Maybe I made it too complicated, so basically here is a simpler version

Just say two people toss a coin, with the geometric distributions of getting a head, respectively

Note: p1 != p2

P(player1)=p1(1-p1)^n-1

P(player2)=p2(1-p2)^n-1

What's probablity player 2 gets a head before player 1 does.

So say, player 2 gets a head at the 10th trial, then player 1 has to be 11.....n

Can anyone help me with this?

Thanks