Alice and Bob play the following game: they roll a die one after the other until a 6 appears. The winner is The first one gets a 6 on the roll. Calculate the probability that the one who starts the game (i.e. the one who rolls first) will also win the game.
B. If it is known that after 20 tosses no one won ( both Alice and Bob tossed 10 times and no one got still 6), calculate the probability that the one who started the game (the one who rolled first) will also win in the game
We begin with the event that bob starts first and wins. For Bob to win in round i, we get $\frac{1}{6} *(\frac{5}{6})^{i-1}*(\frac{5}{6})^{i-1}$ where Bob and Alice fail together the first i-1 round and bob wins on i.
Then we calculate $(\frac{1}{6*\frac{25}{36}})\sum_{i=1}^{\infty}(\frac{25}{36})^{i}$ such that the win might happen on first round or second or third or... we get $\frac 6 {11}$
Now, we want Alice to start first. Which is the same calculation $\frac 6 {11}$
for B, I went into direction of Pr[Who starts first wins | A+B>20 ] where A,B are random variables indicating number of round for alice and bob and they distribute geometry A,B~G(1/6). I thought of memory loss property for geometric distribution but have no idea how to continue