If A and B play a game with a fair coin such that A wins by getting consecutive TT for the first time, while B wins by getting a consecutive TH. Who has higher probability of winning the game?

I tried the following method but not sure:

For A to win,

TT, HTT, HHTT ... that is, 2-k for each game length k

For B to win,

TH, HTH, HHTH ... that is, 2-k for each game length k

So ultimately I am getting same probability of each to win.

• @DonThousand Not true. for instance, HT has a probability of $75\%$ to appear before TT (basically, as long as the first two throws aren't TT, HT will come first). – Arthur Mar 19 at 8:51
• So, they toss the same coin? Or each is tossing her own coin, with A starting first? – Jimmy R. Mar 19 at 9:00
• @Arthur what did you just explain to DonThousand – Sayangdipto Chakraborty Mar 19 at 9:51
• @JimmyR "a game with a fair coin" seems to imply that there is a single coin, and whoever has their sequence appear first wins. – Arthur Mar 19 at 10:51
• @SayangdiptoChakraborty He thought that just because they are two equally long sequences, they are equally likely to appear first. But that argument is flawed, as is evidenced by my example. – Arthur Mar 19 at 11:32