss, he wins. If he tosses tails, he is now in effect the second player, so his probability of winning given that he tossed tails the first time is $1-p$. From this you can derive a simple linear equation in $p$, which you can then solve for $p$.

Added: Of course, you can also do it more straightforwardly but less elegantly. If A wins, he does so on an odd-numbered toss. His probability of winning on toss $2n+1$ is $\left(\frac12\right)^{2n+1}$, since there is exactly one string of $2n+1$ tosses leading to this outcome. With this approach you end up summing a geometric series.

A cute way to look at this is to write everything in binary. The probability that A wins on toss $2n+1$ is

$$0.\underbrace{000\dots000}_{2n}1_{\text{two}}\;,$$

so

$$p=0.1010101010\dots_{\text{two}}\;,$$

and the probability that B wins is

$$1-p=0.0101010101\dots_{\text{two}}\;.$$

Thus, $p$ is just $1-p$ left-shifted one place; a one-place left shift in binary corresponds to what arithmetic operation?

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It will be the same probabilty A's winning against B altho i nned to find 1 of them first :) – serkan Mar 20 '13 at 18:47
@serkan: Let $p$ be A’s probability of winning. As you say, $p$ is also B’s probability of winning. And one of them is certain to win, so $2p=\ldots~?$ – Brian M. Scott Mar 20 '13 at 18:57
Sorry to answer late Proffesor, I forgot to mantion A starts first on the game does this changes it ? – serkan Mar 20 '13 at 19:09
@serkan: Yes, it does; I’ll update my answer. – Brian M. Scott Mar 20 '13 at 19:11
Even i coudnt solve the problem totally yet, I just wanted to thank you for your time and help sir. – serkan Mar 20 '13 at 20:09