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  • Locked thread
504
Feb 2, 2016

by R. Guyovich
What if C A T really spells dog?

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Phyzzle
Jan 26, 2008
Bob has two children. The younger one is a girl. What are the odds that the older one is a boy?

About 50%

Bill has two children. One is a girl. What are the odds that the other one is a boy?

67%. Yes.

HorseLord
Aug 26, 2014
is god's dick so big even he can't suck it

Phyzzle
Jan 26, 2008
His dick extends forever in the dickwise direction, so a wind going that way provides spatially unlimited suction.hth.

GAINING WEIGHT...
Mar 26, 2007

See? Science proves the JewsMuslims are inferior and must be purged! I'm not a racist, honest!

Phyzzle posted:

Bob has two children. The younger one is a girl. What are the odds that the older one is a boy?

About 50%

Bill has two children. One is a girl. What are the odds that the other one is a boy?

67%. Yes.

Explain yourself

Phyzzle
Jan 26, 2008
There are 4 equally likely outcomes when you have 2 kids:
Boy Boy
Boy Girl
Girl Boy
Girl Girl

In Bob's case, we know
B G
G G
Are the only two possibilities.

In Bill's case,
B G
G B
G G
Are three equally likely possibilities.

computer parts
Nov 18, 2010

PLEASE CLAP
Sounds like a good time to introduce the Monty Hall problem.

GAINING WEIGHT...
Mar 26, 2007

See? Science proves the JewsMuslims are inferior and must be purged! I'm not a racist, honest!

Phyzzle posted:

There are 4 equally likely outcomes when you have 2 kids:
Boy Boy
Boy Girl
Girl Boy
Girl Girl

In Bob's case, we know
B G
G G
Are the only two possibilities.

In Bill's case,
B G
G B
G G
Are three equally likely possibilities.

I'm not sure about this. I think throwing in order of birth is just a red herring. B G and G B are the same "possibility" for Bill; each family can either have two girls or one of each. Fifty fifty.

Take it to the extreme; a family has 100 kids and 99 of them are girls, but we don't know the order. Does that make it a 99% chance the other is a boy? No. Each birth is a 50/50 probability.

computer parts posted:

Sounds like a good time to introduce the Monty Hall problem.

This isn't quite the same as Monty Hall, where the fact that there are three distinct options is beyond question, and the fact that we know there IS a goat SOMEwhere.

A Buttery Pastry
Sep 4, 2011

Delicious and Informative!
:3:

GAINING WEIGHT... posted:

I'm not sure about this. I think throwing in order of birth is just a red herring. B G and G B are the same "possibility" for Bill
BG and GB aren't the same outcome though. Yes, you have a both a girl and a boy in both cases, but that is not the sole piece of information counted in this problem. The difference between the two scenarios is that in one case you only get one piece of information (one child is a girl), while in the other you get two (one child is a girl, and it is the youngest). In either case, you start off with all the possible combinations, all of which are equally likely:

BB
BG
GB
GG

And then you eliminate the ones that don't fit the criteria. For Bob that results in:

BB - No girls
BG
GB - The boy is the younger sibling
GG

As there are only two possibilities left, and one has a boy, the odds the other child is a boy is 50%. For Bill on the other hand you get:

BB - No girls
BG
GB

GG

The two out of three equally likely possibilities means there's a 67% chance that the other child is a boy.

e: Of course none of this is actually true in real life, as any given child is more likely to be a boy than a girl.

A Buttery Pastry fucked around with this message at 17:53 on Sep 3, 2016

Phyzzle
Jan 26, 2008

GAINING WEIGHT... posted:


Take it to the extreme; a family has 100 kids and 99 of them are girls, but we don't know the order. Does that make it a 99% chance the other is a boy? No.

Yes.

What are the odds that the last coin flip is heads? What are the odds that one of the hundred coin flips is heads?

PoizenJam
Dec 2, 2006

Damn!!!
It's PoizenJam!!!

GAINING WEIGHT... posted:

I'm not sure about this. I think throwing in order of birth is just a red herring. B G and G B are the same "possibility" for Bill; each family can either have two girls or one of each. Fifty fifty.

No, he's absolutely right about 67%. Trust me when I say statistics are often incredibly unintuitive- but you really need to take a course on Bayesian inference to understand the key difference between Bob and Bill. Prior information is key- which is why this is superficially similar to Monty Hall.

Other weirdly unintuitive applications of Bayes theorem reveals:

-Getting a positive result on a mammogram only results in a 20% chance of actually having cancer (as it is much more likely to be a false positive when the population rate is taken into account)

-DNA evidence is questionable without prior suspicion; given only DNA evidence tying someone to a crime, is far more probable that person belongs to the population of false positives than true positives.

-by extension, this is also why blanket testing policies for whole populations (testing everyone for allergies; screening welfare recipients for drug use) is garbage- specificity of tests are never perfect, and without the benefit of prior knowledge (suspicion of allergy in first case, drug use in the latter) you'll have more false positives than true.

A Buttery Pastry posted:


e: Of course none of this is actually true in real life, as any given child is more likely to be a boy than a girl.

It's still true, and the posterior probability of a BG pair is still higher, it's just not as exact or as clean mathematically when you move away from 50/50.

GAINING WEIGHT...
Mar 26, 2007

See? Science proves the JewsMuslims are inferior and must be purged! I'm not a racist, honest!

Phyzzle posted:

Yes.

What are the odds that the last coin flip is heads? What are the odds that one of the hundred coin flips is heads?

That second question isn't quite the same though. Asking: "if I flip a coin 100 times, what is the probability that any is heads?" is not the same as "if I flip a coin 100 times and 99, in any order, are tails, what is the probability that the other one is heads?" In the second instance, for 99 of the flips, the probability of heads is now 0, which changes the overall total a bit.

JVNO posted:

No, he's absolutely right about 67%. Trust me when I say statistics are often incredibly unintuitive- but you really need to take a course on Bayesian inference to understand the key difference between Bob and Bill. Prior information is key- which is why this is superficially similar to Monty Hall.

I'll admit to being ignorant of the nuances of Bayes' Theorem, but I'm not just intuiting my answer here, I'm at least attempting to run the numbers myself. The way it's being phrased is giving us a false impression of what our possibilities are. Obviously the probability of one child being a boy is (roughly) 50/50. Adding the second child gives us the four combinations. That much I can grant. But in both cases, we know we can eliminate one child from consideration, because we're told the gender. We don't have to consider that one anymore when assessing the probability of the other. That's why I asked the 100 children question; the probability of the other 99 being boys is 0, because we are told they are girls. It's functionally equivalent to "I have 100 children, but let's ignore 99 for a moment - consider just this one child, what is the probability they are a boy?"

Another way to think about it: we have the four combinations - BB, BG, GB, GG - each with a 25% probability, and we have information we can use to eliminate some of our choices. In the first, this is easy, because we are told something about birth order, so we take away BB and GB, leaving us with the other two. In the second, we aren't told order, so it feels like GB and BG are both still on the table, giving us 3 equally likely outcomes. This isn't the case, though; one of those two options is impossible, because there is definitely a girl, we're just not sure where. If the girl is the older sibling, the possibility that the older sibling is a boy is 0, even if we don't know that. True, we can't know which choice between BG and GB is impossible, but knowing that one of them is impossible is enough. We can know that BB and [one of (BG or GB)] are eliminated, giving us 2 options.

It kinda reminds me of Sudoku, where you know the 4 is in one of like three boxes, and you don't know which, but you can still use that knowledge to eliminate the possibility of a 4 being in some other boxes that are still blank.

To again draw the comparison to Monty Hall, this is like saying there's two doors, and both can have either a goat or a car. So there might be two cars, might be two goats, might be one of each. Monty then opens a door to reveal what's behind it, and asks you the odds of the other door being a car. If we specified that he opened Door #1, rather than "either door", that would not affect the 50/50 chance of the car being behind the door that's been left unopened.

The feeling that there's 3 equally likely outcomes in the latter case seems more of a trick of language rather than the quirks of statistical mathematics. Still, this isn't a hill I'm willing to die on; I can certainly be talked out of this.

PoizenJam
Dec 2, 2006

Damn!!!
It's PoizenJam!!!
It's not really a trick of language though. The big key to Monty Hall is he will never, under any circumstance, open the door with the car behind it. Since we know the prior chances (1 car, 2 goats), there are just two possibilities. If you pick a car, he picks a goat randomly; If you pick a goat, he reveals the other goat. This data must be considered: Switching from a goat door to another goat door is not one of the possible outcomes of the second choice. Only switching from car to goat or from goat to car.

This data has an interesting consequence: Choosing to switch is essentially betting on the chance that you picked wrong on the first choice. That happens to be 66%.

PoizenJam fucked around with this message at 03:55 on Sep 4, 2016

Phyzzle
Jan 26, 2008

GAINING WEIGHT... posted:

That second question isn't quite the same though. Asking: "if I flip a coin 100 times, what is the probability that any is heads?" is not the same as "if I flip a coin 100 times and 99, in any order, are tails, what is the probability that the other one is heads?" In the second instance, for 99 of the flips, the probability of heads is now 0, which changes the overall total a bit.

It is the same. For 99 of the flips, the probability of heads is now 0, which does not change the probability of the other flip.

"If I flip a coin 100 times, and 99, in any order, are tails, what is the probability that #1 is heads Or #2 is heads Or #3 is heads Or #4 is heads Or #5 is heads Or . . . "

A Buttery Pastry
Sep 4, 2011

Delicious and Informative!
:3:

GAINING WEIGHT... posted:

Another way to think about it: we have the four combinations - BB, BG, GB, GG - each with a 25% probability, and we have information we can use to eliminate some of our choices. In the first, this is easy, because we are told something about birth order, so we take away BB and GB, leaving us with the other two. In the second, we aren't told order, so it feels like GB and BG are both still on the table, giving us 3 equally likely outcomes. This isn't the case, though; one of those two options is impossible, because there is definitely a girl, we're just not sure where. If the girl is the older sibling, the possibility that the older sibling is a boy is 0, even if we don't know that. True, we can't know which choice between BG and GB is impossible, but knowing that one of them is impossible is enough. We can know that BB and [one of (BG or GB)] are eliminated, giving us 2 options.
That's not how it works. You can't eliminate an option because it's incompatible with another option, because they're all incompatible with each other. It can be either BG, GB, or GG, all of which are mutually exclusive, not just BG and GB, leaving you with only a single (undetermined) combination. Which leads us back to the 1/3 chance for any of them being the right one.

Phyzzle posted:

It is the same. For 99 of the flips, the probability of heads is now 0, which does not change the probability of the other flip.

"If I flip a coin 100 times, and 99, in any order, are tails, what is the probability that #1 is heads Or #2 is heads Or #3 is heads Or #4 is heads Or #5 is heads Or . . . "
I think you're both partially right, but that's because you're arguing slightly different problems. The ones actually being discussed seems to be:

The final flip is heads (Bob scenario):
50%

Any combination with only one heads (Fertile Bill scenario):
HTT...T
THT...T
TTH...T and so on until
TTT...H

Plus a final "no heads combination"
TTT...T

Leaving you with 100 options where you have exactly one heads, and one option with no heads, resulting in a 100/101 chance of there having been 1 heads if 99 of a 100 are tails, or ever so slightly above 99% chance.

Any combination with at least one heads (New scenario):
1-(1/2)^100 = basically 100% chance

GAINING WEIGHT...
Mar 26, 2007

See? Science proves the JewsMuslims are inferior and must be purged! I'm not a racist, honest!

A Buttery Pastry posted:

That's not how it works. You can't eliminate an option because it's incompatible with another option, because they're all incompatible with each other. It can be either BG, GB, or GG, all of which are mutually exclusive, not just BG and GB, leaving you with only a single (undetermined) combination. Which leads us back to the 1/3 chance for any of them being the right one.

I think the main disagreement seems to be between whether our scenario is this:

[1] Three equal choices:
GG
-or-
GB
-or-
BG

or this:

[2] Two equal choices:
GG
-or-
[one of:(GB or BG)]

The reason I think it's 2 and not 1 is this: you're right that GB and BG are equally likely to each other, but they aren't equally likely to GG. It's like a nested 50/50 chance within another 50/50 chance. That's why I was going the "collapse" route. We don't know which is possible, but we know if one is possible (that the girl we're told about is the younger sibling and her older sibling is either a boy or a girl) then the other is impossible (that the girl we're told about it the older sibling).

We could take the opposite route, and instead of collapsing the one-of-each possibility, expand the both-girls possibility. We're told that one sibling is a girl, but not which one. It could be the older or the younger. This actually gives us two outcomes where both siblings are girls: the girl we were told about is the older one, who has a younger sister, -or- the girl we were told about is the younger sibling, and she has an older sister. Which we could show like this:

GG -or- GG

With the underlined G being the sibling we were told about. So there's actually four possibilities, given the knowledge that there are two children, one of whom is a girl:

GG
GG
GB
BG

Thus the probability of us having a boy somewhere in the mix is 2/4, or 50%. That's why I'm saying it's a trick of language: we have GG as one possibility because we see no difference in GG or GG because they are both represented by the word "girl". It's actually two possibilities hidden in one, masked by the word we use to refer to their gender.

quote:

Any combination with only one heads (Fertile Bill scenario):
HTT...T
THT...T
TTH...T and so on until
TTT...H

Plus a final "no heads combination"
TTT...T

Leaving you with 100 options where you have exactly one heads, and one option with no heads, resulting in a 100/101 chance of there having been 1 heads if 99 of a 100 are tails, or ever so slightly above 99% chance.

Phyzzle posted:

It is the same. For 99 of the flips, the probability of heads is now 0, which does not change the probability of the other flip.

"If I flip a coin 100 times, and 99, in any order, are tails, what is the probability that #1 is heads Or #2 is heads Or #3 is heads Or #4 is heads Or #5 is heads Or . . . "

To put a finer point on my above reasoning with the 100 coins example, your treating of "no heads" as one option is mistaken. It's actually 100 options: the coin we didn't know about is the first, and it was tails; the coin we didn't know about was the second, and it was tails; etc. There's 100 cases where the "other coin" was tails, and 100 cases where it was heads, giving us a 100/200 overall chance of heads and a 100/200 overall chance of tails.

Phyzzle
Jan 26, 2008

Phyzzle posted:

"If I flip a coin 100 times, and 99, in any order, are tails, what is the probability that #1 is heads Or #2 is heads Or #3 is heads Or #4 is heads Or #5 is heads Or . . . "

GAINING WEIGHT... posted:

To put a finer point on my above reasoning with the 100 coins example, your treating of "no heads" as one option is mistaken. It's actually 100 options: the coin we didn't know about is the first, and it was tails; the coin we didn't know about was the second, and it was tails; etc.

See the 'Or' in the statement about one heads result? You're statement about no heads has semicolons, but they don't represent 'Or', do they? They represent 'And'. A bunch of 'And's strung together to make one unlikely case, instead of a bunch of cases connected by 'Or's to make one much more likely case.

Buried alive
Jun 8, 2009

GAINING WEIGHT... posted:

...
Another way to think about it: we have the four combinations - BB, BG, GB, GG - each with a 25% probability, and we have information we can use to eliminate some of our choices. In the first, this is easy, because we are told something about birth order, so we take away BB and GB, leaving us with the other two. In the second, we aren't told order, so it feels like GB and BG are both still on the table, giving us 3 equally likely outcomes. This isn't the case, though; one of those two options is impossible, because there is definitely a girl, we're just not sure where. If the girl is the older sibling, the possibility that the older sibling is a boy is 0, even if we don't know that. True, we can't know which choice between BG and GB is impossible, but knowing that one of them is impossible is enough. We can know that BB and [one of (BG or GB)] are eliminated, giving us 2 options.
...

GAINING WEIGHT... posted:

...
[1] Three equal choices:
GG
-or-
GB
-or-
BG

or this:

[2] Two equal choices:
GG
-or-
[one of:(GB or BG)]

You're being inconsistent. Yes, if BG is the case then GB is not the case. However, we don't even know if Bill has a boy at all. If GG is the case, then neither BG nor GB are the case. Similarly, if it turns out that BG is the case, then neither GB nor GG are the case. Two of the three options are impossible, depending on what the result is. Which results in this:
[one of:(GG or GB or BG)]. Which is just scenario 1, phrased differently.

GAINING WEIGHT... posted:

...
With the underlined G being the sibling we were told about. So there's actually four possibilities, given the knowledge that there are two children, one of whom is a girl:

GG
GG
GB
BG

Thus the probability of us having a boy somewhere in the mix is 2/4, or 50%. That's why I'm saying it's a trick of language: we have GG as one possibility because we see no difference in GG or GG because they are both represented by the word "girl". It's actually two possibilities hidden in one, masked by the word we use to refer to their gender.

Which sibling we're told about doesn't matter. The thing we're being asked to consider is which set of circumstances is true? The 1st being a girl and the 2nd being a girl is the same set of circumstances regardless of which sibling we're told about. The 1st being a boy and the 2nd being a boy is a different set of circumstances due to which sibling it actually is, not which one we're told about.

Buried alive fucked around with this message at 16:01 on Sep 4, 2016

A Buttery Pastry
Sep 4, 2011

Delicious and Informative!
:3:

GAINING WEIGHT... posted:

[2] Two equal choices:
GG
-or-
[one of:(GB or BG)]
Would you say the chance of getting two sixes when rolling two dice, is the same as getting a six and a one?

GAINING WEIGHT... posted:

To put a finer point on my above reasoning with the 100 coins example, your treating of "no heads" as one option is mistaken. It's actually 100 options: the coin we didn't know about is the first, and it was tails; the coin we didn't know about was the second, and it was tails; etc. There's 100 cases where the "other coin" was tails, and 100 cases where it was heads, giving us a 100/200 overall chance of heads and a 100/200 overall chance of tails.
No, there is only a single combination of flips that make 100 out of 100 coins become tails, this one:

TTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTT

While the scenario where there is a single heads can have that head be at any point in the chain, resulting in 100 combinations.

Buried alive posted:

You're being inconsistent. Yes, if BG is the case then GB is not the case. However, we don't even know if Bill has a boy at all. If GG is the case, then neither BG nor GB are the case. Similarly, if it turns out that BG is the case, then neither GB nor GG are the case. Two of the three options are impossible, depending on what the result is. Which results in this:
[one of:(GG or GB or BG)]. Which is just scenario 1, phrased differently.
Also the first part of my previous post, phrased differently. :v:

Buried alive
Jun 8, 2009
Edit: I'm a dumb.

Inferior Third Season
Jan 15, 2005

Things not being intuitive to non-mathematicians don't make them paradoxes, everyone.

PoizenJam
Dec 2, 2006

Damn!!!
It's PoizenJam!!!
I dunno- statistical inference is pretty unintuitive, and at least has the appearance of looking paradoxical.

If we were only discussing true paradoxes... Well, they're wouldn't be much to discuss. They're unsolvable by definition.

GyroNinja
Nov 7, 2012
I just realized that Gaining Weight's argument would be correct if Bill were revealing the gender of one of his children at random, boy or girl, and not just telling you that one of his children is a girl. (This same distinction is actually also important for the Monty Hall problem).

GAINING WEIGHT...
Mar 26, 2007

See? Science proves the JewsMuslims are inferior and must be purged! I'm not a racist, honest!

Phyzzle posted:

See the 'Or' in the statement about one heads result? You're statement about no heads has semicolons, but they don't represent 'Or', do they? They represent 'And'. A bunch of 'And's strung together to make one unlikely case, instead of a bunch of cases connected by 'Or's to make one much more likely case.

Of course they mean or?

Buried alive posted:

You're being inconsistent. Yes, if BG is the case then GB is not the case. However, we don't even know if Bill has a boy at all. If GG is the case, then neither BG nor GB are the case. Similarly, if it turns out that BG is the case, then neither GB nor GG are the case. Two of the three options are impossible, depending on what the result is. Which results in this:
[one of:(GG or GB or BG)]. Which is just scenario 1, phrased differently.

I am fairly sure you're the one being inconsistent. We can talk about two possibilities ("contracted" view as above), which are: either both are girls, or there is one of each; or we can talk about four possibilities ("expanded" view), where order matters and GG is two of the possibilities, but we can't talk about 3 outcomes, because then you're expanding/contracting one and not the other.

This is easier to see if neither of the two options for the unknown child match the first. So let's add in a third gender and phrase the problem like this: there are two kids, one is a girl, the other is either a boy or a flibbertyjib. Then it's easy to see that there are four possibilities (taking birth order into account) :

GB
GF
BG
FG

Half of which are B, giving us the 50%

quote:

Which sibling we're told about doesn't matter. The thing we're being asked to consider is which set of circumstances is true? The 1st being a girl and the 2nd being a girl is the same set of circumstances regardless of which sibling we're told about. The 1st being a boy and the 2nd being a boy is a different set of circumstances due to which sibling it actually is, not which one we're told about.

Right, this is the inconsistency I was talking about. Either we're considering birth order or we're not, we can't do it for the second and not the first.

A Buttery Pastry posted:

Would you say the chance of getting two sixes when rolling two dice, is the same as getting a six and a one?

Yes. The second case seems more likely because you have 2 options per die, but you have to subtract out the cases where you get 2 sixes or 2 ones, which is not the outcome you're looking for. Both are 1/36.

quote:

No, there is only a single combination of flips that make 100 out of 100 coins become tails, this one:

TTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTT

This is the trick of language thing. There are 100 cases where a single coin is unknown, then it turns out to be tails. The fact that it matches the state of the 99 coins is confusing, and makes it seem like it's only 1 option. See the "third gender" explanation above and apply it thusly.

GyroNinja
Nov 7, 2012

GAINING WEIGHT... posted:

Yes. The second case seems more likely because you have 2 options per die, but you have to subtract out the cases where you get 2 sixes or 2 ones, which is not the outcome you're looking for. Both are 1/36.

No, the probability of getting a one and a six (in any order) is 2/36, not 1/36. Look, here's the list of (non-repeating) dice combinations;

1 1
1 2
1 3
1 4
1 5
1 6
2 2
2 3
2 4
2 5
2 6
3 3
3 4
3 5
3 6
4 4
4 5
4 6
5 5
6 6

Notice how there's only 21 possibilites? If all of them had a probability of 1/36, then what do you think happens the other 15/36ths of the time you roll the dice? Instead, the six doubles are only 1/36, while the fifteen non-doubles are 2/36, for a total probability of 36/36, which is what you'd expect.

Let me ask a (slightly) different question. Do you think that rolling a 5 and a 6 is as likely as rolling two 6s? And if so, do you realize that you're saying that the odds of getting a total roll of 11 are the same as a total roll of 12, which is not how observable reality works?

Phyzzle
Jan 26, 2008

GAINING WEIGHT... posted:

Of course they mean or?

Oh! Well taking another look, the problem must be this:

quote:

To put a finer point on my above reasoning with the 100 coins example, your treating of "no heads" as one option is mistaken. It's actually 100 options: the coin we didn't know about is the first, and it was tails; the coin we didn't know about was the second, and it was tails; etc.

So the "it" in the sentence refers to coin we don't know about? Then you're assuming that probability of it being tails is 1/2, so that the sum is 100/200 for all coins. But assuming the 1/2 begs the question.

botany
Apr 27, 2013

by Lowtax

GAINING WEIGHT... posted:

Yes. The second case seems more likely because you have 2 options per die, but you have to subtract out the cases where you get 2 sixes or 2 ones, which is not the outcome you're looking for. Both are 1/36.

answer A: can we please play poker some time.

answer B: https://www.geogebra.org/m/UsoH4eNl

GAINING WEIGHT...
Mar 26, 2007

See? Science proves the JewsMuslims are inferior and must be purged! I'm not a racist, honest!

Phyzzle posted:

So the "it" in the sentence refers to coin we don't know about? Then you're assuming that probability of it being tails is 1/2, so that the sum is 100/200 for all coins. But assuming the 1/2 begs the question.

I'm sorry, I don't follow. Why would the probability of a coin being tails being 1/2 be begging the question? We're assuming fair coin right?

botany posted:

answer A: can we please play poker some time.

answer B: https://www.geogebra.org/m/UsoH4eNl

I'll go ahead and concede the dice thing, but it's not really the same question anyway. I'm really only still persisting on the boy/girl problem, and I think the "third gender" explanation is my best articulation of why I don't find the 1/3 answer convincing.

GyroNinja
Nov 7, 2012

GAINING WEIGHT... posted:

Of course they mean or?


I am fairly sure you're the one being inconsistent. We can talk about two possibilities ("contracted" view as above), which are: either both are girls, or there is one of each; or we can talk about four possibilities ("expanded" view), where order matters and GG is two of the possibilities, but we can't talk about 3 outcomes, because then you're expanding/contracting one and not the other.

This is easier to see if neither of the two options for the unknown child match the first. So let's add in a third gender and phrase the problem like this: there are two kids, one is a girl, the other is either a boy or a flibbertyjib. Then it's easy to see that there are four possibilities (taking birth order into account) :

GB
GF
BG
FG

Half of which are B, giving us the 50%

I think I see your problem. You're visualizing it as "Well, I know that one of the siblings is a girl, which means that either I know that the first child is a girl, or that the second child is a girl, and therefore there's only one child who's gender I don't know, so the probability of the unknown child being a boy is 50%", but in reality, you don't know the gender of EITHER child right now. That is to say, the first child could still be a girl or a boy, and the second child could also be a boy or a girl, so it's wrong to act as if one of the child's genders is already "known". They're both unknown.

Let me try a different way of looking at it. Let's not actually talk about whether his "other child" is a son, because I think that's confusing language that makes it sound like we know more than we do. Instead, what we're asking is, "Given that Bill has at least one daughter, what is the possibility he also has one son?" Instead of talking about BG or GB, let's just look at the number of daughters Bill has. With two children, he either has no daughters (1/4th of the time), one daughter (1/2), or two daughters (1/4th again). Notice that with no prior information, there is a 75% chance that Bill has a daughter, a 75% chance that Bill has a son, and that the probability of him having a daughter and a son is twice as high as him having two daughters (or two sons, for that matter).

So what happens if we ask Bill if he has a daughter, and he answers "Yes"? We can now ignore the 25% possibility that he has two sons. But this gives us no information to make us think that he has either one or two daughters; he would answer the same either way. So it would make no sense for his Yes answer to change the fact that he is still twice as likely to have one daughter as opposed to two, which is why the new probabilities become 2/3rds for one daughter, and 1/3rd for two. So learning that he has one daughter reduces the chances of him having a boy from 75% to 66%.

If we have some way of differentiating the two children, and we know that one of them specifically is a girl, then that would only leave one unknown child, so the answer would be 50%. Likewise, if Bill were to pick one of his two children at random, and reveal their gender to us, boy or girl, that would also make the answer 50%, since Bill would be twice as likely to show us a daughter if he had two of them instead of one, so him revealing a daughter would be new information that would counterbalance our original probability of him being twice as likely to have one daughter instead of two. But when we're only asking Bill if either one of his children is girl, the 2/3rds answer is correct.

GyroNinja fucked around with this message at 00:09 on Sep 5, 2016

Phyzzle
Jan 26, 2008

GAINING WEIGHT... posted:

I'm sorry, I don't follow. Why would the probability of a coin being tails being 1/2 be begging the question? We're assuming fair coin right?

A fair coin doesn't always have a 1/2 probability of being tails, any more than a 'fair' child must always have a 1/2 probability of being male.

So, when you say that 100/200 = 1/2 must be the right probability, because the treating of "no heads" as one option is a mistake, because it's actually many options, because 1/2 is obviously the right probability, that's begging the question.

Phyzzle
Jan 26, 2008
You can also just flip a pair of coins.

H H you win

H T or T H you lose

T T you toss the result

If H T and T H are really a *nested* pair with a 50% probability, you should come out even with a lot of flips. Would you play that one in a casino?

MattD1zzl3
Oct 26, 2007
Probation
Can't post for 4 years!
Remember that time doctor who let himself out of that unescapable box-prison by giving his deux-ex-lockpick to his friend after he escaped from it via time travel? Good times.

Mc Do Well
Aug 2, 2008

by FactsAreUseless

MattD1zzl3 posted:

Remember that time doctor who let himself out of that unescapable box-prison by giving his deux-ex-lockpick to his friend after he escaped from it via time travel? Good times.

It seems to me in our 'phase' of existence/perception time travel is completely impossible. We are fundamentally linear in existence, although imagination lets us make illusions and play games with nonlinear time, and glimpse the weirder reality. Organisms are conceived, develop toward one goal or another, and then decay and die. The cosmos might not be a ticking clock but our biology is fairly mechanistic. One thought that has occurred to me is that time as we experience it is completely dependent on gravity

VitalSigns
Sep 3, 2011

Let's brute-force the boy-girl problem. I wrote a MATLAB script
code:
%initialize cases. boy is total outcomes of at least one boy. noboys is total outcomes of no boys (all girls)
boy=0;
noboys=0;
index=0;
while(index<10000)
%loop 10,000 times
     %set each child to a random integer from 0 to 1. 0 will be boys, 1 will be girls.
     c1=randint(1,1,[0 1]);
     c2 = randint(1,1,[0 1])

     if(c1+c2==2)
     %this means both children were girls, increment noboys. Increment loop index.
          noboys=noboys+1;
          index =index+1;
     else if(c1+c2==1)
     %this means one child is a boy, increment boy, increment loop index.
          boy=boy+1;
          index=index+1;
          end
     end
%the other case, c1+c2 == 0 means both children were boys, ignore this case because it's impossible as given by the problem. Do not increment loop index.
end
boy_percent=boy*100/10000;
noboys_percent=noboys*100/10000;


66.98% instances of a boy, 33.02% instances of only girls.

GAINING WEIGHT... you should probably listen to the people telling you why your intuition is wrong.

Edit: alternatively play the coinflip game with me a bunch of times.
Every time you flip both heads, I'll give you a 2.2:1 payout instead of the 2:1 payout that would make it an even game according to you. You can't lose!

VitalSigns fucked around with this message at 03:06 on Sep 5, 2016

GAINING WEIGHT...
Mar 26, 2007

See? Science proves the JewsMuslims are inferior and must be purged! I'm not a racist, honest!

VitalSigns posted:

Let's brute-force the boy-girl problem. I wrote a MATLAB script
code:
%initialize cases. boy is total outcomes of at least one boy. noboys is total outcomes of no boys (all girls)
boy=0;
noboys=0;
index=0;
while(index<10000)
%loop 10,000 times
     %set each child to a random integer from 0 to 1. 0 will be boys, 1 will be girls.
     c1=randint(1,1,[0 1]);
     c2 = randint(1,1,[0 1])

     if(c1+c2==2)
     %this means both children were girls, increment noboys. Increment loop index.
          noboys=noboys+1;
          index =index+1;
     else if(c1+c2==1)
     %this means one child is a boy, increment boy, increment loop index.
          boy=boy+1;
          index=index+1;
          end
     end
%the other case, c1+c2 == 0 means both children were boys, ignore this case because it's impossible as given by the problem. Do not increment loop index.
end
boy_percent=boy*100/10000;
noboys_percent=noboys*100/10000;


66.98% instances of a boy, 33.02% instances of only girls.

Fair enough, I can't argue with the brute numbers. I wonder what was at fault with my "third gender" conception - seems like it should come out the same, yeah? It just uses a different word for "girl that isn't the child the man was talking about", right?

Phyzzle
Jan 26, 2008
A man flips two coins. One comes up H, but the other one rolls under a desk. The other one has a 50% chance of being T.

A man flips two coins, and they both roll under a desk. His wife looks under the desk and says, "It's dark, but I can see one of them, and it's H". The other one now has a 67% chance of being T. For the man. Since he doesn't know which coin came up H, there are three equal possibilities for him:
TT, TH, and HT.

But since his wife is looking directly at one specific coin rather than hearing a report of "at least one H", the other coin still has a 50% chance of being T. But only for her.

Something seems amiss here . . .

wateroverfire
Jul 3, 2010

Phyzzle posted:

A man flips two coins. One comes up H, but the other one rolls under a desk. The other one has a 50% chance of being T.

A man flips two coins, and they both roll under a desk. His wife looks under the desk and says, "It's dark, but I can see one of them, and it's H". The other one now has a 67% chance of being T. For the man. Since he doesn't know which coin came up H, there are three equal possibilities for him:
TT, TH, and HT.

But since his wife is looking directly at one specific coin rather than hearing a report of "at least one H", the other coin still has a 50% chance of being T. But only for her.

Something seems amiss here . . .

When his wife tells him "I can see one of them, and it's H.", TT is eliminated and the possibilities are only TH or HT.

Phyzzle
Jan 26, 2008

wateroverfire posted:

When his wife tells him "I can see one of them, and it's H.", TT is eliminated and the possibilities are only TH or HT.

Or HH.

wateroverfire
Jul 3, 2010

Oh, yup.

Oh, I see what happened..

A man flips two coins. One comes up H, but the other one rolls under a desk. The other one has a 50% chance of being H.

A man flips two coins, and they both roll under a desk. His wife looks under the desk and says, "It's dark, but I can see one of them, and it's H". The other one now has a 67% chance of being T. For the man. Since he doesn't know which coin came up H, there are three equal possibilities for him:
HH, TH, and HT.

But since his wife is looking directly at one specific coin rather than hearing a report of "at least one H", the other coin still has a 50% chance of being H. But only for her.

Something seems amiss here . . .

wateroverfire fucked around with this message at 17:35 on Sep 5, 2016

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VitalSigns
Sep 3, 2011

Phyzzle posted:

Something seems amiss here . . .

No it's 50% for both of them, see GyroNinja's explanation.

It's twice as likely for there to be HT/TH than TT. But if it's TT she's twice as likely to see a tails so they cancel out. Her husband knows she only saw one coin so he has the same information she does.

If the man says "Do you see any tails" and she says "yes" he can only eliminate the HH case, giving him a 33% chance for each of the other three cases.


Edit: to put it in Bayesian terms, the first situation asks "given she happened to see a tails, what is the probability that one is heads" (50%). The second asks "given they aren't both heads, what is the probability that one or the other is heads" (67%)


Edited again: clarity

VitalSigns fucked around with this message at 04:23 on Sep 7, 2016

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