Probability of X being a trick coin (heads every time) after heads is flipped k amount of times 
A magician has 24 fair coins, and 1 trick coin that flips heads
  every time. 
Someone robs the magician of one of his coins, and flips it $k$ times
  to check if it's the trick coin.
A) What is the probability that the coin the robber has is the trick
  coin, given that it flips heads all $k$ times? 
B) What is the smallest number of times they need to flip the coin to
  believe there is at least a 90% chance they have the trick coin, given
  that it flips heads on each of the flips?

Here is my approach:
Let $T$ be the event that the robber has the trick coin
Let $H$ be the event where the robber flips a heads k times in a row
$Pr(T) = 1/25$
$Pr(H|T) = 1$
$Pr(T') = 24/25$
$Pr(H|T') = 1/2$ when $k=1$, $1/4$ when $k=2$, $1/8$ when $k=3$... etc
$Pr(T|H) = (1 * 1/2) / (1 * 1/2 + Pr(H|T') * 24/25) = 1/13, 1/7, 1/4,...$ etc
So the Pr(T|H) answer changes for every k, do I answer with the formula? How can I answer A? How do I make a probability distribution when k can be infinite?
Also is B 8 flips? Since when k = 8, Pr(T|H) = 1/256.
Thanks for any help.

$$P(trick|H_k)=\frac {P(trick \cap H_k)} {P(H_k)}=\frac {P(H_k|trick).P(trick)} {P(H_k)}.$$
Now, 
$$\begin{align}P(trick)&= \frac{1}{24}\\
P(H_k|trick)&=1\\
P(H_k)&=P(H_k|trick)\cdot P(trick)+P(H_k|fair)\cdot P(fair)\\
&=1\cdot\frac{1}{24}+\frac{1}{2^k}\cdot\frac{23}{24}\end{align}$$
Hope this helps
