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.
Best Answer
Perhaps I am missing some subtlety, in which case someone here will tell me. That said:
The coin is tossed repeatedly. Alice and Bob watch bored until the first tail appears. The next toss will settle the game. Each has an equal chance to win.
The game is more interesting if Bob wins on the first "heads-tails". In that game suppose a head appears before anyone has won. Then Bob will win as soon as the first tail appears, which will happen eventually. Since the first flip is heads with probability $1/2$ Bob wins with at least that probability. If the first flip is tails then Alice wins with tails on the second toss. Bob wins eventually if the second toss is heads. So overall Bob wins with probability 3/4.