Bias/Nonbiased Probability Puzzle Question

I got this puzzle that I need help on. I hope its not too easy because I really don't understand it.

A bag contains 100 coins. 99 of them are fair and will give heads or tails with an equal probability. 1 biased coin will always yield heads. I pull a coin out randomly from the bag and toss it up 3 times. In each of the 3 times it lands on heads.

What is the probability that I pulled out the biased coin?

Let $A$ be the situation of picking the loaded coin and let $B$ the situation of getting heads thrice.
By Bayes' Theorem $$P(A|B) = \frac{P(B|A)P(A)}{P(B)} = \frac{1\cdot\frac{1}{100}}{1\cdot\frac{1}{100}+\frac{1}{8}\cdot\frac{99}{100}} = \frac{8}{107}.$$
Imagine repeating the experiment $100000$ times. Then "about" $1000$ times, we will pull out the funny coin, and of course get $3$ heads. About $99000$ times, we will pull out an ordinary coin. About $\frac{1}{8}$ of these times we will get $3$ heads, so it will happen about $12375$ times.
So we get three heads a total of about $13375$ times. In about $1000$ cases, it was the biased coin. So the required probability is $\dfrac{1000}{13375}$.