Who's the first to toss a head? 
Tom, Dick, and Harry toss a fair coin in turn: Tom tosses first, then Dick,
  then Harry. The first to toss a head wins. What is the probability that the winner
  is (a) Tom, (b) Dick, and (c) Harry?

My answers are $(a) \frac12, (b) \frac14, (c) \frac18$?
Are my answers correct?
 A: The probability that player number $i \in \{1,2,3\}$ wins is given by $$\sum_{k=0}^\infty \frac{1}{2^{i+3k}}=\frac{2^{3-i}}{7}$$
The reason for the infinite sum (instead of your reasonable attempt) is that there is a probability that the two players other than player $i$ will get only tails and give player $i$ yet another turn, and this could potentially go on forever (although the probability for that scenario goes to zero, which is why the sum converges). 
Feel free to ask further questions in the comments below if this argument is not yet clear to you.
A: Actually, you are almost correct with $\color{blue}{\frac12}$, $\color{blue}{\frac14}$, and $\color{blue}{\frac18}$. Namely, if they do one round, no one wins, and they give up.
The probability of that happening is the remaining probability $\color{red}{\frac18}$.
But now, if they do another round then it's again $\color{blue}{\frac12}$, $\color{blue}{\frac14}$, and $\color{blue}{\frac18}$. If they give up after the second round, the accumulated winning probabilities are 
$\color{blue}{\frac12} (1 + \color{red}{\frac18})$, 
$\color{blue}{\frac14} (1 + \color{red}{\frac18})$, 
and
$\color{blue}{\frac18} (1 + \color{red}{\frac18})$.
I think you will know how to proceed from here.
A: Another way of handling this, that doesn't involve the infinite sums (but of course comes to the same answer): Let the overall probability that Tom wins be $T$.  Then either Tom can win in the first round (probability $\frac12$) or he can get another chance after everyone has taken their turn (probability $\frac18$) — but since the game is 'memoryless', after everyone has taken their turn the probability that he wins is just $T$ again.  This gives an equation of the form $T=\frac12+\frac18T$.  Likewise, Dick's probability $D$ satisfies $D=\frac14+\frac18D$ and Harry's satisfies $H=\frac18+\frac18H$.  You should be able to solve each of these equations, and verify that $T+D+H=1$.
