This week the #puzzle is: Can You Canoodle at the Coldplay Concert? #probabilities #geometry #integral #estimat #montecarlo

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?

And for extra credit:

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?

Can You Canoodle at the Coldplay Concert?

Highlight to reveal (possibly incorrect) solution:

Program

Information:

• This particular couple canoodles half of the time.
• 1% of couples at the concert will be shown on the jumbotron, randomly chosen.
• Going to n concerts, there’s at most 50% chance that this couple will be on the jumbotron, canoodling. Find n.

My first guess is:

• At any given concert:
  • There’s a 1% chance this couple will be on the jumbotron.
  • There’s a 50% chance they are canoodling.
  • There’s a 1% * 50% = 0.5% chance they are canoodling on the jumbotron.
  • There’s a 100% – 0.5% = 99.5% chance they are not canoodling on the jumbotron.
• At n concerts:
  • There’s a (99.5%)ⁿ chance they are not canoodling on the jumbotron.
  • We want (99.5%)ⁿ > 50%.
  • (99.5%)ⁿ > 50%
  • log(0.995ⁿ) > log(0.5)
  • n * log(0.995) > log(0.5)
  • Divide by negative number on both sides.
  • n < log(0.5)/log(0.995) = 138.282572986
  • n = 138

My second approach was to write a program to confirm this number.

And for extra credit:

Spreadsheet Desmos Program

Information:

• For each couple, each member wants to canoodle t_i of the time, t_i randomly chosen between o and 1.
• The couple then canoodles t = t₁ * t₂ of the time.
• A random canoodling couple will be shown on the jumbotron.
• The chance of being chosen for the jumbotron is proportional to the time spent canoodling.
• Slicing the couples into groups depending on their t, the couples shown most frequently belong to the group that has t = C. Find C.

My first guess didn’t fit the programs I also wrote, so this is my second guess:

• Assume that all the couples are sorted into a grid. With e.g. 10 * 10 couples, there are 10 * 10 cells. In the first row, all the first members of the couples have t₁ = 0.05 +- 0.05, in the next row it’s 0.15 +- 0.05 etc. In the first column, all the second members of the couples have t₂ = 0.05 +- 0.05. This way we get a regular sample of couples.
• For a value a, we can look at all the couples with t = a. Estimating the size of this group, we look at a slice going from a to a + b.
• a = x * y or y = a/x.
• Look at the area A between a/x and (a+b)/x.
• The probability that a couple from this group will be shown is A * a.
• Maximize A * a.
• I use a combination of Desmos and a spreadsheet graph to find this a = 0.368.

My program produced this histogram:

                               * *      *    *   *     *                                            
                     ******************** ****** ***** *   *                                        
                ************************************************ *  *                               
           ******************************************************* *** **                           
        ************************************************************** ***** *                      
      ************************************************************************** *                  
    *********************************************************************************** *           
  **************************************************************************************** *        
 ********************************************************************************************* *    
0---------1---------2---------3---------4---------5---------6---------7---------8---------9---------A
Histogram shows probabilities from 0 through 0.1 and 0.9 to 1 (0, 1, 9, A).
Probability 0.31 appeared 150030 times.
Probability 0.33 appeared 152955 times.
Probability 0.4 appeared 164107 times.
Probability 0.45 appeared 152542 times.
Probability 0.49 appeared 157924 times.
Probability 0.55 appeared 151192 times.
1000 loops, 10000 appearances in a concert, 10000 couples

It confirms the basic shape of the curve, and points towards the same interval of max values, but I couldn’t calculate a with the same precision.