Consider the ways to get heads on both of the first two tosses: you can have the unfair coin, or you can have a fair coin and toss heads twice in a row. The probability of having the unfair coin is $\frac13$. The probability of having a fair coin and tossing heads twice in a row is $\frac23\cdot\frac14=\frac16$. If you know that you’ve tossed heads twice in a row, then you know that you’re in one of these two cases; the probability that you’re in the first case (have the unfair coin) is $$\frac{\frac13}{\frac13+\frac16}=\frac23\;,$$ and the probability that you’re in the second case (have a fair coin) is $1-\frac23=\frac13$.
You know the probability of getting tails with the unfair coin and the probability of getting tails with a fair coin, so you can now calculate the overall probability of getting tails if you’ve tossed heads on your first two tries; just use the same basic approach that you used in the first three questions.
If you began by tossing tails, twice, on the other hand, you know that you have a fair coin.
Not as Elegant a Basic Approach as I Had Foreseen. We show this by induction.
First, for $n = 1$, a single coin. Obviously, then the probability of an even number of heads is simply the probability that this coin flips tails. If this coin is unfair, this probability is clearly not equal to $1/2$. Therefore the coin must be fair. This establishes the basis step.
Now, suppose that the proposition is true for some $n > 0$. Let us now show it for $n+1$. The antecedent is that the probability of an even number of heads in these $n+1$ flips is $1/2$. If (at least) one of the first $n$ coins is fair, then the consequent is true.
If, on the other hand, none of the first $n$ coins is fair, we already know that such a circumstance does not permit the probability of an even number of heads in the first $n$ tosses to be $1/2$. Let us say therefore that this probability is instead $P_n \not= 1/2$, and let the $n+1$th coin have a probability of heads of $q$. Then the probability that the number of heads is even after all $n+1$ tosses is
$$
P_{n+1} = P_n(1-q) + (1-P_n)q = P_n + q(1-2P_n)
$$
But we know, by hypothesis, that $P_{n+1} = 1/2$, so we write
$$
\frac12 = P_n + q(1-2P_n)
$$
which gives us, after some simple algebra,
$$
q = \frac{1/2-P_n}{1-2P_n} = \frac12
$$
This establishes the induction step and the proposition is shown.
Best Answer
Let $O(n)$ and $E(n)$ be the probability to get odd number and even number of heads from $n$ throws. Also let $p$ be the probability to get head from a throw. We have the following relations:
$$ \begin{aligned} O(n) &= O(n-1)\cdot(1-p)+E(n-1)\cdot p \\ E(n) &= E(n-1)\cdot (1-p) + O(n-1)\cdot p \end{aligned} $$
Thus we have the following:
$$ O(n)=E(n) \implies (1-2p)\cdot O(n-1) =(1-2p)\cdot E(n-1) $$
The implication trivially holds for $p=\frac{1}{2}$. Now let's assume $p\neq\frac{1}{2}$ to see if any other $p$ satisfy the implication. From this assumption, $1-2p\neq 0$ so the implication becomes the following:
$$ O(n)=E(n) \implies O(n-1)=E(n-1) \implies ... \implies O(1)=E(1) $$
However, $O(1)=p$ and $E(1)=1-p$ so in the end we again end up with $p=\frac{1}{2}$.