“Time flies, whether you’re wasting it or
not.”
― Crystal Woods, Write like no one is
reading
Prologue
Its truly unbelievable that I haven't been up
with blogging for almost 6 years now! *Phew!* Infact, its astounding that I had
almost forgotten that I even had a blog existing... It was virtually into its
dusky SchrÃ¶dinger cat's phase of being aliveanddead.. until
I received a note from blogger.com that there are over 300 comments that are
requiring my moderation(!). Alas!, a tiny chunk of the author in me died
right there; how did I not find a sense of internal deviance of not having
blogged for such a long time? I wondered, pondered and ruminated. For the
uninitiated, I felt that the aura of the IEDs (Idiotic Emission of Thoughts)
have died upon me.. but, as expected, the utter idioticity in
me sprung up and I decided to write some nonsensical stuffs backed by some
spurious math.
So, let's get the inkling
rolling...
Well, yes! while, 6 *long*
years seem eons ago, it does also mean that I have grown eonsover. Not to
mention that the packets of advice that are showered upon me have moved from
being "Choosing the path of right espouse in life" to "Choosing
the right spouse in life". Needless to say that, I have been overwhelmed
by people showering advices over the intricacies of how one's partner needs to
be chosen. So, in this "comeback" blog, I thought I should discuss
my IED, which I term  "The Partner Theory".
Let's start with some basics.. From a thoughts of
a birdbrained mind(like mine), finding the best partner seems
a very "apprentice"ing term, the whole concept of this ritual is
quite equivocal.. It’s much stronger than simple good platonic
friendship, but at the outset doesn’t seem to be anywhere near a romantic
relationship. It seems alleasy with a simple "matchthefollowing"
algorithm, but, at the same time, there is a deeper "Theory" in it
that lies unexplored for a sobel soul like mine. So, yesterday night, as I was
putting myself to sleep, I was trying my best to somehow mathematically slot
the pareto of this process; Urgh, yes. my thoughts around this
might not be perfect (So, let me place a pleasedonttryitathome caution)–but
lest assured, I'd say that my solution would be pretty good for such a really
difficult problem.
Ok, so for the sane sake of defining the problem
 let me first define my notion of "best partner problem",
essentially, the aim is to find the "person" whom you think would
meet your criteria for a "propose" action, but, that's not it.. the
action also requires a mandatory reaction phase of "reciprocation".
Simply put, your aim is to search for your “best partner” by dating
various people. Your only goal is to find the best person willing to “nod” for
your “ayes” and any thing less is a failure.
Hmm. So, let's move on and define the assumptions
and constraints for this problem
 Let C = {c_1,c_2,c_3,...c_k} be the set
of potential candidates (not a bad term at all, I think. Candidate(?) )
for the "best partner" tag for a person P
 It goes without saying that C must be a
type of nonconcurrent idempotence,wherein, P is allowed
to date only one candidate at a time (not a bad assumption, I suppose)
 For each pair (P, c_i), there is exactly two possible
outcome  {0,1}. 0 for rejection and 1 for selection.
 The outcomes are irreversible and nonrepetitive. So, if
there's a rejection (either way) the decision cannot be reversed
 The set C is countablyfinite. Which
means that you can date only a finite set of people during your lifetime
(quite an important clause, I think :) )
 For each pair (P,c_i) there exist a score function S(c_i
 P) = S_(i,P) which is a grade that the person P assigns to the candidate
c_i
 Goes without saying that the grading is relative, which
means, as you date people, you can only tell relative grades and not true
grade. This means you can tell the second person was better than the first
person, but you cannot judge whether the second person is your best
partner. After all, there are people you have not dated yet.
Now, let's see the plausible naive strategies
that can creepin.
1.
Early picking : Well, its definitely not a case of "early bird
getting the worm" See, if you pick someone too early, you are making a
decision without checking out your options. Sure, you might get lucky, but it’s
a big risk.
2.
Lazy stalling : Again, if you wait too long, you leave yourself with only a
few candidates to pick from. yes, surely a risky strategy.
So, as always. It boils down to a static two
player game [1] . The optimal strategy of a this would
be to lock yourself into the search for ordinately finite interval and
then hold the best match as (s)he comes along. But, there's a bigger trick
here! its quite ambiguous as to how many people should you
reject?
Well, with a simply math, it turns out to be
proportional to how many people you want to date, so let’s investigate
this issue.
To make this concrete, let’s look at an example
for someone that wants to date three candidates, so C = {c1,c2 and c3}. A
naive approach is to select the first relationship. What are the odds the first
person is the best? Yes! It is equally likely for the first candidate to
be the best, the second best, or the worst. This means by pure luck you have a
1/3 chance of finding best partner if you always pick the
first person. You also have a 1/3 chance if you always pick the last person, or
always pick the second.
Can you do better than pure luck?
Yes, you can.
Consider the following strategy: get to
know–but always reject–the first candidate. Then, select the next candidate
judged to be better than the first person.
How often does this strategy find the best
overall partner? It turns out it wins 50 percent of the time!
For the specifics, there are 6 possible dating
orders, and the strategy wins in three cases.
(The notation First date=c3, second date= c1 and
third date = c2 means you dated the worst candidate first, then the best, and
then the second best. I marked the candidate that the strategy would pick in
bold and indicated a win if the strategy picked the best candidate overall.)
First Date

Second Date

Third Date

Outcome

C1

C2

C3

Loss

C1

C3

C2

Loss

C2

C1

C3

Win

C2

C3

C1

Win

C3

C1

C2

Win

C3

C2

C1

Loss

You increase your odds by learning information
from the first candidate. Notice that in two of the cases that you win you do
not actually date all three candidates.
As you can see, it is important to date people to
learn information, but you do not want to get stuck with fewer options.
So do your odds increase if you date more people?
Like 5, or 10, or 100? Does the strategy change?
The answer is both interesting and surprising.
From the example, you can infer the best strategy is to reject some
number of people (k) and then select the next person judged better than the
first k people.
When you go through the math, the odds do not change
as you date more people. Although you might think meeting more people helps
you, there is also a lot of noise since it is actually harder to determine
which one is the best overall.
So, essentially, to cut the long story short.
Here's the summary of the best approaches
(i) Keep the candidates size as small as
possible. (greater than 3 preferably, but, be aware that the as the size
increases although the odds doesn't change, it does make rejectionselection
problem much difficult as you need to holdin a lot of prior information
(ii) If both of you are firsttimers. Its better
to talk a lot and decide on nothing. Perhaps, even if you really like the
person, its always good to let them know in true Schwarzenegger style
that.. "You'd be back" after assessing few other candidates (its
better this way for both the folks, you see)
(iii) If only one of you is a firsttimer. Its
better to go with the choice of the nonfirst timer (this shoots your chances
to over 60% when you do the calculations right)
(iv) Don't worry if you are an "early
victim"
(v) If you are experiencedbee, while the better
strategy for you seems to be to ensure that you keep the firsttimer to
nonfirst timer ration to as close to 2/3 as possible.
(vi) Mostly importantly, any one of you could reach out to me to seek more
details on the calculations.. :D