
Parrondo's Paradox: How two ugly parents can make a beautiful baby - pavanlimo
http://www.datagenetics.com/blog/august22012/index.html
======
GavinB
Isn't it a little sketchy to have one of the "games" be a game that takes into
account outside information like your balanace?

Here's another "paradox":

Game A: You lose a dollar every time.

Game B: If the last game you played was game A, you win a million dollars.
Otherwise, you lose a dollar.

AAAAAAA... loses, BBBBBBB... loses, but ABABABABA... makes you rich!

Suddenly it doesn't seem so paradoxical to me.

~~~
unreal37
Your balance isn't really outside information. In a game like Texas Holdem
Poker, your balance is a key factor in how much you bet.

But if we change the game so that "if you flip more than two heads in a row,
your chances of flipping a third head are only 10%; if you flip two tails in a
row, your chances of flipping a third tail are 90%", the result is the same,
the odds turn more negative over time than Game A and you should switch to A
after two consecutive flips.

~~~
Dylan16807
It's not your balance that matters there, it's the amount of money you
_choose_ to bring into the game at the start. If I could boost my odds by
bringing in only $498 dollars of my five hundred then I would do so every
time.

Your actual balance is ridiculous to include in a game's calculations.

~~~
roblev
In texas hold'em your actual balance is a very significant factor in your
optimal behaviour in a particular hand. Short-stack tactics are different to
big stack tactics.

~~~
Dylan16807
Okay, I didn't make myself clear enough. Let's say you have a thousand dollar
bill in your hand. You decide to convert $600 into chips because that's a
multiple of three, and then you put $400 into your room safe. Your _real_
balance is 1000, but your balance _for the game_ is 600. Your stack tactics
are based on the 600. Your chance of winning is based on the 600 you chose to
bring. Anything that counted the 400 in your room vault would be ridiculous.
Yet the game in the blog would.

------
viggity
I was expecting something more along the lines of actual genetics, in which
case two ugly parents make stunning children all the time. It is called hybrid
vigor. The overwhelming amount of our produce is bred this way. They will
inbreed corn like made so that it is 100% homozygous, but do it in five
different pools. The resulting offspring of any two inbreds from two different
pools is amazing. Inbred corn plants are about 4-5 feet tall and ugly as sin,
but their offspring are 8 feet tall and near perfect.

~~~
Scriptor
And if the parents aren't homozygous it could be attributed to recessive
genes. Same reason why two brown-haired parents can have a blond child.

------
adamc
It didn't seem like much of a paradox to me, since game B is really two
different games, and interleaving plays of game A changes the likelihood of
playing B1 vs. B2. Interesting, perhaps, but not terribly unintuitive.

~~~
dclowd9901
Right. When I got to the part about Game B, my first thought was, "Wait, this
isn't what you described in the story above..."

What a crock. I think the Monty Hall Problem
(<http://en.wikipedia.org/wiki/Monty_Hall_problem>) is far more vexxing and
interesting.

~~~
pulplobster
I think the Monty Hall Problem is also extremely misleading in its
formulation.

Take Wikipedia's formulation for example: Suppose you're on a game show, and
you're given the choice of three doors: Behind one door is a car; behind the
others, goats. You pick a door, say No. 1 [but the door is not opened], and
the host, who knows what's behind the doors, opens another door, say No. 3,
which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it
to your advantage to switch your choice?

What's not being explicitly said here is that the host knows what's behind the
doors (this part is said) but that he also always chooses the goat. With that
information, it's pretty clear that when the contestant first made the choice
of a door, the probability of getting the right one was 1/3. The probability
that the right door is among the other two is 2/3. NOW however, the host
removes the one of (or the only) wrong option among those two doors. The thing
to realize is that the host opening one of those doors does not give us ANY
new information that would change the distribution. Therefore, the other door
that the contestant didn't pick must have the probability 2/3.

~~~
dclowd9901
It's true that that's the crux of the Monty Hall problem, but it's also the
understood nature of it. You just don't realize how much impact that would
have on the outcome unless you actually start delving into the outcomes.

------
K2h
In the other examples at the bottom tucked in there is a great example of how
seemingly negative things can actually turn out to be a positive thing, this
example is talking about the board game chutes and ladders.

    
    
      I explained the seemingly paradoxical situation where it is 
      possible to add additional chutes to a game board and 
      reduce the average length of a game! How come? Well, if you 
      have a long ladder in the game (such that landing on one 
      will propel you far up the board for a strong advantage), 
      and you miss it; If the added additional chutes on the   
      board send you back to before this long ladder, then you 
      have a second attempt to hit the long ladder!
    

wow- I love this, it supports the notion to 'fail early' in business

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cperciva
As much as I like coin-flipping examples, I think the Brownian ratchet
(<http://en.wikipedia.org/wiki/Brownian_ratchet>) is a much clearer
explanation of this phenomenon.

------
brilee
Uh... Something's fishy with the first graph of the 'drunken man's walk'. It's
well known that the random walk veers away from the zero-line at a rate
sqrt(N), where N is number of flips. Normalized by the number of flips, the
random walk converges to the zero-line at the rate 1/sqrt(N). The graph does
neither, so I'm not quite sure how it was generated...

~~~
brilee
Oops, I misinterpreted how the graph was generated... but the complaint still
holds. The averaging of a million trials will cut the amplitude of the
fluctuations by a factor of 1000, but the sqrt(N) effect should still be
visible. (Note that if you multiply the Y-axis by sqrt(million) = 1000, then
you get a scale of about 20, which is approx sqrt(500))

~~~
yeahboats
I'm not very good at this stuff, but I don't see anything wrong with the
graph. It hovers around 0. A Brownian motion starting which starts at 0, will,
at time T be normally distributed with mean 0 and variance T.

<http://en.wikipedia.org/wiki/File:Random_Walk_example.svg>

Some of them go up, some go down, but all will cross 0 infinitely many times,
and all the paths averaged will equal 0.

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intellegacy
Can someone explain the uptown / downtown train story to me?

If the downtown train arrives and departs a few minutes after the uptown train
arrives and departs, respectively, why would he take the downtown train more?
It seems to me that if he missed both trains, then he would take the uptown
train, as it would arrive earlier.

~~~
brucehart
He is saying the opposite. The downtown train arrives on the hour and the
uptown train arrives at 5 minutes after the hour.

~~~
n09n
No, he does not say the opposite:

>even though the trains arrive with the same frequency, the downtown train
departs just a few minutes after the uptown train has departed. Because of
this, there is only a narrow time window in which the man is able to get on
the uptown train.

He _should_ say it like you suggested though.

~~~
brucehart
You're right. I was looking at the diagram. Clearly the author accidentally
swapped the two names in that sentence.

------
slug
Babies are made through love, love is beautiful, ergo babies are beautiful.
Q.E.D. (I read the whole post, it was an interesting read)

~~~
sigkill
Extending your discussion, love is nothing but chemicals firing in our brain.
Maybe human beings are evolved to trigger release chemicals in the brain when
certain conditions are met, which could possibly be deterministic? I agree
that each person is different, but we're all same in the same way. Think of
the attraction to the opposite sex as a seeded random number. Sure, we don't
know the seed but if you know a person well enough you're able to make fair
judgement and confidently claim that "X isn't Y's type".

I'd love to continue discussion on this philosophical point. "Love" is
afterall nothing but an abstract construct which has proven beneficial in
furthering the species when we were being chased by lions and bears.

~~~
andreasvc
Love is "just chemicals" in the same way that anger is "just chemicals". The
chemicals are the substrate, but to say that is not very informative.
Analogously you also wouldn't say computer software is "just electrons".
Beware of "nothing buttery"; reduction is a great thing, but only if we truly
understand the reduction (as in, heat is movement of molecules).

Re: determinism, I expect the whole process to be non-deterministic through
and through, such that say what you had for breakfast might as well have some
amount of influence on your social interactions.

~~~
Dove
Indeed. Chemicals are the implementation. That doesn't mean they are the
essence.

In fact, I think the design of a thing is closer to its essence than the
implementation. _Moonlight Sonata_ is neither just bits, nor just vinyl, nor
just vibrating strings. In any implementation, it is _Moonlight Sonata_. The
essence of it is the music, the information, Beethoven's intent and insight.

Love is the same way. Of course consciousness and will are implemented in
chemicals. They have to be implemented in _something_. That does not make them
less real.

------
blameless
I'm unable to reproduce the results. Balance is always negative. What's wrong
with my code?

    
    
      // for a graph
      var balances = [];
    
      // constants
      var winnings =  1, 
          losses   = -1, 
          epsilon  = 0.05;
    
      function play(probOfWinning) {
          return Math.random() < probOfWinning ? winnings : losses;
      }
    
      for (var experiment = 0; experiment < 100; experiment++) {
          var balance = 0;
          for (var flip = 0; flip < 1000000; flip++) {
              if (flip % 3 == 0) { // game A
                  balance += play(0.5 - epsilon);
              } else { // game B
                  balance += Math.round(balance) % 3 == 0 ? play(1/10 - epsilon)
                                                          : play(3/4  - epsilon)
              }
          }
          balances.push(balance);
      }

~~~
SoftwareMaven
The problem is you are using a rounding of the balance across the million
tries to decide on game B. The choice of which game B to use need to not be
dependent on the other experiments.

------
tborg
> (I think I’ve probably lost a good number of people in this post already, so
> if you’re still reading this, I assume you’re familiar with Eigenvectors).

Actually, no! I was kind of hoping that by reading this section, I might find
out about them. Why write a blog post section specifically targeted at those
readers for whom it holds no new information? The responsible thing to do
here, if you really didn't feel like explaining some key component of your
discussion, would have been to provide a reference to someone else's
explanation. Now it feels like you were wasting my time -- 'oh, hey kid,
wasn't talking to you!'

Also, as others above have mentioned, the generalized introduction (a way to
make 2 losing games into a winner!) does not lead intuitively into the
extremely case-specific exposition.

------
unreal37
Imagine if instead of Game B being dependent on your cash balance, let's say
Game B is blackjack and your chances of winning depend on the number of
10s/face cards that have already been played.

If the collective number of 10s/face cards already played is <= 25%, you have
a positive chance of winning and you play blackjack.

If the collective number of 10s/face cards already played is >= 35%, you have
a negative chance of winning. You switch to roulette until the balance of
10s/faces works back in your favor.

I think that could be a winning strategy when playing blackjack and roulette
together. LOL

Edited to add: I think the author's point is that there are games where you
can calculate your chances of winning "this hand" even though your chances
over time are negative, and you should avoid playing (switch to something
where your chances are better) when your chances are low. Which is a bit of an
obvious point for such a long article filled with graphs.

~~~
Dylan16807
_I think that could be a winning strategy when playing blackjack and roulette
together._

Or you can get even more money by s/roulette/taking a nap/. You're just
changing your bet size (including $0) in blackjack based on the current odds.
No fancy 'combining two losing strategies'.

------
hcarvalhoalves
"The key to understanding this paradox is that the two games are not
independent."

Well... can it really be called a paradox then? It's more of a classical
failure at defining the problem since the probability distribution of B
definitely depends on A - the description is just more convoluted, but it's
not a contradiction.

------
pastaking
I read the article and understood what he's saying - but how exactly do two
ugly parents make a beautiful baby?

~~~
unreal37
They don't. But two losing games (the parents) can be combined to make a
winning strategy (the beautiful baby).

(I read the whole article til the end waiting to see what this had to do with
genetics. It doesn't.)

------
lunchbox
What does the "mixed nuts" example at the bottom have to do with this
phenomenon?

------
guilloche
I am not sure whether the conclusion is right. But the explanation is lame
since the second game depends on the your cash status which could be affected
in a favored way to make the second game a winning one.

------
softbuilder
Struggling with this one. Help me out: If I play roulette and bet black, red,
red, black, red, red, etc., I'm going to win?

~~~
lotharbot
One way to think about it is that there are really three games: game A (a
slight loser), and games which I'll call BL (losing) and BW (winning). If you
win a couple rounds of BW, the casino changes you to playing BL for a round.
But if you don't play BL and instead go play A for a round, then when you come
back to the table you'll be back to game BW. So you use game A, a slight
loser, to avoid BL, a bad loser.

Playing only A is a slightly losing strategy. Playing a mix of BW and BL is a
losing strategy, because BL is so harsh. But if you play BW mixed with A, _you
combine big wins with small losses_ , and therefore come out ahead.

This doesn't work in roulette because, no matter what color you play, it's a
slight loser. Black and red are both examples of game A, so no matter how you
mix them you're just playing AAAAAAAA.

~~~
5partan
But you don't loose on two tables equaly, so what if you play on two tables A
and B, play ABBABB... and chose colors randomly?

