Player A and B alternate when flipping a coin. If the number of heads is K more than the number of tails, A wins, if the number of tails is K more than heads, B wins. What is the probability that the game is not over after 100 coin tosses?
I started by considering simpler cases like winning if there are 2 more heads/tails than tails/heads in 100 tosses and it is clear the the game finishes at most on the 3rd toss with probability 1, hence it makes sense that with 50 more heads/tails than tails/heads in 100 tosses the game finishes in at most 99 tosses? For more than that I would need to compute the permutations of the different scenarios for winnings and divide by the total number of posibilities for each K, not feasible. I was thinking maybe I could solve it via recursion or some sort of conditioning
Need help with ideas, hints and/or intuition please!
Also how would that change if the players toss the coins (no alternation) separately and the one to have k more heads/tails than the other wins?