Tossing Coin Expected Value Against Intuition I have the following two coin toss games:
Game 1:
A and B tosses a coin. At first the coin is unbiased. Through the game, if heads comes A wins and game stops. If tail comes, the coin is swapped with another coin which is biased such that the probability of heads showing up becomes 1/3 and game continues. For kth turn the probability of heads showing up is 1/(k+1), if tails comes the coin is swapped with another biased coin for which probability of heads showing up becomes 1/(k+2). What is the expected value of number of turns in which the game ends ie. heads shows up.
Game 2:
The game is same, but the biases of coins are different. This time, for kth turn, if probability of heads is 1/(2^k), the next turn it is 1/(2^(k+1)). So it goes 1/2, 1/4, 1/8, 1/16, … .What is the expected value of number of turns in which the game ends.
After calculating I found the expected value in game 1 as divergent, and in game 2 as smaller than 2 which feels very counter-intuitive since A seems to be have a greater chance of winning in game 1. 
So my question is what does a divergent expected value actually mean and do you think I made a mistake with the calculations or does the problem really act against intuition.
 A: Some quick calculations, to showcase (presumably) what the OP has done.
For the first game, the probability that the game ends on the $k$th turn is $\frac{(k-1)!}{(k+1)!}=\frac{1}{k(k+1)}$. Thus when $X$ is the number of turns, $$E(X)=\sum_{i=1}^\infty\frac{1}{i+1}=\infty$$
which is known to diverge.
For game two, the probability that the game ends on the $k$th turn is strictly less than $\frac{1}{2^k}$ (well, equal for $k=1$, but strictly less otherwise). Thus when $X$ is the number of turns, $$E(X)<\sum_{i=1}^\infty \frac{i}{2^k}=2$$
as you said.
Now why is this?
Well, think about what we assumed. We assumed that the game would end. In 1, the probability that the game lasts at least $k$ moves is $\frac{1}{k}$ so the game has to end. But for game 2, the probability that the game lasts forever is...
$$\prod_{i=1}^\infty\frac{2^i-1}{2^i}\approx 0.288$$
(Oh and an OEIS for this product too, because why not?)
Thus, the second game actually has a chance that you'll never finish! What does this tell you about the calculated expected probability?
