Probability of Twisting a Phone Cord During a Call I invented this problem and am unable to solve it. It is not a homework problem.
I make a phone call on a standard handset (with a coiled cord). I start with the
phone on my right ear. With probability p I talk long enough that I transfer the
phone to my left ear, putting a twist in the cord. If I hang up at that point
the cord remains twisted. But at probability p^2 I talk even longer and transfer
the phone back to my right ear, removing the twist. At p^3 I put the twist back,
and at p^4 I remove the twist again, and so on. If the call can be unbounded,
how do I compute the probability P(p) that a call will put a twist in the cord?
Here is a diagram of the call, where the (possibly infinite) call is on the
horizontal and the possible hang-ups on the vertical. 1 means a twist, and 0
means no twist.
       p^1       p^2       p^3       p^4
   0---------1---------0---------1---------0...  Phone call -->
   |         |         |         |
   | 1-p^1   | 1-p^2   | 1-p^3   | 1-p^4         Hangups |
   |         |         |         |                       |
   0         1         0         1                       v

There are scenarios that result in a 1:

Q1 = p^1 * (1 - p^2)             Twist and no untwist
Q3 = p^1 * p^2 * p^3 * (1 - p^4) Twist, untwist, twist again and no untwist
Q5 = p^1 * p^2 * p^3 * p^4 * p^5 * (1 - p^6) Two aborted twists and a twist
...

For any given call, there can be at most one Q. But that seems to mean the
exclusive-or of an infinite number of Qs! How can that be done? Or is that the
wrong approach?
I'm looking for: How to calculate P(p)?
What is the limit of P(p) as p approaches 1.0? (if it exists)
(Graphs of P(p) and (P(p) - p) might be interesting)
Update:
When p is 1.0 my Q scenarios all go to 0 because of the last term. It
is impossible to avoid untwisting.

You do not XOR an infinite number of probabilities, you add them. Given your probability tree:
$$P(p)=p-p^3+p^6-p^{10}+\cdots=\sum_{n=1}^\infty(-1)^{n+1}p^{n(n+1)/2}$$
which does not have a closed-form solution. (Not even Jacobi theta functions as seen here will help.) However, $\lim_{p\to1}P(p)=\frac12$ as can be intuitively derived from the twisting and untwisting process on an infinite-length call.
This is a plot of $P(p)$:

