Rick wins if it is HT and Tick wins if it is TT. Who wins? A fair coin is being tossed. Whenever a tail follows a head, Rick wins the game and whenever a tail follows a tail, Tick wins. Who has more probability to win?
Let's say the coin has been tossed more than twice. For Tick to win, the last three tosses should be HTT, but Rick will have won it already. So if the first two flips are not TT, Tick has no chance of winning the game. 
I do not know how to proceed from here. I initially planned to brute force since we can obviously compute the probability for $n$ tosses (using a simple recursion). But then this observation above kind of made this computation a bit meaningless. 
 A: Suppose the first flip is heads. Now, no matter how many more flips it takes, eventually it will flip tails and Rick will win. So, if the first flip is Heads, Rick wins. That is $\dfrac{1}{2}$ of the time.
Suppose the first flip is tails, but the second flip is heads. No matter how many more flips it takes, eventually, it will flip tails a second time, and as soon as it does, you will have $HT$ ending the sequence with Rick winning. So, that is an additional $\dfrac{1}{2}\cdot \dfrac{1}{2}$ chance that Rick wins.
Suppose the first two flips are tails. Then Tick wins.
So the outcomes:
$$\begin{array}{c|c|c}\text{Flip pattern} & \text{Winner} & \text{Probability} \\ \hline H & \text{Rick} & \dfrac{1}{2} \\ TH & \text{Rick} & \dfrac{1}{4} \\ TT & \text{Tick} & \dfrac{1}{4} \\ \hline \text{Total} & \text{Rick} & \dfrac{3}{4} \\ \text{Total} & \text{Tick} & \dfrac{1}{4} \end{array}$$
A: As you surmised yourself, the only way for Tick to win is if the very first two throws are tails ... if any of those first two throws is heads, then Rick wins. So, Tick wins with a probability of $\frac{1}{4}$, and therefore Rick with a probability of $\frac{3}{4}$
