-1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ @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
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.