Who has more probability of winning the game?

Question

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.

Answer 1

To calculate $E=E[Y]$:

We work from states. There's $\emptyset$, the starting state, or the state in which you've thrown nothing but $H$, there's $T$ in which the string has been $H^aT^b$ with $b>0$, and of course there's the end state.

From the start, you either stay in the starting state or you move to state $\mathscr S_T$. Thus $$E=\frac 12\times (E+1)+\frac 12\times (E_T+1)$$

From state $\mathscr S_T$ we either stay in $\mathscr S_T$ or we end. Thus $$E_T=\frac 12\times (E_T+1)+\frac 12\times 1\implies E_T=2$$

It follows that $$E=4$$

Just to stress: this certainly does not imply that $B$ has a greater chance of winning. The intuitive failure comes from the fact that if you have a $T$ already, and you throw an $H$ next, it takes you at least two turns to get $TT$ whereas if you have a $T$ and throw another $T$ you can still get your $TH$ on the next turn. Indeed, the two have an equal chance of winning, since the first toss after the first $T$ settles the game (as was clearly explained in the post from @EthanBolker).

Answer 2

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.

Answer 3

There is 1 fundamental problems with the attempted solution by math maniac:

The comparison of expected values may give a hint, but can be totally misleading and is not an equivalent of who is more likely to win first.

Consider another game, again played with a sequence of fair coin tosses. Alice wins after 5 tosses, no matter what actually comes up. Bob wins after 2 coin tosses if those first 2 tosses are $HH$, otherwise he wins after 6 coin tosses. Who will win first?

The expected number of coin tosses for Alice to win is easy:

$$E(W_A)=5$$

For Bob, there are two cases: The first 2 tosses are $HH$ (probability $\frac14$), or not (probability $\frac34$). With the given number of tosses until the win, this means

$$E(W_B)=\frac14\times2 + \frac34\times 6 = \frac{2+18}4 = 5$$

So the expected number of tosses until the win is $5$ in both cases. Nevertheless, Alice will win in $\frac34$ of all duels, Bob only in $\frac14$. That's because Bob only wins if the sequence of coin tosses starts with $HH$.

The calculation of the expected value for Bob is weighting the outcomes (2 tosses or 6 tosses) with the probabilities ($\frac14,\frac34$). Because $2$ tosses is much smaller than $6$ tosses, the expexted value for Bob is reduced by $1$ from the $6$ tosses is has for the 'majority case' that the sequence does not start with $HH$.

But for the calculation of who will likely win first, it doesn't matter that Bob wins in only 2 tosses if he wins at all. The fact that $2$ is a much smaller number than $5$ (the number of tosses Alice will always need to win) is irrelevant here.

In other words, the fact that if Bob wins, he will use a much smaller number of tosses than Alice is only relevant for the expected value of coin tosses, but not for the win probability itself. That's the reason why the expected value of coin tosses until the win is not the arbiter of who wins first.