Parrondo's Paradox: How two ugly parents can make a beautiful baby
datagenetics.com
datagenetics.com
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.
If you word it as "a combination of losing strategies becomes a winning strategy" many people will be surprised and ask you to explain.
If you word it as "losing in A adds to the prize in B, so playing both beats the house" people aren't going to be impressed. note: used a simpler A/B mechanic than the blog post for illustration purposes
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.
Your actual balance is ridiculous to include in a game's calculations.
Game A: If you have an even number of chips, gain one. Otherwise, lose two.
Game B: If you have an odd number or chips, gain one. Otherwise, lose two.
The point being that game A (or B) always leaves you with an odd (or even) number of chips, so that if you keep playing it you lose two every turn, but if you alternate, you win one every turn (after the first, possibly). For simplicity, I'm ignoring the behavior near zero chips, but this still seems to capture the essential properties of the "paradox" in a much simpler fashion.
\integral_{-\inf}^{inf} \integral_{-\inf}^{inf} (x + y) f_X(x)f_Y(y) dx dy = \integral_{-\inf}^{inf} f_Y(x) \integral_{-\inf}^{inf} x f_X(x) dx dy + \integral_{-\inf}^{inf} f_Y(y) \integral_{-\inf}^{inf} x f_X(x) dy dx = \integral_{-\inf}^{inf} f_Y(y) dy \integral_{-\inf}^{inf} x f_X(x) dx + \integral_{-\inf}^{inf} f_X(x) dx \integral_{-\inf}^{inf} y f_Y(y) dy
\integral_{-\inf}^{inf} f_Y(y) dy \integral_{-\inf}^{inf} x f_X(x) dx + \integral_{-\inf}^{inf} f_X(x) dx \integral_{-\inf}^{inf} y f_Y(y) dy = \integral_{-\inf}^{inf} x f_X(x) dx + \integral_{-\inf}^{inf} y f_Y(y) dy = E(X) + E(Y)
Therefore, if two games are independent, the expected loss is the sum of the expected losses. For real-valued expected losses, it is not possible to add two real numbers of the same sign and get a real number of the opposite sign, and so the 'paradox' is therefore impossible for independent games.
But you are right, for non-independent games, it doesn't seem that surprising, so it doesn't really meet the definition of a paradox.
As that is something that can be manipulated then it isn't a major leap, I think, to the idea that combinations of the two games would produce different outcomes - and that some outcomes would be positive.
Whilst fun; I rather preferred the latter part of the post - it reminded me of a brilliant book I had as a kid with many such puzzles in it :)
What a crock. I think the Monty Hall Problem (http://en.wikipedia.org/wiki/Monty_Hall_problem) is far more vexxing and interesting.
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.
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!
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.
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.
>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.
"The solution is that, 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."
s/after/before/ to fix the explanation.
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.
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.
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.
Again, although I'm forced to concede your second paragraph that there might be some influence by what you've eaten yesterday, I'd like to believe that they're not that large. Because, if that were true everyone who'd eat (for instance) spicy things would constantly be in a state of irritation and anger.
People's temperament, moods, general thoughts might as well be a pre-seeded RNG.
// 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);
}The graph in the article shows the average state of each round over 1 million games.
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.
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.
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'.
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.
(I read the whole article til the end waiting to see what this had to do with genetics. It doesn't.)
The same odds as always choosing black or red.
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.
They had a spotter at each table that would count cards and wait until the odds were in the player's favor before calling in the big money player. Assuming that the bets placed by the spotter are negligible, the the big money player could choose from the following games:
A : Do nothing. E[x] = 0 (break-even)
BL: Play blackjack when the deck favors the casino, E[x] < 0
BW: Play blackjack when the deck favors the player, E[x] > 0