This isn't a riddle. It's a statistics question.

You're probably thinking of the Monty Hall problem where contestants were presented with 3 doors and one door had a nice prize behind it. You pick a door at random then Monty, who we all like, would proceed to tell you one door which didn't have the prize behind it. He then gives you the option of switching to another door or sticking with your original choice.

The point of the Monty Hall problem is if presented with 2 doors, and you have to choose one, the chance of winning is 0.5. But since Monty eliminates one door for you, you should take this into consideration when making the choice about whether to switch door or not, i.e. the probability is conditional upon a given situation. The chance of winning if you switch doors is then 2/3.

So in a way there is a tenuous relationship between this question and the Monty Hall question. Both use some given information leading to a conditional probability. Intuitively, it would be something along the lines of: since you don't know the true probability of getting heads, you take as given some probability value p (of getting heads) and flip the coin an infinite number of times. But you know the first outcome is a heads, so it's just like discarding all the results which have tails as the first outcome. For low values of p the probability of getting heads is lower, which means you'll be discarding more results when p is less than 0.5. Clearly, this implies the expected value of getting heads is above 50%.