Bayes theorem for fair and unfair coins I'm working on a multi-parted bayes theorem problem for coin flipping.

Suppose you take a coin out of your pocket and you flip it 100 times and observe all heads. A) Would you believe me if I told you this was a fair coin? B) Now suppose I toss the coin $N$ times instead of 100, and it comes up heads. At which $N$ will you start to get suspicious that it's a double headed coin? e.g., when would you be 99% sure that the coin is a double headed coin? C) Now what if the coin I drew gave an alternating sequence, i.e., HTHTHTHTHTHTHTHT. What can you say about the coin?

I think these are open ended questions, and we have insufficient information to come up with actual numbers.
For A), I think this is a simple application of Bayes theorem. Specifically define $D$ to be the event that we drew a double headed coin.
Let $p$ be the probability of choosing a double headed coin.
$$\begin{align}
P(D|100H) &= \frac{P(100H|D)P(D)}{P(100H)}\\ &= 
\frac p {p + (1 - p)2^{-100}} 
\end{align}$$
For $p \geq 10^{-28}$, we observe that $P(D|100H) > 0.99$. So I think we can say with almost certainty that the coin is a double headed coin. However, I am not sure if this is a rigorous justification for why I do not believe the person that this is a fair coin. What would a rigorous justification look like? Hypothesis testing with p-values?
For B), I am stuck because my above expression would look like
$$\begin{align}
P(D|NH) &= 0.99 = \frac{P(NH|D)P(D)}{P(100H)}\\ 
&= \frac p  {p + 2^{-N}(1 - p)}
\end{align}$$
So here we have 2 unknowns in $p$ and $N$. Is there any further simplification that can be made in this situation to determine $N$? I think we would have to assume a value for $p$.
For C), I have no idea what kind of coin this would be. Using my imagination, I can say that this could be some fair coin such that every subsequent toss is the reverse of the previous, but I don't know of any physical phenomena that could guarantee this.
 A: A better presentation of part (b) would be : there are two coins in a box, one fair and the other double headed.  one of the coins is chosen at random and then flipped n times.
The coin comes up all heads.
Then, it becomes reasonable to plug in your formula, with $(p=1/2)$, and you obtain $\frac{2^n}{2^n + 1},$ as the chance that the coin chosen at random was the double headed coin.
The key difference between this presentation and yours is that in this presentation, before you start flipping the coin, you know that the chance of it being a fair coin is precisely $(p)$.
This means that if you assume that $p$ is some fixed rational number between $0$ and $1$, you can assume that you have a box with coins, where the chance of picking a fair coin at random from the box is $p$.
With your presentation of part (b), it was unclear what the probability of the coin being fair was, absent any other information.
In fact, you suggested the same idea in your query:

I think we would have to assume a value for $p$.


For what it's worth, in part (c), for any fixed value of $p$, you know immediately that the coin chosen had to be a fair coin, because it is impossible for a double headed coin to produce such a sequence.

An alternate presentation of part (c) would be that you have two coins in the box, one fair, and one lopsided, that comes up heads with probability $r$, where $s = (1-r).$
Then, with $r,s$ fixed, when asked what the chances of the coin being fair is, you would have $(1/2)^{2n}$ versus $(rs)^n.$
Here, for any value of $r \neq (1/2)$, the chances are better than $(1/2)$ that the coin was the fair coin.  This is because $(rs)$ must be $< (1/4).$
