Let $H$ be the event that we get heads twice in a row, and let $C_1$ and $C_2$ be the events that we picked coin $C_1$ and coin $C_2$, respectively. We want to compute the probability of $C_1$ given $H$.

Using Bayes’ theorem, we have:

$P\left(C_1 \mid H\right)=\frac{P\left(H \mid C_1\right) P\left(C_1\right)}{P(H)}$

We can compute the probabilities on the right-hand side as follows:

• $P(H | C_1)$ is the probability of getting heads twice in a row given that we picked coin $C_1$. Since $C_1$ gives head with probability $1/2$, this probability is $(1/2)^2 = 1/4$.
• $P(C_1)$ is the probability of picking coin $C_1$ in the first place, which is $1/2$.
• $P(H)$ is the probability of getting heads twice in a row, which we can compute using the law of total probability:

$\begin{aligned} P(H) & =P\left(H \mid C_1\right) P\left(C_1\right)+P\left(H \mid C_2\right) P\left(C_2\right) \ & =\frac{1}{4} \cdot \frac{1}{2}+\left(\frac{1}{3}\right)^2 \cdot \frac{1}{2} \ & =\frac{1}{12}+\frac{1}{18} \ & =\frac{5}{36}\end{aligned}$

Substituting these values into Bayes’ theorem, we get:

$P\left(C_1 \mid H\right)=\frac{(1 / 4) \cdot(1 / 2)}{5 / 36}=\frac{9}{20} \approx 0.45$

Therefore, the probability that we picked coin $C_1$ given that we got heads twice in a row is approximately $0.45$.

Let $X$ be the number of heads obtained from 90 throws of the biased coin. We have $X \sim \mathrm{Bin}(n=90,p=1/3)$, where $\mathrm{Bin}(n,p)$ denotes a binomial distribution with $n$ trials and success probability $p$. We want to find $P(X \geq 80)$.

We will use the Chebyshev’s inequality, which states that for any random variable $X$ with finite mean $\mu$ and variance $\sigma^2$, and for any positive constant $k$, we have:

$P(|X-\mu| \geq k \sigma) \leq \frac{1}{k^2}$

Applying this inequality to our problem, we have:

$P(X \geq 80)=P(X-\mu \geq 80-\mu) \leq \frac{\operatorname{Var}(X)}{(80-\mu)^2}$

Note that $\mu = np = 90 \cdot 1/3 = 30$ and $\mathrm{Var}(X) = np(1-p) = 90 \cdot 1/3 \cdot 2/3 = 20$. Substituting these values, we get:

$P(X \geq 80) \leq \frac{20}{(80-30)^2}=\frac{1}{25}=0.04$

Thus, the probability of obtaining 80 heads or more from 90 throws is not more than 0.04. However, this is an upper bound obtained by Chebyshev’s inequality, and it may be loose in practice. To get a tighter bound, we can use the normal approximation to the binomial distribution, which is valid when $np$ and $n(1-p)$ are both large enough. In our case, $np=30$ and $n(1-p)=60$, so the normal approximation is reasonable. Using the continuity correction, we have:

$P(X \geq 80) \approx P\left(Z \geq \frac{80+0.5-n p}{\sqrt{n p(1-p)}}\right)$

where $Z$ is a standard normal random variable. Substituting the values, we get:

$P(X \geq 80) \approx P(Z \geq 4.45) \approx 2.4 \times 10^{-6}$

Thus, the probability of obtaining 80 heads or more from 90 throws is extremely small, and certainly not more than 0.16.

We will prove the statement using the Pigeonhole Principle, which states that if $n$ pigeons are placed into $m$ pigeonholes, and $n>m$, then there exists at least one pigeonhole with more than one pigeon.

Consider the set $S={1,2,\ldots,100}$ and the set $T={101,102,\ldots,199}$. Note that $S$ and $T$ are disjoint, and their union is ${1,2,\ldots,199}$.

Suppose for the sake of contradiction that $C$ is a subset of $T$ with $|C|=51$ that does not contain two consecutive integers. Then, $C$ and $C+1={c+1: c\in C}$ are disjoint subsets of $S$ with $|C|=|C+1|=51$. This means that $S$ contains two disjoint subsets $A$ and $B$ with $|A|=|B|=51$.

Now, consider the set of differences $D={b-a: b\in B, a\in A}$, which consists of all the possible differences between an element in $B$ and an element in $A$. Note that $D$ is a subset of ${-99,-98,\ldots,99}$, since the largest difference between an element in $B$ and an element in $A$ is $99$ (when the largest element in $B$ is subtracted from the smallest element in $A$).

Since $|D|=51^2>199$, there must be two distinct pairs $(a_1,b_1)$ and $(a_2,b_2)$ in $A\times B$ with the same difference $b_1-a_1=b_2-a_2$. But then, $b_2-a_1=b_2-a_2+(a_2-a_1)$ is a difference between an element in $B$ and an element in $A$, and $b_2-a_1$ is equal to $b_1-a_1$, which means that $C$ contains two consecutive integers, namely $b_1$ and $a_1+1$. This is a contradiction, so we conclude that any subset $C$ of $T$ with $|C|=51$ must contain two consecutive integers.