Support NeoGAF

[TheExodu5]

If both children were randomly selected from the world's population of children, then yes, the probability would be 2/3 that if one child is a boy, the other would be a girl.

If you already know that one is a boy and then randomly select the other from the global population, then you're back to 50/50. It's the same as asking what the probability would be that the second child is a girl (always 50/50 independant of the first child).

You sure? I would assume each combination (BB, BG, GB) would have slightly differing weights now in accordance to the ratio of G:B in the world, so the probabilities would be tweaked somewhat.

 

[Kazerei]

In honour of tomorrow's release, here's a little puzzle to get everyone warmed up:

"I was buying some presents for an old friend I hadn't seen in ages whom I was about to visit. I knew that my friend had 2 children, and I remembered that one of them was a boy, but I couldn't remember what the other one was. My good friend, Professor smartypants, who was with me offered some sage advice."

Did Professor smarty pants say:

1. The other child is most likely another boy
2. Chances are the other child is a girl
3. There's a 50/50 chane that the other child is either gender

I still think there's a 50/50 chance ... but I'm not sure. This is very similar to Bertrand's box problem but with four boxes. I think we can re-word this problem like so. I'm just going to rip off the Wikipedia article because I'm lazy.

There are four boxes:
a box containing two gold coins
a box with two silver coins
a box with one of each
a box with one of each

After choosing a box at random and withdrawing one coin at random, if that happens to be a gold coin, what is the probability that the remaining coin is gold?

There are two arguments.

Probability 1⁄3:
Originally, all four boxes were equally likely to be chosen.
The chosen box cannot be box SS
So it must be from box GG or GS or GS
The three remaining possibilities are equally likely, so the probability the box is GG is 1⁄3.

Probability 1⁄2:
Originally, all eight coins were equally likely to be chosen.
The chosen coin cannot be from drawer S of either box GS, or from either drawer of box SS.
So it must come from the G drawer of either box GS, or either drawer of box GG.
The four remaining possibilities are equally likely, so the probability the drawer is from box is GG is 1/2.

[To clarify, the four possibilities are drawer of box GG, other drawer of box GG, drawer of box GS, drawer of other box GS]

The problem can be shown to be equivalent to asking the question "What is the probability that I will pick a box with coins of the same colour?" This can be easily shown to be a probability of 1/2. When the first coin is shown to be gold, the probability that the second coin must be gold must also be 1/2, because the value of the first coin is independent of the second.

Bertrand's point in constructing this example, was that merely counting cases is not always proper. Instead, one should sum the probabilities that the cases would produce the observed result; and the two methods are equivalent if this probability is either 1 or 0 in every case. This condition is true in the second solution method, but not in the first. The correct way to solve the problem by treating the boxes as the individual cases is:

Originally, all three boxes were equally likely to be chosen.
The probability the GG would produce a gold coin is 1.
The probability the SS would produce a gold coin is 0.
The probability the GS would produce a gold coin is 1⁄2.
The probability the GS would produce a gold coin is 1/2.

So the probability the chosen box is GG is (P(GG) ⁄ [P(GG)+P(SS)+P(GS)+P(GS)]) = (1 ⁄ [1+0+1/2+1/2]) = (1⁄2).

Alternatively, the chosen box has two coins of the same type 1⁄2 of the time. So regardless of what kind of coin is in the chosen drawer, the box has two coins of that type 1⁄2 of the time.

 

[Kazerei]

Note: If you haven't heard of Bertrand's box problem, you should probably read that Wikipedia article first. It is a slightly simpler than this problem (three boxes instead of four).

 

[marjo]

You sure? I would assume each combination (BB, BG, GB) would have slightly differing weights now in accordance to the ratio of G:B in the world, so the probabilities would be tweaked somewhat.

I was making the simplifying assumption that randomly picking one child (or giving birth for that matter) would result in a 50% chance that you would get a child of either sex.

In truth, slightly more boys are born then girls.

 

[TheExodu5]

Kazerei the important thing here is that since you know he picked a box with a gold coin, the probability of him picking a box with two identical coins becomes 1/3, because you remove SS from the list of options.

 

[marjo]

I just called my local gamestop and they've got it in! I'm in Kitchener/Waterloo.

 

[Kazerei]

Kazerei the important thing here is that since you know he picked a box with a gold coin, the probability of him picking a box with two identical coins becomes 1/3, because you remove SS from the list of options.

Removing SS from the list of options does NOT change the original probability of picking a box with two identical coins. That is the "trick" with these types of problems.

For example, the Monty Hall problem. You're given the choice of three doors: behind one is a car, behind the other two are goats. So you have a 2/3 chance of picking a goat. But then the host opens one of the other doors, revealing a goat, which eliminates one of the possibilities.

So now that we're down to two doors, it seems like you have a 1/2 chance of getting a goat. But that's not true, there's still a 2/3 chance your door has a goat. Which means there's a 2/3 chance the remaining unopened door has a car.

 

[Brandon F]

Already? Damn I still haven't purchased the last Layton game yet.

 

[TheExodu5]

Removing SS from the list of options does NOT change the original probability of picking a box with two identical coins. That is the "trick" with these types of problems.

For example, the Monty Hall problem. You're given the choice of three doors: behind one is a car, behind the other two are goats. So you have a 2/3 chance of picking a goat. But then the host opens one of the other doors, revealing a goat, which eliminates one of the possibilities.

So now that we're down to two doors, it seems like you have a 1/2 chance of getting a goat. But that's not true, there's still a 2/3 chance your door has a goat. Which means there's a 2/3 chance the remaining unopened door has a car.

That's the thing! Your probabilities change in the Monty Hall problem because an unknown is revealed. With the box problem, you know that there is a gold coin in the box he picked, so that's also an unknown that's revealed.

If he randomly picks a box and gets SS, the problem basically doesn't satisfy the initial condition. It's a throwaway case. In reality, SS is not even in the list of options at all, and therefore you're only left with 3 boxes: GG, GS, SG.

The important thing is you're picking after this unknown is revealed to you. If you were to pick before the unknown is revealed, then the probabilities don't change, you're right, and that's why your original pick in the Monty Hall problem is less likely to be correct...because it was original picked from a larger set.

 

[Tiktaalik]

Already? Damn I still haven't purchased the last Layton game yet.

According to the 1up review this is a prequel, so I suppose you could pick up this one out of order without playing the one before and not have any sort of storyline spoiler issues.

Speaking of that, is there any storyline spoiler issues in the Layton series in general? I've only ever played the first game.

 

[TheExodu5]

I would definitely pick up this game first. Take it from someone who played the first two games over 2 years time: you'll suffer from Layton fatigue. Might as well go with this one first, since it includes Layton's Life.

 

[sorijealut]

where is this one on the timeline?
also what's the mini-game for this one?

sho excited.

 

[Chairhome]

I just called my local gamestop and they've got it in! I'm in Kitchener/Waterloo.

it might be worth mentioning that i think GS has it for $35, BB has it for $40, and TRU has it for $30 (if they get it in...)

 

[Kazerei]

That's the thing! Your probabilities change in the Monty Hall problem because an unknown is revealed. With the box problem, you know that there is a gold coin in the box he picked, so that's also an unknown that's revealed.

If he randomly picks a box and gets SS, the problem basically doesn't satisfy the initial condition. It's a throwaway case. In reality, SS is not even in the list of options at all, and therefore you're only left with 3 boxes: GG, GS, SG.

The important thing is you're picking after this unknown is revealed to you. If you were to pick before the unknown is revealed, then the probabilities don't change, you're right, and that's why your original pick in the Monty Hall problem is less likely to be correct...because it was original picked from a larger set.

You're right, I just found the Wikipedia article on this boy/girl problem, heh. It's all in how you phrase the problem, and whether the fact that one is a boy is known before or after you pick.

From all families with two children, at least one of whom is a boy, a family is chosen at random. This would yield the answer of 1/3.
From all families with two children, one child is selected at random, and the sex of that child is specified. This would yield an answer of 1/2.

 

[whatsinaname]

Well done, this is absolutely correct.

If I had said the first is a boy, or the second is a boy, then the chance of the other being either a boy or a girl would have been 50/50. But saying that one is a boy makes the chance of the other being a girl 2/3.

It really is counter intuitive as most people with a basic understand of probabilities misapply the principals of independent events in this case.

For those of you that still have a hard time believing this, think of it this way. You are twice as likely to have one child of each gender then you are of having two boys (because there are twice as many combinations that produce that outcome).


Here's another way of putting it that gets people even more confused:

If I were to flip two identical pennies in secret, and show you that one of them was heads, there would be a 2/3 chance that the other was tails (for the same reasons as the problem above). However, if one the pennies had a mark on it, and you saw that mark when I revealed that it was heads, the chances of the other being tails is now 50%.

When presented with this, most people think it sounds ridiculous. How can the fact that there is a mark on one penny change the odds of what the other can be?

The answer of course, is that it can't. What has changed, in a subtle way, is the question. In the second example, the question is reduced to simply, "what was the result on the penny without the mark", which is clearly 50/50, completely independent of what happened to the one with the mark. In the first case, the question can be thought of "what are the chances he got 2 heads"?

In the second case, we have complete information about one of the coins, but none about the other, whereas in the first example we have some information about the combination of the coins.

Like I said, very unintuitive.

Sorry, I don't see how age enters into the picture in your question.

http://en.wikipedia.org/wiki/Boy_or_Girl_paradox

The question as you framed it is very ambiguous, and even the original poser of the question (from 1959) feels that way.

Within the question you framed, I see three possible choices {BB, BG = GB, GG}. As GG is ruled out, your P = 1/2.

 

[Boney]

I can't wait for the iOS game. Can never get enough of this series.

it's a shame cause I don't have an ios device, but the game looks rad

 

[Oni Link 666]

Did anyone pick it up today?

 

[GSR]

As far as I can tell today was the ship date, not the "available in stores" date. A few people have their hands on it, but not many.

 

[Ratrat]

where is this one on the timeline?
also what's the mini-game for this one?

sho excited.

It happens before the originally trilogy and is the story of how Layton first meets Luke.

 

[Boney]

It happens before the originally trilogy and is the story of how Layton first meets Luke.

how is that story divided in 3 games though jesus

 

[GSR]

Last Specter has him meet Luke and take him on as assistant, the movie and Mask of Miracle (and presumably the sixth one whenever that gets made) deal with the antagonist introduced in Last Specter as well as other standalone plots.

 

[Stumpokapow]

Sorry, I don't see how age enters into the picture in your question.

http://en.wikipedia.org/wiki/Boy_or_Girl_paradox

The question as you framed it is very ambiguous, and even the original poser of the question (from 1959) feels that way.

Within the question you framed, I see three possible choices {BB, BG = GB, GG}. As GG is ruled out, your P = 1/2.

BG isn't GB--are you saying that you being a boy and your sister being a girl is the same thing as you being a girl and your brother being a boy? Imagine that both children are hidden behind a curtain. One is standing to (your) left of the other.

If you know the left one is a boy, there is a 50% (+- accounting for birthrates) chance the right one is a girl. If you know that at least one child is a boy without knowing which one is, there is a 2/3 chance that the other is a girl because it could either be that they are both boys, the left is a boy and the right is a girl, or the right is a boy and the left is a girl, each possibility with equal probability.

Age is introduced just as I introduced position there, to stress that your knowledge that one is a boy refers to the knowledge that at least one is a boy, without knowing which is, or if both are.

 

[marjo]

Did anyone pick it up today?

I got it from the Gamestop in Conestoga mall in Waterloo, Ontario. It was $30.

I've played through the first 7 puzzles so far, and my first impressions are positive. I'm looking forward to playing some more tonight.

 

[TheExodu5]

BG isn't GB--are you saying that you being a boy and your sister being a girl is the same thing as you being a girl and your brother being a boy? Imagine that both children are hidden behind a curtain. One is standing to (your) left of the other.

If you know the left one is a boy, there is a 50% (+- accounting for birthrates) chance the right one is a girl. If you know that at least one child is a boy without knowing which one is, there is a 2/3 chance that the other is a girl because it could either be that they are both boys, the left is a boy and the right is a girl, or the right is a boy and the left is a girl, each possibility with equal probability.

Age is introduced just as I introduced position there, to stress that your knowledge that one is a boy refers to the knowledge that at least one is a boy, without knowing which is, or if both are.

It's not the GB/BG that's in question here, it's the GG.

Two ways to pose the question:

From all families with two children, at least one of whom is a boy, a family is chosen at random. This would yield the answer of 1/3.
From all families with two children, one child is selected at random, and the sex of that child is specified. This would yield an answer of 1/2.

In the first case, your sample set ignores GG. In the second case, your sample set includes GG. That's what effects the probabilities.

 

[marjo]

Okay, in the interest of not derailing this thread any further, I will attempt to make this my last post on this subject.

I read through the article on the Gardener paradox, and it took a while, but I think I finally understand what it's getting at.

To put the argument being made in the context of my initial question, if the gift shopper happened to know the sex of one of his friends kids for no particular reason (IE. the child known to him was determined randomly), then the fact that it was a boy and not a girl that he knew about meant that there was a higher likelihood of neither child being a girl. The chance of the other child being a boy then works out to 1/2.

If instead, I had said that the reason he knew his friend had a son was because he had met them at a father/son picnic (IE. the sex of the known child was not randomly determined), then the chance of the other child being a boy would be 1/3.

I'm not sure I completely agree with the premise of the paradox, but it certainly is interesting.

 

[TouchMyBox]

I just wanted to say that as soon as I loaded the OP, the banner took a long time to load and it looked like really good 8 bit pixel-art for a brief moment.

I wish I took a screencap.

 

[ConradCervantes]

Picked my copy up from Best Buy. For some reason they have it listed as $40, so I had them price match it with Target and used two RZ coupons to bring it down to $10. Haven't played it yet, attempting to slog through The Walking Dead premiere right now.

 

[TehOh]

Did anyone pick it up today?

Local Best Buy had it, but was charging an absurd $40.

Gonna just wait and go grab it from Toys R Us later this week (they didn't have it in yet), since they're having a b2g1 sale.

 

[Smellycat]

The game got a 9 from IGN. They said that it is the best Layton game yet. It also got an A- from 1UP.

I would definitely pick up this game first. Take it from someone who played the first two games over 2 years time: you'll suffer from Layton fatigue. Might as well go with this one first, since it includes Layton's Life.

Layton fatigue??? There is no such thing.

Every time I finish a Layton game I get sad knowing that Japan already 1 or 2 new Layton games already out, and I have to wait at least one year.

Also, I would definitely play the games in order of release because they keep adding more and more content with each release.

 

[TheExodu5]

Puzzle 4 - is there really any logical way of working this out or is it trial and error? I got it, but it was just tedious trial and error.

 

[SaintMadeOfPlaster]

I decided earlier today to buy this game, but I couldn't find it anywhere in town. It's a real bummer because this will be my first Layton game to play, and I was very excited to try.

 

[SteveWinwood]

Will buy tomorrow with the coupon from Kmart for batman. So excite.

 

[duckroll]

Yay! Layton 4 GET!!! This should be good.

 

[Evershade]

In honour of tomorrow's release, here's a little puzzle to get everyone warmed up:

"I was buying some presents for an old friend I hadn't seen in ages whom I was about to visit. I knew that my friend had 2 children, and I remembered that one of them was a boy, but I couldn't remember what the other one was. My good friend, Professor smartypants, who was with me offered some sage advice."

Did Professor smarty pants say:

1. The other child is most likely another boy
2. Chances are the other child is a girl
3. There's a 50/50 chane that the other child is either gender

3.
The possible outcomes could be:
bb
bg
gb
gg

as one child is a boy and the order of the children is not in question, there are only 2 possible choices:
bb
bg

Giving a 50:50 split as to whether its a boy or a girl.

 

[Skellig Gra]

3.
The possible outcomes could be:
bb
bg
gb
gg

as one child is a boy and the order of the children is not in question, there are only 2 possible choices:
bb
bg

Giving a 50:50 split as to whether its a boy or a girl.

The whole thing is a cop out (as are a lot of puzzles). The answer is what the creator decided it should be. Which makes both answers correct really.

 

[marjo]

I just unlocked the train mini game. While the first puzzle was kind of simple, I can see it becoming really interesting later on.

 

[Oni Link 666]

I just went and got it. I tried Best Buy and Target, but they didn't have it so I ended up getting it at GameStop. Last time I try to get a game at anywhere but GS at launch. I probably wasted more money on gas than I would have saved running around town.

 

[Boogiepop]

Amazon still hasn't shipped my copy.

 

[Skellig Gra]

Just picked it up! Super stoked, I have a long flight out to LA to see the Zelda symphony so this will come in real handy.

 

[Salsa]

Got it. Trying to play with my under-the-screen dust particle but IT HURTS ;_;

 

[ShaneB]

Holy crap, I had no idea this was out. Surprise release! Awesome.

 

[Like the hat?]

cancelled my still-not-shipped amazon order and picked it up at a local store.

london life is so adorable.

 

[Smellycat]

Where can I find the password for the hidden door in The Unwound Future?? It is usually in the white boxes in the hidden door section of the new game. But it is not there.

edit: WOW, I completely misunderstood what the screen was saying. The password is available in the Diabolical Box. haha

 

[watershed]

Just picked it up. Haven't played it yet but will do so soon.

 

[SecretMoblin]

A question for anybody in the know:

From the Layton's London Life section on the Wikipedia page (I know, I know):

Professor Layton's London Life (レイトン教授のロンドンライ&#12501 is a 100 hour RPG developed by Brownie Brown. The game features many of the Professor Layton Series characters, from all four games released as of 2009, and three downloadable characters from Professor Layton and the Eternal Diva.[13] The game allows the player to create his/her own avatar that they can have live in London. Specific Londoners will ask them to do certain tasks, unlocking more places in London to visit. The game has been described to have graphics very similar to Mother 3, and it has been said that London Life is a sneak peek for a new Level-5 game, Fantasy Life launching in Japan during 2011. Level-5 announced that the US localization will include London Life, and as a bonus will, as opposed to the Japanese version, be available before the completion of the main game. However, London Life will not be included in the European version. London Life has now also been confirmed for Australia.

I'm confused about the meaning of 'downloadable'; does this mean it has DQIX-style "DLC", where you connect to the internet and unlock content on the cart, or is this Pokémon-style stuff that requires attending certain events or stores? Either way, I'm not really a fan of being required to connect to the internet to get all the features of my DS game -- especially if it's already on the cart.

 

[duckroll]

A question for anybody in the know:

From the Layton's London Life section on the Wikipedia page (I know, I know):



I'm confused about the meaning of 'downloadable'; does this mean it has DQIX-style "DLC", where you connect to the internet and unlock content on the cart, or is this Pokémon-style stuff that requires attending certain events or stores? Either way, I'm not really a fan of being required to connect to the internet to get all the features of my DS game -- especially if it's already on the cart.

Have you actually played Layton's London Life yet? If you have, it's obvious what they're referring to. There are online aspects of the game which requires you to connect online to access certain social and trading features. When you do so, it also results in certain characters showing up in the town. It's just part of the game.

 

[SecretMoblin]

Have you actually played Layton's London Life yet? If you have, it's obvious what they're referring to. There are online aspects of the game which requires you to connect online to access certain social and trading features. When you do so, it also results in certain characters showing up in the town. It's just part of the game.

I haven't, no. Thanks for the info.

 

[duckroll]

I don't see anyone talking about it here, but my friend code for London Life is 1163-2773-6028. I wonder how many people on GAF will even end up playing this...

 

[Winnie the Pimp]

is London Life ACTUALLY worth it though?

I am considering importing just for this but if it's just a pointless little mini game in the end then there is no use and i could just get the euro version hmmm...

 

[slaughterking]

I will play it. My US copy just takes a while to find its way to Europe.