Is there a mathematical reason why chocolate chip cookies have 37% (1/e) chocolate in them?

37% refers to the amount of dry cocoa solids in the chocolate chips. These percentages are how chocolate is sold.[1][2] In the US and EU you need more than 35% cocoa solids in order to refer to your product as semisweet chocolate and chocolate (respectively).[3] The chocolate bar recommeneded in this review[4] as "splurge worthy" is labeled 61% dark chocolate. This is not because the bar is 61% chocolate and 39% air/other. It is a 61% chocolate bar because the chocolate contains 61% dry cocoa solids.

[1] http://www.thestoryofchocolate.com/Savor/content.cfm?ItemNumber=3454&navItemNumber=3376

[2] http://www.21food.com/products/brick---2-lbs---milk-37-51823.html

[3] https://en.wikipedia.org/wiki/Types_of_chocolate#Classification

[4] http://www.cookinglight.com/cooking-101/essential-ingredients/taste-test-chocolate-00400000058842/

Businesses prefer round or familiar numbers -to make it easier to compute by mind. So I am willing to bet that the the company would have been more inclined to have, say, a round 40% content (or 8 gr in the 20 gr overall that each cookie weighs according to the blogger), rather than $1/e$. Or even, 33% (which is familiar in the sense that is translated as $1/3$ or a $2:1$ ratio). So why did they end up with $1/e$? I conjecture it has to do with production cost considerations.

This is a consumer's product, produced in large quantities. For legal and commercial reasons, it needs to be as homogeneous as possible. One way to achieve this would be to create each cookie one-by-one. It doesn't take much thought to realize that this would be a very costly process...

So cost-considerations mean that we do not create the dough-chocolate mix cookie-per-cookie, but we create one large mass of dough-chocolate and then we separate it in parcels of same weight and send to bake, in order to arrive at the ~20gr ready cookie. Given this production process, in order to achieve homogeneity among final cookies, we need to have the chocolate chips as homogeneously distributed as possible in the dough mass. Therefore the machines in the factory need to "mimic" a "random selection" process (selection of the 3D point that each chocolate chip will occupy inside the dough mass), so that they are scattered ~uniformly in this mass.

Now the cookie dough is not dry, but watered. So it is less messy (and so less costly) to move around the chocolate chips in transfer-belts instead. Then it looks like they prepare the cookie dough in some kind of large containers, then pour in the chocolate chips, then "shake" (the correct English word escapes me) in order to make the mix as homogeneous as possible... And here comes another cost issue: how long do they need to shake the mix in order for it to become homogeneous? The longer it takes, the costlier it is. So a cost-sensitive business would opt for the minimum possible amount of "shaking time", that would guarantee 3D-homogeneity of the mix as regards the distribution of the chocolate chips.

And this is the answer "why $1/e$": if the weight proportions of the total dough mass and the total chocolate chips are such that conform with the theoretical end-result of the theoretical random selection process already described in the question, it will take the minimum amount of time to effectively deterministically mimic this random selection process -i.e. to make the whole mix homogeneous.

As to why "if this is so, why not everybody is doing it?", the answer has to do with the fact that not all businesses are primarily cost-driven (a firm may want to have a 50% chocolate content and charge a higher price), and that history matters -a cookie may have traditionally contained 20% chocolate, and be commercially successful, and they didn't want to jeopardize that.

Suppose we are making $c$ cookies (where $c$ is a large number). For simplicity, let's approximate by assuming each cookie has $k$ discrete locations where chocolate chips might go (where $k$ is a positive integer). Then there are a total of $n = ck$ locations where chocolate chips might go. So we start with $n$ chocolate chips and randomly throw them down. Each one independently lands in one of the $n$ locations. If two or more land in the same location, they all bounce into the garbage.

So at the end of this process, a location has a single chocolate chip if and only if exactly one chip landed there. Now given any particular location (of the $n$ possible ones), the probability that a particular chocolote chip attempts to land there is $q = 1/n$. The probability that there is a chocolate chip there after this process is done is: $$ nq(1-q)^{n-1} = (1-1/n)^{n-1} $$ When $n$ gets large (so that $n\rightarrow\infty$) we get: $$ (1-1/n)^{n-1} \rightarrow 1/e $$ and so the fraction of chocolate chips that are not thrown away converges to $1/e \approx 0.367879$.