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Hours You Won't Get Back
Jul. 8th, 2010 01:20 pmI spent several hours on and off yesterday and today pondering the issues raised in this post:
http://alexbellos.com/?p=725
which asks:
"I have two children. One is a boy born on a Tuesday. What is the probability I have two boys?"
and comes up with the answer 13/27.
As is usually the case with "probability" questions, it isn't a question of probability at all, but one of logic and set theory (or what I call set theory). The numbers part is trivial.
Bellos also throws in a few red herrings, which can lead you down the wrong paths. Most of the comments were fairly wrong-headed, I felt.
Like most people, my immediate response was, "well, since, without the added information, the answer is 1/3, and since the added information is nothing to do with anything, the answer must still be 1/3.".
However, I did a "frequentist" analysis, and the 13/27 answer seemed to hold good, so I looked for a flaw in my analysis.
Eventually the question came down to "can a moderator ("born on a Tuesday") change the overall probability, even though the moderator has nothing to do with the initial probability?"
And the answer is, yes.
Here are some more interesting examples of this.
I throw two dice. One is a six thrown on a Tuesday. What is the probability that I throw two sixes.
Answer 13/70.
The next one assumes that children like one of two colours, red or yellow, but that they only develop this preference when they reach the age of five.
"I have two children. At least one is a boy who prefers the colour Red. What is the probability that I have two boys?"
Answer, 3/7.
I throw two dice. One is a six that prefers the colour Red. What is the probability that I throw two sixes?
Answer, 3/20.
The last just seems ridiculous, until you realize that what is happening is that the "universal set" is changed by the modifier. So it doesn't matter whether the modifier has anything to do with the initial statement, or indeed whether the modifier is physically possible (as in the last case). All that matters is how it alters the constituency of the universal set.
The easiest case is the example:
"I have two children, one is a boy who prefers the colour Red. What is the probability I have two boys".
As we know, without the modifier, the probability is 1/3*
BB
BG
GB
If we add in the colour preferences, we get a sample size of 12 fathers (could be mothers, of course).
They have the following pairs of children.
BR BR
BR BY
BY BR
BY BY
_________
GR BR
GR BY
GY BR
GY BY
BR GR
BR GY
BY GR
BY GY
Of this sample, 8 are GB or BG, while 4 are BB, retaining the probability of 1/3.
But now, apply the "modifier" and ask all of the fathers whose children DO NOT fit into the statement to "step aside"
BR BR
BR BY
BY BR
..................BY BY
________________
GR BR
..................GR BY
GY BR
..................GY BY
BR GR
BR GY
.................BY GR
.................BY GY
Notice that only one of the BB group steps aside, while four of the GB/BG group steps aside.
This reduces the sample size from 12 to 7, of whom three were originally in the BB group and four were in the BG/GB group.
So the answer to the question:
"I have two children, at least one of whom is a boy who prefers the colour Red. What is the probability I have two boys"? is 3/7 (up from pre-modifier 1/3)
Similarly:
"I have two children, at least one of whom is a boy who prefers the colour Yellow. What is the probability I have two boys"? is 3/7
"I have two children, at least one of whom is a boy who prefers the colour Yellow. What is the probability I have 1 girl and 1 boy?" is 4/7 (down from pre-modifier 2/3).
Note how The third statement is less counterintuitive. "Hell", you say, "that must reduce the chances of there being GB/BG."
_________________
* For those who disagree with this premise, I refer you to:
http://www.jesperjuul.net/ludologist/?p=1048
“a) We keep flipping two coins simultaneously.
b) If both coins are tails, we flip the coins again.
c) Otherwise, you give me $15 if there is one head, and I give you $20 if there are two heads.
If the probability is 1/2, you will be making money. If it’s 1/3, I will.
Any takers?”
This also gives another way of explaining why the 13/27 (about 48%) answer is correct. The "cocktail party" explanation cleverly gets the thinker's move away from "moving down from 50%" towards "moving up from one-third".
Thus nearly 48% of fathers having a son who claim a son born on Tuesday have two sons, rather than one-third, because the fathers with two sons have two votes. They fail to reach 50% because a small proportion (1/49), just over 2% have two sons both born on a Tuesday, so only raise their hand once.
______________________
http://alexbellos.com/?p=725
which asks:
"I have two children. One is a boy born on a Tuesday. What is the probability I have two boys?"
and comes up with the answer 13/27.
As is usually the case with "probability" questions, it isn't a question of probability at all, but one of logic and set theory (or what I call set theory). The numbers part is trivial.
Bellos also throws in a few red herrings, which can lead you down the wrong paths. Most of the comments were fairly wrong-headed, I felt.
Like most people, my immediate response was, "well, since, without the added information, the answer is 1/3, and since the added information is nothing to do with anything, the answer must still be 1/3.".
However, I did a "frequentist" analysis, and the 13/27 answer seemed to hold good, so I looked for a flaw in my analysis.
Eventually the question came down to "can a moderator ("born on a Tuesday") change the overall probability, even though the moderator has nothing to do with the initial probability?"
And the answer is, yes.
Here are some more interesting examples of this.
I throw two dice. One is a six thrown on a Tuesday. What is the probability that I throw two sixes.
Answer 13/70.
The next one assumes that children like one of two colours, red or yellow, but that they only develop this preference when they reach the age of five.
"I have two children. At least one is a boy who prefers the colour Red. What is the probability that I have two boys?"
Answer, 3/7.
I throw two dice. One is a six that prefers the colour Red. What is the probability that I throw two sixes?
Answer, 3/20.
The last just seems ridiculous, until you realize that what is happening is that the "universal set" is changed by the modifier. So it doesn't matter whether the modifier has anything to do with the initial statement, or indeed whether the modifier is physically possible (as in the last case). All that matters is how it alters the constituency of the universal set.
The easiest case is the example:
"I have two children, one is a boy who prefers the colour Red. What is the probability I have two boys".
As we know, without the modifier, the probability is 1/3*
BB
BG
GB
If we add in the colour preferences, we get a sample size of 12 fathers (could be mothers, of course).
They have the following pairs of children.
BR BR
BR BY
BY BR
BY BY
_________
GR BR
GR BY
GY BR
GY BY
BR GR
BR GY
BY GR
BY GY
Of this sample, 8 are GB or BG, while 4 are BB, retaining the probability of 1/3.
But now, apply the "modifier" and ask all of the fathers whose children DO NOT fit into the statement to "step aside"
BR BR
BR BY
BY BR
..................BY BY
________________
GR BR
..................GR BY
GY BR
..................GY BY
BR GR
BR GY
.................BY GR
.................BY GY
Notice that only one of the BB group steps aside, while four of the GB/BG group steps aside.
This reduces the sample size from 12 to 7, of whom three were originally in the BB group and four were in the BG/GB group.
So the answer to the question:
"I have two children, at least one of whom is a boy who prefers the colour Red. What is the probability I have two boys"? is 3/7 (up from pre-modifier 1/3)
Similarly:
"I have two children, at least one of whom is a boy who prefers the colour Yellow. What is the probability I have two boys"? is 3/7
"I have two children, at least one of whom is a boy who prefers the colour Yellow. What is the probability I have 1 girl and 1 boy?" is 4/7 (down from pre-modifier 2/3).
Note how The third statement is less counterintuitive. "Hell", you say, "that must reduce the chances of there being GB/BG."
_________________
* For those who disagree with this premise, I refer you to:
http://www.jesperjuul.net/ludologist/?p=1048
“a) We keep flipping two coins simultaneously.
b) If both coins are tails, we flip the coins again.
c) Otherwise, you give me $15 if there is one head, and I give you $20 if there are two heads.
If the probability is 1/2, you will be making money. If it’s 1/3, I will.
Any takers?”
This also gives another way of explaining why the 13/27 (about 48%) answer is correct. The "cocktail party" explanation cleverly gets the thinker's move away from "moving down from 50%" towards "moving up from one-third".
Sample 1, cocktail party: All dads with a boy are asked to raise a hand, and 750 of them do so. We know that of these, 250, or 1/3, have two boys. Next, the 750 dads whose hands are up are told to tell a neighbor which day of the week their son was born. Key part: If they have two sons, they should use the birthday of only one of their sons (randomly chosen). 1/7 of the 500 dads with one son (and one daughter) say Tuesday, and 1/7 of the 250 dads with two sons say Tuesday. So of those who say Tuesday, just 1/3 have two sons. And that’s true of any other day of the week.
Sample 2, raising hands by day of week: Dads with a son born on a Sunday are asked to raise their hands. Next dads with a son born on a Monday are asked to raise their hands. Then dads with a son born on a Tuesday are asked to raise their hands, etc. Note that most dads with two sons will raise their hands twice, except for those both sons are born on the same day of the week. This means on any given day of the week, dads with two sons are over-represented relative to dads with one son, since they have two opportunities (one from each son) to have a son born on that day. The diagram demonstrates this over-representation– 13/27th, or 48%, of the dads claiming a son born on a Tuesday have two sons. And that’s true of any other day of the week.
The key difference in cocktail party sampling is that each dad says the birthday of one son, even if they actually have two. In the second sampling, the dads with two sons get to raise their hands on two days of the week, one for each son (unless the sons were born on the same day of the week).
Thus nearly 48% of fathers having a son who claim a son born on Tuesday have two sons, rather than one-third, because the fathers with two sons have two votes. They fail to reach 50% because a small proportion (1/49), just over 2% have two sons both born on a Tuesday, so only raise their hand once.
______________________
