In the Seinfeld episode ‘The Kiss Hello’ (S06E17), Jerry’s Nana remembers that Uncle Leo was supposed to give Jerry’s mom 50 dollars 53 years before. When Morty (Jerry’s father) finds out, he starts to calculate the amount that Uncle Leo would owe his wife, Helen:

MORTY: Do you know what the interest on that fifty dollars comes to over fifty-three years?

HELEN: Oh, Morty, please.

MORTY: Six hundred and sixty-three dollars and forty-five cents. And that’s figuring conservatively at five percent interest, over fifty-three years, compounded quarterly. #

Using the 5% interest rate that Morty mentions, the best case scenario for the $50 over 53 years would be for continuous compounding:

Total = P * e ^ rt, where P=50, r=0.05 and t=53, so Total ~= $707

The value that Morty came up with – $663.45 is extremely close to what he would have received for a yearly compound:

Total = P * ( 1 + r/n) ^ nt, where P=50, r=0.05, t=53 and n=1, so Total ~= $663.75

However he does mention “compounded quarterly“, so the final expected amount should be:

Total = P * ( 1 + r/n) ^ nt, where P=50, r=0.05, t=53 and n=4, so Total ~= $696.17

In conclusion, Morty underestimated the amount that Helen should have received by $32.42. Quarterly compound is better than yearly compound, and the continuous compound is the best. Also, $696.17 in 1995 (when the episode aired) is worth $1,451.14 today.

Intuition about the e number

Imagine an initial investment of $1, earning 100% annual interest. So P=1, r=1, t=1.

• If interest is compounded once per year, you end up with: (1+1)^1 = 2
• If interest is compounded twice per year, you end up with (1 + 1/2) ^2 = 2.25
• Quarterly compounding: (1 + 1/4) ^ 4 = 2.44
• Monthly compounding: (1 + 1/12) ^ 12 = 2.61
• Weekly compounding: (1 + 1/52) ^ 52 = 2.69
• Continuous compounding: e = 2.718

Discrete vs continuous growth

At its core, e measures the idea of continuous growth, which is different from the discrete growth. Discrete growth means that a given population only grows at certain moments (example at the hour mark). Continuous growth is different: instead of waiting until the end of the hour, the population is growing all the time, at every moment, like a plant that grows gradually instead of in sudden jumps.

The discrete growth is modelled by the (1 + r)^t formula, while the continuous growth is modeled by the e^rt formula. The second one results in a slightly higher number because you’re adding a little bit of growth continuously.

Assuming a population of 1000 astrophage, and a growth rate of 5% per hour, here is how it will grow over a 10-hours period:

You have 8 teams. They will be drawn one against each other, so 4 pairs in total.

Question 1: how many distinct pair sets are possible?

105. I got to this number after running a large number of simulations. Then I did a little bit of research and I also found the formula:

Question 2: if 4 of the 8 teams are from England, what is the probability that all 4 of them will be drawn together?

Again, after analyzing the 105 distinct pair sets, I found that only 9 of them have all-English pairs. The full probability set is:

• two English pairs: 9/105 or 8.57%
• exactly one English pair: 72/105 or 68.57%
• no English pair: 24/105 or 22.86%

CL draws probabilities – follow up for the quarterfinals:
8.57% – 2 English QF
68.57% – exactly 1 English QF
22.86% – no English QF

cc @emi_nga @julienguyon1977 @iamsircolin @BassTunedToRed @joel_archie @paul_tomkins @DanKennett

— Dorin M. (@dorinmoise) March 14, 2019

A/B vs Multivariate Testing

A/B testing: two versions (A and B) are compared, which are identical except for one variation that might affect a user’s behavior. Total number of variations: 2. More
Multivariate testing: multiple variables are modified for testing a hypothesis. The goal of multivariate testing is to determine which combination of variations performs the best out of all of the possible combinations. [Total # of Variations] = [# of Variations on Element A] X [# of Variations on Element B] ... More

Permutations, Arrangements, Combinations

Given a set of n elements (ex – for n=3, the set is A, B, C)
Permutations: each ordered set of n elements P(n) = n!
In our example with n=3, P(3)=3!=6: АВС, АСВ, ВАС, ВСА, САВ, СВА
Arrangements: each ordered set of k elements A(n,k) = n! / (n-k)!
In our example with n=3, ordered pairs of 2, A(3,2)=3!/(3-2)!=6: AB, BA, AC, CA, BC, CB
Combinations: each unordered set of k elements C(n,k) = n! / k! (n-k)!
In our example with n=3, un-ordered pairs of 2, C(3,2)=3!/2!*1!=3: AB, AC, BC
And the relationship between P, A, C: C=A/P
Remember that for the permutations you don’t need a k! More

About learning

Learning isn’t done to you, it’s something you do. You need to take responsibility of your education. There will always be a new technology to learn, but this is not that important. Is the constant learning that counts.
Andy Hunt – Pragmatic Thinking and Learning

We all tend to learn best by doing and teaching. Active learning is a much more effective way to learn than any other way.
It seems a bit strange, but it should really be no surprise that play is a powerful mechanism for learning. […] This simple process that comes natural to us all, but somehow gets “taught” out of us, is the simplest and purest way to learn.
John Sonmez – Soft Skills

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