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Parrondo's Paradox: How two ugly parents can make a beautiful baby

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Re: Parrondo's Paradox: How two ugly parents can make a beautiful baby

#61

Post by Dylan16807 »

roblev wrote: post #60
Dylan16807 wrote: post #52

Earlier quoted context omitted.

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.

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.

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.

Re: Parrondo's Paradox: How two ugly parents can make a beautiful baby

#62

Post by Evbn »

Dylan16807 wrote: post #57
legaultmarc wrote: post #26

Earlier quoted context omitted.

The author does say that the Parrondo's paradox only works if the games are not independent. It still is paradoxal that "A combination of losing strategies becomes a winning strategy".

It's only 'paradoxical' when the details of the games are sufficiently obscured. 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

That's what "paradox" is: when a simple model or explanation seems to show a contradiction, and a more sophisticated model is needed to understand the situation.

Re: Parrondo's Paradox: How two ugly parents can make a beautiful baby

#64

Post by SoftwareMaven »

blameless wrote: post #28

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()

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.

Re: Parrondo's Paradox: How two ugly parents can make a beautiful baby

#65

Post by pulplobster »

dclowd9901 wrote: post #34
adamc wrote: post #6

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.

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.

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.

Re: Parrondo's Paradox: How two ugly parents can make a beautiful baby

#66

Post by 3pt14159 »

tocomment wrote: post #15
GavinB wrote: post #13

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.

I was thinking the same thing. Is there a version that makes it a paradox?

No.

Re: Parrondo's Paradox: How two ugly parents can make a beautiful baby

#67

Post by 5partan »

lotharbot wrote: post #41
softbuilder wrote: post #38

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?

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. Playin…

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?

Re: Parrondo's Paradox: How two ugly parents can make a beautiful baby

#68

Post by Dylan16807 »

Evbn wrote: post #62
Dylan16807 wrote: post #57

Earlier quoted context omitted.

It's only 'paradoxical' when the details of the games are sufficiently obscured. 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

That's what "paradox" is: when a simple model or explanation seems to show a contradiction, and a more sophisticated model is needed to understand the situation.

But in this case the 'simple model' is less 'simple' and more 'misleading'. Something should be a paradox without trickery. Someone hiding invalid division to make 1=2 is not a paradox. "Set of sets that don't contain themselves" is a paradox.

Re: Parrondo's Paradox: How two ugly parents can make a beautiful baby

#69

Post by A1kmm »

GavinB wrote: post #13

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.

If they were independent, it would be impossible: E(X + Y) = E(X) + E(Y) if X and Y are independent. (where E(X) = E(x | X=x) is the expected value of a random variable). This is easily provable: E(x | X = x) = \integral_{-\inf}^{inf} x f_X(x) dx, and by integration by parts:

  \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
Since p.d.f.s f_X(x) integrate to 1 over their domain,

\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.

Re: Parrondo's Paradox: How two ugly parents can make a beautiful baby

#70

Post by dclowd9901 »

pulplobster wrote: post #65
dclowd9901 wrote: post #34

Earlier quoted context omitted.

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.

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…

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.
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