2.6.11 problem 70

Let $F$ be the event that the coin is fair, and let $H_i$ be the even that the $i$-th toss lands Heads.

(a)

Both Fred and his friend are correct. Fred is correct in that the probability of there being no Heads in the entire sequence is very small. For example, there are $\left(\frac{92}{45}\right)$ sequences with $45$ Heads and $47$ Tails, but only $1$ sequence of all Heads.

On the other hand, Fred’s friend is correct in his assessment that any particular sequence has the same likelihood of occurance as any other sequence.

(b)

$$P\left(F\right|H_{1\le i\le 92})=\frac{P\left(F\right)P\left(H_{1\le i\le 92}\right|F)}{P\left(F\right)P\left(H_{1\le i\le 92}\right|F)+P(F^c\left)P\right(H_{1\le i\le 92}\left|F^c\right)}=\frac{p{\left(\frac{1}{2}\right)}^{92}}{p{\left(\frac{1}{2}\right)}^{92}+(1-p)}$$

(c)

For $P\left(F\right|H_{1\le i\le 92})$ to be larger than $\frac{1}{2}$, $p$ must be greater than $\frac{2^{92}}{2^{92}+1}$, which is approximately equal to $1$, where as for $P\left(F\right|H_{1\le i\le 92})$ to be less than $\frac{1}{20}$, $p$ must be less than $\frac{2^{92}}{2^{92}+19}$, which is also approximately equal to $1$. In other words, unless we know for a fact that the coin is fair, $92$ Heads in a row will convince us otherwise.