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.