Solution¶
The distribution of actual canoodling times across couples' individual preferred canoodling times is shown below.
This week’s puzzle comes from David Polidori:
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?
The camera operators show just $1\%$ of couples during the concert. You and your partner only spend half the time canoodling.
The probability of you being shown canoodling and not canoodling for a show is thus $0.5\%$ and $99.5\%$, respectively.
Solving for the complement, that is, the number of shows that you will have at least a 50 percent chance of never being shown canoodling,
\begin{align*} (0.995)^n &\ge \frac{1}{2} \\ n\ln 0.995 &\ge \ln \frac{1}{2} \\ n &\le \frac{\ln \frac{1}{2}}{\ln 0.995} \approx 138.2826 \end{align*}The total number of shows is $$\boxed{138}$$
The corresponding probabilities are $49.93\%$ for being shown and $50.07\%$ for not being shown.
Now, everyone at the concert spends at least some time canoodling. In particular, each member of a couple wants to spend some fraction of the time canoodling, where this fraction is randomly and uniformly selected between 0 and 1. This value is chosen independently for the two members of each couple, and the actual time spent canoodling is the product of these values. For example, if you want to canoodle during half the concert and your partner wants to canoodle during a third of the concert, you will actually canoodle during a sixth of the concert.
Meanwhile, the camera operators love to show canoodling couples. So instead of randomly picking couples to show on the jumbotron, they randomly pick from among the currently canoodling couples. (The time shown on the jumbotron is very short, so a couple’s probability of being selected is proportional to how much time they spend canoodling.)
Looking around the concert, you notice that the kinds of couples who most frequently appear on the jumbotron aren’t constantly canoodling, since there are very few such couples. Indeed, the couples who most frequently appear on the jumbotron spend a particular fraction C of the concert canoodling. What is the value of C?
The distribution of actual canoodling times across couples' individual preferred canoodling times is shown below.
In order to determine the most frequently appearing canoodling group of fraction $C$ depends on the value of $C$ and the number of couples with value $C$.
The product of the arc length above and $C$ is to be maximized.
Starting with arc length:
\begin{align*}
L &= \int\limits_a^b \sqrt{1 + \left(\frac{dy}{dx}\right)^2}dx \\
L &= \int\limits_C^1 \sqrt{1 + \left[-\frac{C}{x^2}\right]^2}dx \\
L &= \int\limits_C^1 \frac{\sqrt{x^4 + C^2}}{x^2}dx \\
\end{align*}
This form is indeterminate. Let's define weight $W$ as that product.
\begin{align*} W &= C \cdot L \\ W &= C \cdot \int\limits_C^1 \frac{\sqrt{x^4 + C^2}}{x^2}dx \\ \end{align*}I used code to iterate 10 significant figures of $C$.
