Applications/How it relates to EA

This article's main application would be to help you decide whether you should:

1. Ask for help with a project you're working on.
2. Ask someone else to do the project for you.
3. Allocate tasks across a team of people, whether it be for a school project, a new startup, or anything else.
4. Other things that I haven't thought of. (If you do think of anything, please mention it in the comments.)

This ties to EA since many people in EA collaborate with and distribute work across co-workers, and since a key part of EA is trying to maximize good done, and one of the ways to do this is to choose the best person to do a given task.

Disclaimer

Although I do have a lot of experience in math, this article has not yet been thoroughly peer-reviewed yet (if you think you should peer-review it, please do.), and is subject to change. Do not take the things I say in this article at face value unless the stakes aren't that high.

How Much More Valuable is Your Time, Compared to Others?

TL;DR

This article is about how to decide who does what. (e.g. Should you send an email to your co-worker, who works faster than you, but is much busier?)

To figure out who should do what, with the objective of increasing the amount of good done, we look at:

the expected amount of good done by person 1^[1] if person 1 does the task (a), 

the expected amount of good done by person 2 if person 1 does the task (b),

the expected amount of good done by person 1 if person 2 does the task (c), and

the expected amount of good done by person 2 if person 2 does the task (d). (we only look at the amount of time where person 1 and 2 are affected by who does said task.)

Since the amount of good done (in the allotted time slot) if person 1 does the task is equal to $a+b$, and the amount of good done (in the allotted time slot) if person 2 does the task is $c+d$,

person 1 does the task if:

$a+b>c+d.$

person 2 does the task if:

$a+b<c+d.$

It doesn't matter who does the task if:

$a+b=c+d.$

Example and Details (Feel Free to Skip This Section)

Prerequisites

This article assumes that you know that $E[someconstant\cdot X]$=$someconstant\cdot E\left[X\right]$^[2]

If you don't know this, a simple proof is as follows:

for some random variable X, E(X) is what you'd expect X to be. ^[2]

Now, for some constant C, we just need to show that $E[c\cdot X]=c\cdot E\left[X\right]$.

This can be shown by imagining that we randomly generate some value for X, which we'll call ${\hat{X}}_1$. Now, let's multiply ${\hat{X}}_1$ by c. We'll call this new variable ${\dot{X}}_1$. Then, we'll generate another value for X, which we'll call ${\hat{X}}_2$. Now, let's multiply ${\hat{X}}_2$ by c. We'll call this new variable ${\dot{X}}_2$. We repeat this process a bunch of times.

We can then take the arithmetic mean of  all the $\dot{X}$'s, resulting in $\frac{{\dot{X}}_1+{\dot{X}}_2+{\dot{X}}_3+...+{\dot{X}}_n}{n}$, as $n$ trends towards infinity. By definition, this is $E\left[\dot{X}\right]$.

Since we can rewrite each ${\dot{X}}_i$ as $c\cdot {\hat{X}}_i$, we can rewrite $\frac{{\dot{X}}_1+{\dot{X}}_2+{\dot{X}}_3+...+{\dot{X}}_n}{n}$ as$\frac{c\cdot {\hat{X}}_1+c\cdot {\hat{X}}_2+c\cdot {\hat{X}}_3+...+c\cdot {\hat{X}}_n}{n}$, and, since $ab+ac=a(b+c)$, and $\frac{a\cdot b}{c}=a\cdot \frac{b}{c}$, $\frac{c\cdot {\hat{X}}_1+c\cdot {\hat{X}}_2+c\cdot {\hat{X}}_3+...+c\cdot {\hat{X}}_n}{n}=\frac{c\cdot ({\hat{X}}_1+{\hat{X}}_2+{\hat{X}}_3+...+{\hat{X}}_n)}{n}=c\cdot \frac{({\hat{X}}_1+{\hat{X}}_2+{\hat{X}}_3+...+{\hat{X}}_n)}{n}$.

and, since, if $a=b$ and $b=c$, then $a=c$, $\frac{{\dot{X}}_1+{\dot{X}}_2+{\dot{X}}_3+...+{\dot{X}}_n}{n}=c\cdot \frac{({\hat{X}}_1+{\hat{X}}_2+{\hat{X}}_3+...+{\hat{X}}_n)}{n}.$

And, since, by definition, $\frac{{\dot{X}}_1+{\dot{X}}_2+{\dot{X}}_3+...+{\dot{X}}_n}{n}=E\left[\dot{X}\right]=E[c\cdot X]$, and $\frac{({\hat{X}}_1+{\hat{X}}_2+{\hat{X}}_3+...+{\hat{X}}_n)}{n}=E\left[X\right]$.

And, since, if $a=b$ and $b=c$, then $a=c$, then:

 $E[c\cdot X]=c\cdot E\left[X\right]$.

Example/premise

For the sake of this example, let's say you and your friend are running an ice cream stand. (Bare with me!)

Details:

1. It takes you:
   1. 🍫1 seconds to make chocolate ice cream you can't make chocolate ice cream. (Which I'll be referring to as 🍫 from now on) .
   2. ${🍦}_1$ seconds to make vanilla ice cream (Which I'll be referring to as 🍦 from now on), 
   3. and ${🍓}_1$ seconds to make strawberry ice cream (Which I'll be referring to as 🍓 from now on).
2. It takes your friend:
   1.  ${🍫}_2$ seconds to make 🍫,
   2. 🍦2 seconds to make  🍦 your friend can't make 🍦.
   3. and ${🍓}_2$ seconds to make 🍓.
3. The price of each flavor X can be expressed as 💵(X). (e.g., the price of 🍫 is  💵(🍫).)
4. When there's a choice between serving someone who wants 🍓 and serving someone who wants  🍫OR^[3]🍦, for whatever reason^[4], they must choose 🍓.
5. All you and your friend care about is money.

Currently, you're handling a large line of people who want 🍦, and your friend is handling a large line of people who want 🍫.

A group of $n$ people come in, saying they want 🍓.

The question is: should you serve the 🍓-ordering customers, or should your friend?

(From now on, I will write "serving the 🍓to the people who ordered 🍓" as "🟥".

Your friend quickly puts together a table, showing the pros and cons of each option.

Now, we'll simplify both of these formulas. Feel free to skip this part.

Simplifying

$n\cdot E\left[💵\right(🍓\left)|1\right]$$-n\cdot E[\frac{{🍓}_1}{{🍦}_1}\cdot 💵(🍦\left)|1\right]+$$E\left[\right(💵\left(🍦\right)\cdot time\left(inseconds\right)youspendserving🍦\cdot {🍦}_1|1]$$-E\left[\right(💵\left(🍦\right)\cdot time\left(inseconds\right)youspendserving🍦\cdot {🍦}_1|2]$

$=n\cdot \left(E\right[💵\left(🍓\right)|1]-E[\frac{{🍓}_1}{{🍦}_1}\cdot 💵\left(🍦\right)|1\left]\right)+$$E\left[\right(💵\left(🍦\right)\cdot time\left(inseconds\right)youspendserving🍦\cdot {🍦}_1|1]$$-E\left[\right(💵\left(🍦\right)\cdot time\left(inseconds\right)youspendserving🍦\cdot {🍦}_1|2]$.

$n\cdot E\left[💵\right(🍓\left)|2\right]-$$n\cdot E[\frac{{🍓}_2}{{🍫}_2}\cdot 💵(🍫\left)|2\right]$$+E\left[\right(💵\left(🍫\right)\cdot time\left(inseconds\right)yourfriendspendsserving🍫\cdot {🍫}_2|2]$

$-E\left[\right(💵\left(🍫\right)\cdot time\left(inseconds\right)yourfriendspendsserving🍫\cdot {🍫}_2|2]$

$=n\cdot \left(E\right[💵\left(🍓\right)|2]-E[\frac{{🍓}_2}{{🍫}_2}\cdot 💵\left(🍫\right)|2\left]\right)$$+E\left[\right(💵\left(🍫\right)\cdot time\left(inseconds\right)yourfriendspendsserving🍫\cdot {🍫}_2|2]$

$-E\left[\right(💵\left(🍫\right)\cdot time\left(inseconds\right)yourfriendspendsserving🍫\cdot {🍫}_2|2].$

Final formula

You should 🟥 if 

$n\cdot \left(E\right[💵\left(🍓\right)|1]-E[\frac{{🍓}_1}{{🍦}_1}\cdot 💵\left(🍦\right)|1\left]\right)+$$E\left[\right(💵\left(🍦\right)\cdot time\left(inseconds\right)youspendserving🍦\cdot {🍦}_1|1]$$-E\left[\right(💵\left(🍦\right)\cdot time\left(inseconds\right)youspendserving🍦\cdot {🍦}_1|2]$

^[13]>

$n\cdot \left(E\right[💵\left(🍓\right)|2]-E[\frac{{🍓}_2}{{🍫}_2}\cdot 💵\left(🍫\right)|2\left]\right)$$+E\left[\right(💵\left(🍫\right)\cdot time\left(inseconds\right)yourfriendspendsserving🍫\cdot {🍫}_2|2]$

$-E\left[\right(💵\left(🍫\right)\cdot time\left(inseconds\right)yourfriendspendsserving🍫\cdot {🍫}_2|2].$

Your friend should 🟥 if

$n\cdot \left(E\right[💵\left(🍓\right)|1]-E[\frac{{🍓}_1}{{🍦}_1}\cdot 💵\left(🍦\right)|1\left]\right)+$$E\left[\right(💵\left(🍦\right)\cdot time\left(inseconds\right)youspendserving🍦\cdot {🍦}_1|1]$$-E\left[\right(💵\left(🍦\right)\cdot time\left(inseconds\right)youspendserving🍦\cdot {🍦}_1|2]$

<

$n\cdot \left(E\right[💵\left(🍓\right)|2]-E[\frac{{🍓}_2}{{🍫}_2}\cdot 💵\left(🍫\right)|2\left]\right)$$+E\left[\right(💵\left(🍫\right)\cdot time\left(inseconds\right)yourfriendspendsserving🍫\cdot {🍫}_2|2]$

$-E\left[\right(💵\left(🍫\right)\cdot time\left(inseconds\right)yourfriendspendsserving🍫\cdot {🍫}_2|2].$

It doesn't matter who 🟥 if

$n\cdot \left(E\right[💵\left(🍓\right)|1]-E[\frac{{🍓}_1}{{🍦}_1}\cdot 💵\left(🍦\right)|1\left]\right)+$$E\left[\right(💵\left(🍦\right)\cdot time\left(inseconds\right)youspendserving🍦\cdot {🍦}_1|1]$$-E\left[\right(💵\left(🍦\right)\cdot time\left(inseconds\right)youspendserving🍦\cdot {🍦}_1|2]$

=

$n\cdot \left(E\right[💵\left(🍓\right)|2]-E[\frac{{🍓}_2}{{🍫}_2}\cdot 💵\left(🍫\right)|2\left]\right)$$+E\left[\right(💵\left(🍫\right)\cdot time\left(inseconds\right)yourfriendspendsserving🍫\cdot {🍫}_2|2]$

$-E\left[\right(💵\left(🍫\right)\cdot time\left(inseconds\right)yourfriendspendsserving🍫\cdot {🍫}_2|2].$

Real-life interpretation

🍦= what you do on a day-to-day basis.

${🍦}_1$=the rate at which you complete tasks.

💵(🍦) = the amount of good done per task you complete.

🍫= what [the person who could do🍓instead of you] (2) does on a day-to-day basis.

${🍫}_2$ = the rate at which (2) completes tasks.

💵(🍫) = the amount of good done per task (2) completes.

Serving🍓= 🟥= some task that you could do, but you're considering having (2) do it instead.

${🍓}_1$=the rate at which you would do 🍓.

${🍓}_2$=the rate at which (2) would do 🍓.

(💵(🍓)|1) = the amount of good done if you do 🍓.

(💵(🍓)|2) = the amount of good done if (2) does 🍓.

$n$ = the amount of 🍓tasks (assuming each 🍓task is the same)

$time\left(inseconds\right)youspendserving🍦$= time (in seconds) you spend working on 🍦AND^[14] being affected by who does🟥.

sidenotes

1. I just realized that the table doesn't say anything about how person (2) doing 🟥 affects ${🍦}_1$ or 💵(🍦), but, since the formulas used all have a "|1" or "|2", the formulas still account for this automatically. oops but not oops!
2. Similarly, there's no mention of the negative side effects of 🟥, but (💵(🍓)|1) and  (💵(🍓)|2) both account for this anyways.
3. If you're deciding between who should do what with more than 2 people, just do Whoever is better (whoever is better (person 1, person 2), person 3), or whoever is better (Whoever is better (whoever is better (person 1, person 2)), person 3), person 4), and so on.

Closing statement

If you have any questions, comments, suggestions, corrections, or feedback, please feel free to put them in the comments!

Also, please don't strong-downvote without telling me what I did wrong so I can fix it.

1. ^{^}

   This includes the expected amount of good generated by them being happy or sad or productive or whatever you value.

2. ^{^}

   

3. ^{^}

   In this context, $aORb$ is:

   True if $a$ is True.

   True if $b$ is True.

   False if $a$ and $b$ is False.

^{^}

Some reasons might be:

1.  If they have to meet a daily quota for people who ordered 🍓 served.

2. If the price of 🍓 is bigger than the price of  🍫and 🍦.

^{^}

The reason behind the claims on the right is the same as the reasons stated in the footnotes of the cell to the left. (i.e.,

Has reasoning behind its claims in the footnotes

You are here

)

^{^}

In this context, (B|A) = B, given A, and, more specifically, in this scenario, $E\left[💵\right(🍓\left)|1\right]=E\left[💵\right(🍓\left)\right],$ given that $n$ people ordered 🍓, and you (1) do 🟥.

^{^}

The reason we include a "|1" is because, in real life, instead of dealing with 💵(🍓), we often deal with things that vary depending on the number of beneficiaries 

(such as how [QALYs of (placing malaria nets in a village) per malaria net] changes depending on what % of the village has malaria nets. (e.g., if there's enough malaria nets to surround the outer parts of the village, (I think) less mosquitos will settle into the village than in a village whose nets only partly surround the village.)), and things that are dependent on who does them.

(such as how a statistician would be better at promoting EA than a worm.)

 

 

^{^}

The reason for the number $n\cdot E\left[\frac{{🍓}_1}{{🍦}_1}|1\right]$ is that it takes $n\cdot E\left[\frac{{🍓}_1}{theamountoftimeinasecond,inseconds}|1\right]$ seconds for you to serve $n$ people who are ordering 🍓(i.e., [Time lost doing 🟥]), and if we change the unit of time from seconds to {the counterfactual number of people who are served 🍦}, (i.e., ${🍦}_1$), we get 

 {the counterfactual number of people who are served 🍦} = $E[n\cdot \frac{{🍓}_1}{{🍦}_1}|1]$, and, in this case, since $n$ is a known, non-random variable, $E[n\cdot \frac{{🍓}_1}{{🍦}_1}|1]=n\cdot E\left[\frac{{🍓}_1}{{🍦}_1}|1\right].$

^{^}

Here, we're assuming that this is true to properly reflect how, in real life, if you don't do some task (in this case, serving people who order 🍦) during some time slot (Time you spend serving people who order 🍦), you can never do said task.

Another reason we're assuming this to be true is that, if we didn't, there would be no consequences to serving people who ordered 🍓 first, regardless of whether you serve them or if your friend serves them. (since, in both cases, the same amount of money would be made ($n\cdot 💵\left(🍓\right)+$the amount of money that comes from the large lines of people).)

^{^}

By "Time, in seconds, you spend serving  🍦", I mean "Time, in seconds, you spend serving  🍦and are affected by who does 🟥". (the reason for this is because, if there's some time you spend not affected by  who does 🟥, then it doesn't need to be accounted for.)

^{^}

(Time, in seconds, you spend serving  🍦|1) = [Time, in seconds, you spend affected by who does 🟥] - [Time lost doing 🟥], and [Time lost doing 🟥] is established to be^[8] $E[n\cdot \frac{{🍓}_1}{{🍦}_1}|1]$, 

and (Time, in seconds, you spend serving  🍦|2) = Time, in seconds, you spend serving🍦and are affected by who does 🟥. 

^{^}

BEDMAS is an order of operations, and it stands for:

1. Brackets $\left(B\right)$

2. Expontentials${}^E$

3. Multiplication and Division $\frac{DM}{M}$

4. Addition and Subtraction $A+-S=A-S$

Synonyms include:

BODMAS, PEDMAS, PEMDAS, BIDMAS, POMDAS, and PODMAS, as well as others.

I left a link to an explanation of what it means here.

^{^}

Just to clarify, in real life, when there are multiple "🍓"s 

(tasks that you're considering having person 2 do), 

(A|🍓) would be (A|the🍓task that this equation refers to, AND some combination of the allocation of different 🍓s.), and is not to be confused with (A|the🍓task that this equation refers to, all else being equal).

^{^}

In this context, $aANDb$ is:

True if $a$ is True at the same time as $b$ is True.

False otherwise.

^{^}

Since, if $ab>ac$, and a is a negative real number, $b<c$, 

if $a>b$, then $b<a$, and

since we're assuming $n$ is a positive real number (which is a valid assumption, since only a non-negative real integer amount of people can order 🍓, and $n$ isn't equal to $0$ since, if it was, that would defeat the point of this example), $-n$ should be a negative number, so:

if $-nb>-nc$, then $b<c$, and

if $-nb<-nc$, then $b>c$.

Show all footnotes