Expected value of a card game, with a fixed probability of playing it again. A person has to select one of three cards. One card would get him a reward of 1 dollar. Another one would not give him any return. The third one would give him a return of 2 dollars and also compels him to select one of the 3 cards once again- all similar to the first set of 3 cards. The person has no way of identifying the cards beforehand and is equally likely to take any of the 3 cards at any stage. Find the expected value.

If we had only the first two cards, the expected value would be $$E[X] = \frac{1}{2}(1) + \frac{1}{2}(0) = 0.5$$ With the third card, it becomes: $$E[X] = \frac{1}{3}(1) + \frac{1}{3}(0) + \frac{1}{3}\left(2 + E[X]\right).$$ Writing it like this allows you to avoid the geometric series that is lurking behind that third card -- provided that you know $E[X]$ exists. In this case it does, but if $E[X] = \infty$, this is invalid, and you need to expand this into a geometric series.

**Edit:** I fixed this, I was previously treating it as though the first card were worth 1 dollar, the second card 2 dollars, and the third worth a chance to pick another card, and to play again as well.