Calculate probability from infinite tree Jacob is contemplating what ice cream flavor he wants for he can only choose one of the three flavors chocolate, orange or pear. Jacob is very indecisive; he wants chocolate from the beginning but there is a $50$% chance that he will change his mind to another flavor before making the final decision. If he changes his mind, he will choose one of the other two flavors at random. Say he changes to pear, then there will once again be a $50$% chance that he changes his mind before making the final decision (i.e. this process could repeat infinitely). What is the chance that in the end, he will have chosen chocolate flavor?
My Solution:
Due to symmetry I would argue that the probability that he chooses orange is the equal to choosing pear, hence we will only try and compute the probability that he chooses pear and get our answer out of that.
Now if we look at a fraction of the probability tree drawn out below. Let brown, yellow and green circles represent chocolate, orange and pear respectively. Black horizontal line means that he changes his mind and red crosses is the final decision. Furthermore, we will refer to the distance between two adjacent objects (connected with a line segment) as a “path”. Note that every path in the tree describes a $1/2$ chance of choosing that specific route.

We observe that, if we stand on an arbitrary green circle, we can make it to another green circle if we move over 4 paths. If we choose point $P_1$ as our starting point, we may describe the possible routes probability as $(1/(2^{3+4n}))2^n$ for any non-negative integer $n$. That is because, from the initial chocolate point, it takes $2$ moves to the first green $P_1$, then $4n$ moves more to get to any other green point on that side of the tree, and finally $1$ more move to reach the final decision. We then multiply with $2^n$ because the tree splits and creates more routes leading to green points. Similarly, if we choose $P_2$ as our starting point, then we have $(1/(2^{5+4n}))2^n$ for some non-negative integer $n$. We calculate the limit of the sum:
$(1/(2^{5+4n}))2^n + (1/(2^{3+4n}))2^n = 0.17857142857142855$.
We get that the chance of Jacob choosing chocolate is $0.6428571428571429$.
Question: Is the answer correct/reasonable and does there exist more sophisticated methods to calculate the answer?
 A: I got a different answer. 
Here’s a simpler method. Let $p_c$ be the probability that Jacob ends up choosing chocolate; notice that the probability of him ending up at either of the other two flavors are identical and equal to $\frac{1-p_c}{2}$.
Jacob has a $1/2$ probability of not switching at all, so we have
$$p_c=\frac{1}{2}+\frac{1}{2}\mathbb{P}(\text{chocolate | his mind changes at least once})$$
If he changes his mind once, we are now in the same situation as before, but with one of the other two flavors. Given that he switches once, the flavor he switches to will then have a probability $p_c$ of finally being selected, and chocolate will have the probability $\frac{1-p_c}{2}$ of being selected. This implies that
$$p_c=\frac{1}{2}+\frac{1}{2}\cdot \frac{1-p_c}{2}$$
and, solving this equation, we have
$$p_c=\frac{3}{5}$$

Consider a more general version of this problem in which there are $n$ ice cream flavors (hopefully, none of them is pear flavored) and a probability $p$ that Jacob changes his mind after each choice. Using the same method as above, I calculated the following formula for the probability that his final choice is the same as his initial choice:
$$p_c=1-\frac{p(n-1)}{p+n-1}$$
