LessWrongWiki: What the hell are they talking about?

[Mesozoic Mister Nigel]

Both questions, as they are worded, are "what is the probability that the other child is a girl?"

Yes, and when you're given the information that one of them is a girl, the question moves from "what are the chances that a child is born a girl," and becomes "What are the chances that two children are both born girls?" 

Or, in other words, you're asked about two coin flips, not just one.

If I flip a coin four times, the sequence of flips has 1/16 chance of happening.  But if I ask what the chances of any given flip, then it's 1/2.

And if I ask the odds of the total amount of heads v tails, that's another probability entirely.

No, you have it backward. The first coin has already been flipped, there is no remaining probability for the first coin.

[Mesozoic Mister Nigel]

Look at it this way: You have flipped a coin, and it came up heads. What is the probability that it came up tails?

[Bebek Sincap Ratatosk]

http://www.awfulfinance.com/a-girl-named-florida/

This actually pulls some specifics from the book where the problem is found:

Although the statement of the problem says that one child is a girl, it doesn't say which one, and that changes things...The new information – one of the children is a girl – means that we are eliminating from consideration the possibility that both children are boys....That leaves only 3 outcomes in the sample space: (girl, boy), (boy, girl), and (girl, girl).

The variant is this: in a family with two children, what are the chances, if one of the children is a girl named Florida, that both children are girls?...I picked (Florida) rather carefully, because part of the riddle is the question, what, if anything, about the name Florida affects the odds?...are the chances of two girls still 1 in 3?...The fact that one of the girls is named Florida changes the chances to 1 in 2.

There are some interesting charts and then a comment that:

The answer isn't 50%, it's 49.9999875%

*twitch*

[LMNO]

Note: my viewpoint on this has been changing, so each subsequent post may not be fully in line with the previous.

For the record, that is the kind of question that is given in statistics classes to train students to disregard extraneous data. As this thread illustrates perfectly, one of the challenges in statistics is to answer the question that is actually being posed and not the question that you think is being posed.

I see what you're saying, and I think I am ignoring extraneous data?

Pardon if I keep doing it as coin flips.

Situation: two flips.

Probability of both flips being tails: 1/4

Known: One flip is tails.

Probability of both flips being tails: still 1/4.

The question being asked is the probability of two tails.  Telling me that one of the flips is tails changes nothing, as that's what I'm looking for.

[LMNO]

Also, I'm sorry if I keep pushing the "wrong" answer.  Please don't get frustrated with me.

[Nephew Twiddleton]

http://www.awfulfinance.com/a-girl-named-florida/

This actually pulls some specifics from the book where the problem is found:

Although the statement of the problem says that one child is a girl, it doesn't say which one, and that changes things...The new information – one of the children is a girl – means that we are eliminating from consideration the possibility that both children are boys....That leaves only 3 outcomes in the sample space: (girl, boy), (boy, girl), and (girl, girl).

The position is irrelevant though. It doesn't say what's the likelihood that the youngest one is a girl. It says what's the likelihood that they are both girls.

The variant is this: in a family with two children, what are the chances, if one of the children is a girl named Florida, that both children are girls?...I picked (Florida) rather carefully, because part of the riddle is the question, what, if anything, about the name Florida affects the odds?...are the chances of two girls still 1 in 3?...The fact that one of the girls is named Florida changes the chances to 1 in 2.

There are some interesting charts and then a comment that:

The answer isn't 50%, it's 49.9999875%

*twitch*

49.9999875% is more or less the same as 50%

[Mesozoic Mister Nigel]

I'd like to hear more on the explanation that the name of the girl affects the odds of the sex of the other child. It sounds like they're inserting an otherwise meaningless cultural bias and insisting that it has statistical meaning.

[Mesozoic Mister Nigel]

Note: my viewpoint on this has been changing, so each subsequent post may not be fully in line with the previous.

For the record, that is the kind of question that is given in statistics classes to train students to disregard extraneous data. As this thread illustrates perfectly, one of the challenges in statistics is to answer the question that is actually being posed and not the question that you think is being posed.

I see what you're saying, and I think I am ignoring extraneous data?

Pardon if I keep doing it as coin flips.

Situation: two flips.

Probability of both flips being tails: 1/4

Known: One flip is tails.

Probability of both flips being tails: still 1/4.

The question being asked is the probability of two tails.  Telling me that one of the flips is tails changes nothing, as that's what I'm looking for.

You can't do probability on an event that has already happened, so if one of the flips has already been performed and we know the outcome of that flip, and both flips are independent of one another (which they have to be for this example, otherwise we would be performing a much more complex ANOVA calculation) we can only calculate the odds of the remaining flip.

[Nephew Twiddleton]

Note: my viewpoint on this has been changing, so each subsequent post may not be fully in line with the previous.

For the record, that is the kind of question that is given in statistics classes to train students to disregard extraneous data. As this thread illustrates perfectly, one of the challenges in statistics is to answer the question that is actually being posed and not the question that you think is being posed.

I see what you're saying, and I think I am ignoring extraneous data?

Pardon if I keep doing it as coin flips.

Situation: two flips.

Probability of both flips being tails: 1/4

Known: One flip is tails.

Probability of both flips being tails: still 1/4.

The question being asked is the probability of two tails.  Telling me that one of the flips is tails changes nothing, as that's what I'm looking for.

You can't do probability on an event that has already happened, so if one of the flips has already been performed and we know the outcome of that flip, and both flips are independent of one another (which they have to be for this example, otherwise we would be performing a much more complex ANOVA calculation) we can only calculate the odds of the remaining flip.

Let's put it this way-

Moe: I think there's 25% chance that two girls are standing outside.

Larry: Have you looked outside?

Moe: No.

LArry: There's two people, and at least on of them is a girl.

Moe: There's a 50% chance that two girls are standing outside.

[Mesozoic Mister Nigel]

Once again, think of a fair coin that you have just flipped, and it came up heads. What is the chance that it came up tails? None, because it came up heads.

When you have not yet flipped the coin, it has a 50% chance of coming up heads. So does the second coin. To determine the probability that both will come up heads, you can multiply .50 x .50 and find that the probability is .25, right?

Once it has been flipped, there is no more probability, or, you could state it as the probability that it came up heads is 100%.

The second coin flip still has a 50% probability of coming up heads. 1 x .50 = .50. The probability that both coins will be heads, once the first coin has been flipped and came up heads, is 50%, the same as the probability for a single coin flip.

[Mesozoic Mister Nigel]

http://www.awfulfinance.com/a-girl-named-florida/

This actually pulls some specifics from the book where the problem is found:

Although the statement of the problem says that one child is a girl, it doesn't say which one, and that changes things...The new information – one of the children is a girl – means that we are eliminating from consideration the possibility that both children are boys....That leaves only 3 outcomes in the sample space: (girl, boy), (boy, girl), and (girl, girl).

The variant is this: in a family with two children, what are the chances, if one of the children is a girl named Florida, that both children are girls?...I picked (Florida) rather carefully, because part of the riddle is the question, what, if anything, about the name Florida affects the odds?...are the chances of two girls still 1 in 3?...The fact that one of the girls is named Florida changes the chances to 1 in 2.

There are some interesting charts and then a comment that:

The answer isn't 50%, it's 49.9999875%

*twitch*

I might argue that the author of that book isn't doing statistics, he's doing economics. 

[Nephew Twiddleton]

Once again, think of a fair coin that you have just flipped, and it came up heads. What is the chance that it came up tails? None, because it came up heads.

When you have not yet flipped the coin, it has a 50% chance of coming up heads. So does the second coin. To determine the probability that both will come up heads, you can multiply .50 x .50 and find that the probability is .25, right?

Once it has been flipped, there is no more probability, or, you could state it as the probability that it came up heads is 100%.

The second coin flip still has a 50% probability of coming up heads. 1 x .50 = .50. The probability that both coins will be heads, once the first coin has been flipped and came up heads, is 50%, the same as the probability for a single coin flip.

This is reminding me of the time that my stats professor said he doesn't usually play the lottery because, statistically, there's no point, but when he feels like it, he'll go up and say loudly enough for the other numbers players to hear that he's playing 0000, just to piss them off. Each ball has 10% chance of coming up a 0, and on the off chance that 0000 does come up, he'll have been the only spag to play that number.

[Mesozoic Mister Nigel]

Once again, think of a fair coin that you have just flipped, and it came up heads. What is the chance that it came up tails? None, because it came up heads.

When you have not yet flipped the coin, it has a 50% chance of coming up heads. So does the second coin. To determine the probability that both will come up heads, you can multiply .50 x .50 and find that the probability is .25, right?

Once it has been flipped, there is no more probability, or, you could state it as the probability that it came up heads is 100%.

The second coin flip still has a 50% probability of coming up heads. 1 x .50 = .50. The probability that both coins will be heads, once the first coin has been flipped and came up heads, is 50%, the same as the probability for a single coin flip.

This is reminding me of the time that my stats professor said he doesn't usually play the lottery because, statistically, there's no point, but when he feels like it, he'll go up and say loudly enough for the other numbers players to hear that he's playing 0000, just to piss them off. Each ball has 10% chance of coming up a 0, and on the off chance that 0000 does come up, he'll have been the only spag to play that number.

[Bu🤠ns]

I hear there's 50% chance that there's a million dollars in the house next door. However, there is a 50% chance that there is no house next door. 

Well, then let's build one!


(Not to detract...I just wanted to have this thread pop up in my new replies.)

[Mesozoic Mister Nigel]

LMNO, this way of thinking about it might also be helpful:

A family has two children, and at least one of the children is a boy. What are the chances that both children are girls?