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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.

mckelvey
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    I would claim that a standard phone handset does not have a cord and has not for some time. – Ross Millikan Feb 21 '23 at 02:11
  • I do not understand your model. The probability that you transfer the phone twice should be p^2, not p^3. It doesn't make sense that the probability of transferring actually decreases each time. – user21820 Feb 21 '23 at 07:01
  • The original idea was that the longer I talk the less I want to talk. I agree now that that is unrealistic. Also, probability is used instead of time. – mckelvey Feb 21 '23 at 17:08

1 Answers1

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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 $P(p)$:

Parcly Taxel
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    Yes, you are right. I didn't read the posting closely enough. I have therefore deleted my answer. I wrongly assumed the alternate interpretation, simply because I noticed that $q \times (1 + p + p^2 + \cdots) = 1.$ So, the (wrong) alternate interpretation made sense to me. – user2661923 Feb 20 '23 at 23:09