Alice and Bob play a coin tossing game. A fair coin (that is, a coin with equal probability of landing heads and tails) is tossed repeatedly until one of the following happens.
$1.$ The coin lands "tails-tails" (that is, a tails is immediately followed by a tails) for the first time. In this case Alice wins.
$2.$ The coin lands "tails-heads" (that is, a tails is immediately followed by a heads) for the first time. In this case Bob wins.
Who has more probability of winning the game?
My attempt $:$
Let $X$ be the random variable which counts the number of tosses required to obtain "tails-tails" for the first time and $Y$ be the random variable which counts the number of tosses required to obtain "tails-heads" for the first time. It is quite clear that if $\Bbb E(X) < \Bbb E(Y)$ then Alice has more probability of winning the game than Bob$;$ otherwise Bob has more probability of winning the game than Alice. Let $X_1$ be the event which denotes "the first toss yields heads", $X_2$ be the event which denotes "tails in the first toss followed by heads in the second toss", $X_3$ be the event which denotes "tails in the first toss followed by tails in the second toss". Then $X_1,X_2$ and $X_3$ are mutually exclusive and exhaustive events. Let $\Bbb E(X) = r.$ So we have $$\begin{align} r & = \Bbb E(X \mid X_1) \cdot \Bbb P(X_1) + \Bbb E(X \mid X_2) \cdot \Bbb P(X_2) + \Bbb E(X \mid X_3) \cdot \Bbb P(X_3). \\ & = \frac {1} {2} \cdot (r+1) + \frac {1} {4} \cdot (r+2)+ 2 \cdot \frac {1} {4}. \\ & = \frac {3r} {4} + \frac {3} {2}. \end{align}$$ $\implies \frac {r} {4} = \frac {3} {2}.$ So $\Bbb E(X) = r = 6.$
But I find difficulty to find $\Bbb E(Y).$ Would anybody please help me finding this?
Thank you very much for your valuable time.