Coldplay Canoodling

All the many attendees at a particular Coldplay concert are couples. As the CEO of Astrometrics, Inc., you are in attendance with your romantic partner, who is definitely not the head of HR at Astrometrics, Inc. During the concert, the two of you spend half the time canoodling.

The camera operators love to show people on the jumbotron during the concert, but time is limited and there are many attendees. As a result, the camera operators show just $1$ percent of couples during the concert. Couples are chosen randomly, but never repeat at any given concert.

You and your partner are shy when it comes to public displays of affection. While you don’t mind being shown on the jumbotron, you don’t want to be shown canoodling on the jumbotron. How many Coldplay shows can the two of you expect to attend without having more than a $50$ percent chance of ever being shown canoodling on the jumbotron?

For the classic problem, we first want to find what the probability of not being shown on the screen canoodling during any individual concert would be. Let's say that $p_s = 1\%$ is the probability of being shown on screen during a show and $p_c=50\%$ is the probability of canoodling at any moment during a show. So the probability of not being shown while canoodling at any individual show is $P=1-p_cp_s=99.5\%.$

The probability of not being shown canoodling in any of $N$ consecutive shows is $P^N,$ so the maximum number of shows that can be attended without there being a greater than $50\%$ chance of being shown is \begin{align*}N^* &= \max \{n \in \mathbb{N} \mid P^n \geq \frac{1}{2}\}\\ &= \max \{ n \in \mathbb{N} \mid n \ln P \geq - \ln 2\}\\ &= \left\lfloor -\frac{\ln 2}{\ln P}\right\rfloor = 138\end{align*} Coldplay shows.