E is for Eleven Madison Park: an unlikely mathematical journey
A few weeks ago, I was traveling on business to New York when I was fortunate enough to secure last minute dinner reservations at Eleven Madison Park, a three-star Michelin reviewed establishment and generally considered one of the finest restaurants in the world. What elevates it above its stuffier contemporaries is its refreshing frivolity, perhaps singularly typified by their famous “name that milk” dessert guessing game.
The game is simple. You are given four anonymized chocolate bars, each made from a different animal’s milk — cow, sheep, goat, and buffalo. Your job is to guess which chocolate corresponds to which animal and to notate the answer on a provide answer sheet.
My dinner companion David, a longtime friend and maybe the most intellectually curious person I’ve ever met, innocently wondered aloud what the odds were to ace the game — that is, to guess all four correctly. The math on this is pretty straightforward, and so I answered him matter-of-factly that there are 4! or 24 ways to fill out the form. Obviously, there is only 1 correct answer, so the answer is 1/24.
So there we were. Smelling. Tasting. Smelling again. Note that we are far from experts on how exactly buffalo milk differs from goat milk, but anyone only half paying attention would have mistaken us for the finest of milk connoisseurs. Before long, there were four confident graphite strokes on each of our answer sheets.
Then came the next course, the answer key, and a crushing blow to our budding careers in the dairy industry. We had each gotten exactly zero correct. Zero. Zilch. Nada. Bupkis.
After we laughed out loud at our lactose ineptitude, David asked the question that was lingering in the air — well, what were the odds of getting zero right?
Now Pascal and Bernoulli we are not, but David and I are both reasonably capable when it comes to math, probabilities, and calculations. “This won’t take long” I’m sure we both casually thought to ourselves as we began to reason about the problem. However, all of our initial forays surprisingly hit immediate dead ends:
“It must just be some combinatoric permutation of 4, just like before. Hmm wait actually it’s not that simple. So.. the first chocolate can be any of the 3 other possibilities, and then the next can have uh 2..? But what if the first chocolate was matched against the second’s milk, then it could be matched against any of the 3….”
“How about we just count the number of ways? Hm there’s actually quite a few and we don’t really have any paper to list them out….”
“Let’s do it backwards. Let’s count the ways to get 4 right (1/24), 3 right (0/24), and 2 right.. wait what’s the chance you get 2 right….”
Then we tried something different.
“Let’s reduce the problem. What if there were only 3 chocolates, not 4. We’ll solve it for n=3 and then we can scale the answer back to n=4.”
Perfect. Finally we were getting somewhere. Relief replaced the initial onset of stymied frustration, so much so that David insisting that we additionally solve the problem generically for n chocolates seemed like a trivial add-on.
Imagining the game with 3 chocolates, the possibilities become easy to count. There are 6 possible ways to fill out the answer sheet, and 2 of them would be completely wrong. Well, 2 is (n-1)! so maybe that’s the answer! Unfortunately, I quickly pointed out that I could name more than 6 ways to fill out 4-chocolate answer sheet to get them all wrong, so that wasn’t it.
Back to square one.
The pitter and patter of the incessant rain drops on the lumbering cab ride home was drowned out by our voices. Our stomachs had been satiated, but our minds had been left wanting. Theories were being discarded just as quickly as they were being formulated. Maybe even more quickly.
You could hear the growing incredulity in our voices. This shouldn’t be this hard. What are we missing?
I hadn’t even taken off my water laden jacket and damp shoes in the hardwood floor entryway, when I looked up and saw David already fully immersed in a notepad, diligently charting out all the possibilities.
My calls for similar tools left unheeded, I instinctively reached for my backpack, crumpled against the nearby wall, and pulled out my dependable, silver Macbook pro. Screw this. I’m going for the nuclear option. I brought up a text editor and swiftly coded up a computer program to calculate the possibilities. Within a few minutes, the glowing terminal before me had outputted the following:
2 chocolates, 2 total possibilities, 1 with no right answers
3 chocolates, 6 total possibilities, 2 with no right answers
4 chocolates, 24 total possibilities, 9 with no right answers
5 chocolates, 120 total possibilities, 44 with no right answers
6 chocolates, 720 total possibilities, 265 with no right answers
7 chocolates, 5040 total possibilities, 1854 with no right answers
8 chocolates, 40320 total possibilities, 14833 with no right answers
Now that I had all the answers per se, deriving the answer should be child’s play, right? Wrong. 9, 44, 265. Hmm. The numbers didn’t instantly resonate with me. What’s the pattern? Minutes passed as I recited them over and over in my head searching for some commonality.
Finally, my first insight! Each number was divisible by n-1; 44 was divisible by 4, 265 by 5, 1854 by 6, and so on and so forth. But the quotients equally made no sense to me. 3, 11, 53, 309?
Now, I’d love to tell you we dug in and white-boarded equations all night until we hit that hallelujah moment, but time was dragging on, and my eyes were getting heavier with every passing moment. In an act of defiance, I flippantly typed “9 44 265” into Google not expecting to see anything remotely helpful. How wrong I was.
The very first result, in fact the first several pages of results, shouted the word “derangements” at me. What’s a derangement? Well it turns out, David and I had accidentally stumbled onto a famous and well-known mathematical problem.
Derangements are “permutations of the elements of a set such that no element appears in its original position”. A simple way to think of it is to imagine you have a deck of cards in pristine order and you shuffle it such that no card is in its original position. That’s a derangement.
It was nothing short of surreal, reading an expertly detailed Wikipedia entry that had seemingly materialized out of thin area about the exact problem we had been trying to crack for the past two hours. So surprised was I, that I didn’t even bother to utter a word to my friend about this unexpected discovery, until that is, I read the following sentence:
The problem of counting derangements was first considered by Pierre Raymond de Montmort in 1708; he solved it in 1713.
Yes, you read that right. What had taken a noted professional mathematician 5(!) years to solve, we had tried to solve over a meal. Fifteen minutes later, when we had finally managed to stop laughing at our accidental hubris, we read the rest of the article.
It turns out we were tantalizingly close to solving the problem on our own, not intuitively understanding the semantics behind the solution necessarily, but simply deriving the correct formula: the number of derangements for n items, or d(n) = (n-1) * (d(n-1) + d(n-2)).
So what were those mysterious quotients 3, 11, 53, 309 in the end? They were simply the sums of the previous two derangement values:
3 = 1+2
11 = 2+9
53 = 9+44
309 = 44+265
Mystery solved. Sort of.
Let’s try a brief thought exercise. Earlier, I told you that my laptop terminal outputted the probability of getting no right answers when playing the “name that milk” game with 4 chocolates. The answer is 9/24.
Well, what if instead there were 100 chocolates? Would you be more likely or less likely to get zero right?
Think about it and make a guess. Seriously.
Got it? Sure? Okay good, read on.
You’re wrong.
Well, to be fair it’s not your fault. It was a trick question. The answer is neither. It’s a fact that still blows my mind: no matter how many chocolates you play the game with, you have the same probability of getting zero right.
Whether you’re shuffling a deck of 52 cards, returning winter coats to 10 people at the end of a housewarming party, or serving dinner entrées to 4 people, you have basically the same chance of getting it all entirely jumbled up.
And what is that probability exactly? What is the answer to the original question David raised during dinner which kickstarted our wild good chase?
The answer is approximately 1/e (9/24 ≈ 44/120 ≈ … ≈ 1/e. who knew!)
1/e? Seriously? Whiskey. Tango. Foxtrot.
It turns out that there are multiple ways to solve the derangements problem. The formula I wrote out earlier is dynamic, but if you want to arrive at a non-dynamic solution, one of the more popular approaches is to use the inclusion-exclusion principle.
By applying this principle and some clever math, we find the answer is reduced to a summation of terms, which happens to be an almost exact match to the Taylor series expansion of 1/e, hence why that ends up being the answer we were looking for.
With our admittedly strange, nerdy mathematical journey at an end, I was reminded of a quote I had recently read:
“In order to have a learning moment, you have to be open to the fact that you don’t know” — Sarah Tavel, Greylock Partners
She calls it a moment of humility. I would add that it requires a moment of awareness. There are so many aspects of our surrounding day-to-day world we take for granted, so much so that they are invisible to us. Aspects that are begging to be delved into, uncovered, and better understood.
How many Uber riders understand the underlying mechanic of surge pricing? How many startup employees take the time to understand how their options work until it’s time to exercise?
Knowledge is power. Declare ignorance. Ask questions. Pull the thread, as it were. You’ll be surprised what unexpected journeys a little intellectual curiosity and some free time will lead you down.
