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.

  • $\begingroup$ @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). $\endgroup$ – Arthur Mar 19 at 8:51
  • $\begingroup$ So, they toss the same coin? Or each is tossing her own coin, with A starting first? $\endgroup$ – Jimmy R. Mar 19 at 9:00
  • $\begingroup$ @Arthur what did you just explain to DonThousand $\endgroup$ – Sayangdipto Chakraborty Mar 19 at 9:51
  • $\begingroup$ @JimmyR "a game with a fair coin" seems to imply that there is a single coin, and whoever has their sequence appear first wins. $\endgroup$ – Arthur Mar 19 at 10:51
  • $\begingroup$ @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. $\endgroup$ – Arthur Mar 19 at 11:32

As the first T happens, none of them have won yet, and on the next throw after the first T, one of them will win. If that next throw is a T, then A wins, and if it's a H then B wins. That makes it the same probability for either of them to win.


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