I'm not usually a fan of rap, but this struck a chord:
Monday, July 26, 2010
JournoList
How ironic that the vast right-wing conspiracy turned out to be a vast left-wing conspiracy. I'm sure we're all very surprised.
Friday, July 23, 2010
Happy Belated Anniversary to Athwart History
Here I'm two weeks late to the first anniversary of Athwart History, which saw its first post on July 9, 2009.
Since then, we've seen the passage of the historic, and historically awful, health care bill (which I commented on extensively), a not-quite-so-bad-but-still-not-good financial regulation bill (which I haven't commented on at all until now), a shocking Democratic defeat in Massachusetts, a Gulf oil spill, and a bunch of random other items of interest.
The summer tends to be slow, and Athwart History is no exception. But as the mid-term election swing into gear, no doubt the commentary will accelerate again.
One bit of personal news: I am in the midst of changing jobs, and also moving across the country. Currently I'm in New Jersey (one state struggling with a fiscal crisis) and I'll be moving to California (another one). So this is an exit frying pan, enter fire sort of thing. I'm excited about California, if for no other reason than the weather, and the new job should be a blast, even though it may impede my ability to post.
Since then, we've seen the passage of the historic, and historically awful, health care bill (which I commented on extensively), a not-quite-so-bad-but-still-not-good financial regulation bill (which I haven't commented on at all until now), a shocking Democratic defeat in Massachusetts, a Gulf oil spill, and a bunch of random other items of interest.
The summer tends to be slow, and Athwart History is no exception. But as the mid-term election swing into gear, no doubt the commentary will accelerate again.
One bit of personal news: I am in the midst of changing jobs, and also moving across the country. Currently I'm in New Jersey (one state struggling with a fiscal crisis) and I'll be moving to California (another one). So this is an exit frying pan, enter fire sort of thing. I'm excited about California, if for no other reason than the weather, and the new job should be a blast, even though it may impede my ability to post.
The Tuesday Child Problem, Solved
At long last, the solution.
As in the case with flipping two coins, the problem revolves around figuring out what options have been eliminated by the statement of the problem. To review, the statement is that I have two children, one of whom is a born born on a Tuesday. The question is what is the probability that my other child is a boy?
First, I'll do this formally, to show that formal mathematics works just fine to solve problems like this. That is, you don't have to stoop to writing out the various possibilities and then tabulating them, as we did in the coin problem.
There is a formula in probability theory that comes in handy here:
(1) P(X|Y) = P(X and Y) / P(Y)
(The notation P(X|Y) means: the probability that X occurs given that Y has occurred, or in short-hand: the probability of X given Y.)
From (1), by the way, you can get another useful formula:
(2) P(X and Y) = P(Y) P(X|Y)
And since P(X|Y) = P(X) if X and Y are independent, that last formula can be written P(X and Y) = P(X) P(Y) in the case of independence.
Finally, since the "and" operation is commutative (i.e., the order of its operands is irrelevant), we have:
(3) P(Y and X) = P(X) P(Y|X) = P(Y) P(X|Y)
In this case, we want to determine:
P(both my children are boys | one child is a boy born on a Tuesday)
By (1), that's equal to P(both my children are boys and one child is a boy born on a Tuesday) / P(one of my children is a boy born on a Tuesday).
The first probability is clearly equivalent to P(both my children are boys and one of them was born on Tuesday), which by (2) is P(both my children are boys) P(one of my children was born on Tuesday). The first of these is easy to calculate: 1/4. The second is trickier, but the easy way to see it is to imagine that neither was born on Tuesday. Since the likelihood for either not being born on Tuesday is 6/7, the chance that neither was born on Tuesday is just 6/7 squared, or 36/49. That means the chance that at least one was born on Tuesday is 1 - 36/49 = 13/49. So that first factor in our equation above is (1/4)(13/49) = 13/196.
Now let's look at the second factor, P(one of my children is a born born on Tuesday). We can apply similar logic by imagining that neither is. The chance that a single child is a boy born on Tuesday is clearly 1/2 (boy) times 1/7 (born on Tuesday) = 1/14. So the chance that a single child is NOT a boy born on Tuesday is 13/14. Hence the chance that neither of two children is a boy born on Tuesday is 13/14 squared, or 169/196. The chance that at least one child is a boy born on Tuesday is then 1 - 169/196 = 27/196.
Plugging these values back into our original formula, we get the final probability as (13/196) / (27/196) = 13/27. And that is the answer.
Another way to do it (a bit less formally, but still perfectly accurate) is more like the two-coins problem. Let's write B for boy, G for girl, T for Tuesday, and !T for not-Tuesday. Then all possible outcomes for one child can be written like this:
G
BT
B!T
These are not equally likely, of course. P(G) = 1/2, P(BT) = 1/14, and P(B!T) = 3/7. But these add up to 1, and are mutually exclusive, so they must cover every possibility.
So for two children, there are nine possible outcomes:
G,G
G,BT
G,B!T
BT,G
BT,BT
BT,B!T
B!T,G
B!T,BT
B!T,B!T
Again, though, these are not equally likely. But their probabilities are easy to compute since the outcomes are independent. Furthermore, some of these outcomes are not possible since we know that one child is a boy born on Tuesday. So the only outcomes remaining are:
G,BT => (1/2)(1/14) = 1/28
BT,G => (1/14)(1/2) = 1/28
BT,BT => (1/14)(1/14) = 1/196
BT,B!T => (1/14)(3/7) = 3/98
B!T,BT => (3/7)(1/14) = 3/98
The sum of these is 1/28 + 1/28 + 3/98 + 3/98 + 1/196 = 27/196. Two boys show up with probability 3/98 + 3/98 + 1/196 = 13/196. The ratio is 13/27, as before.
As in the case with flipping two coins, the problem revolves around figuring out what options have been eliminated by the statement of the problem. To review, the statement is that I have two children, one of whom is a born born on a Tuesday. The question is what is the probability that my other child is a boy?
First, I'll do this formally, to show that formal mathematics works just fine to solve problems like this. That is, you don't have to stoop to writing out the various possibilities and then tabulating them, as we did in the coin problem.
There is a formula in probability theory that comes in handy here:
(1) P(X|Y) = P(X and Y) / P(Y)
(The notation P(X|Y) means: the probability that X occurs given that Y has occurred, or in short-hand: the probability of X given Y.)
From (1), by the way, you can get another useful formula:
(2) P(X and Y) = P(Y) P(X|Y)
And since P(X|Y) = P(X) if X and Y are independent, that last formula can be written P(X and Y) = P(X) P(Y) in the case of independence.
Finally, since the "and" operation is commutative (i.e., the order of its operands is irrelevant), we have:
(3) P(Y and X) = P(X) P(Y|X) = P(Y) P(X|Y)
In this case, we want to determine:
P(both my children are boys | one child is a boy born on a Tuesday)
By (1), that's equal to P(both my children are boys and one child is a boy born on a Tuesday) / P(one of my children is a boy born on a Tuesday).
The first probability is clearly equivalent to P(both my children are boys and one of them was born on Tuesday), which by (2) is P(both my children are boys) P(one of my children was born on Tuesday). The first of these is easy to calculate: 1/4. The second is trickier, but the easy way to see it is to imagine that neither was born on Tuesday. Since the likelihood for either not being born on Tuesday is 6/7, the chance that neither was born on Tuesday is just 6/7 squared, or 36/49. That means the chance that at least one was born on Tuesday is 1 - 36/49 = 13/49. So that first factor in our equation above is (1/4)(13/49) = 13/196.
Now let's look at the second factor, P(one of my children is a born born on Tuesday). We can apply similar logic by imagining that neither is. The chance that a single child is a boy born on Tuesday is clearly 1/2 (boy) times 1/7 (born on Tuesday) = 1/14. So the chance that a single child is NOT a boy born on Tuesday is 13/14. Hence the chance that neither of two children is a boy born on Tuesday is 13/14 squared, or 169/196. The chance that at least one child is a boy born on Tuesday is then 1 - 169/196 = 27/196.
Plugging these values back into our original formula, we get the final probability as (13/196) / (27/196) = 13/27. And that is the answer.
Another way to do it (a bit less formally, but still perfectly accurate) is more like the two-coins problem. Let's write B for boy, G for girl, T for Tuesday, and !T for not-Tuesday. Then all possible outcomes for one child can be written like this:
G
BT
B!T
These are not equally likely, of course. P(G) = 1/2, P(BT) = 1/14, and P(B!T) = 3/7. But these add up to 1, and are mutually exclusive, so they must cover every possibility.
So for two children, there are nine possible outcomes:
G,G
G,BT
G,B!T
BT,G
BT,BT
BT,B!T
B!T,G
B!T,BT
B!T,B!T
Again, though, these are not equally likely. But their probabilities are easy to compute since the outcomes are independent. Furthermore, some of these outcomes are not possible since we know that one child is a boy born on Tuesday. So the only outcomes remaining are:
G,BT => (1/2)(1/14) = 1/28
BT,G => (1/14)(1/2) = 1/28
BT,BT => (1/14)(1/14) = 1/196
BT,B!T => (1/14)(3/7) = 3/98
B!T,BT => (3/7)(1/14) = 3/98
The sum of these is 1/28 + 1/28 + 3/98 + 3/98 + 1/196 = 27/196. Two boys show up with probability 3/98 + 3/98 + 1/196 = 13/196. The ratio is 13/27, as before.
Thursday, July 8, 2010
World Cup
So it's Spain v. Holland in the final. I have to admit I'm torn on this one. The Dutch, with their centuries-long ties to England, are the closest to being my countrymen, so sheer familial respect prefers them. But the Spanish side, La Roja, is so great to watch.
Often in games like this I have to start watching and see where my heart finds itself rooting. It usually makes some pre-rational decision of its own. Let's just hope it's a good game... and that it doesn't end in penalty kicks.
Often in games like this I have to start watching and see where my heart finds itself rooting. It usually makes some pre-rational decision of its own. Let's just hope it's a good game... and that it doesn't end in penalty kicks.
Wednesday, July 7, 2010
Math Surprises
Some of the best math problems are the simplest to state. But they can be fiendishly difficult to understand, precisely because their domains are so commonplace. A good example is fairly well known: Suppose I have two coins. One of them is showing heads. What is the probability the other one is also showing heads?
The answer most people jump to is 1/2, since we've had it drilled into us that coins are independent, so the outcome of one shouldn't affect the other. But the phrasing of the question contains a subtle trick. By stating that one of the coins is showing heads, I have eliminated the possibility that both coins are showing tails. But there are actually three remaining possibilities: heads/tails, tails/heads, or heads/heads. In only one of these three are both coins heads, so the answer is in fact 1/3.
The trickiness of the phrasing comes from the fact that when we say that one of the coins is heads, we deliberately don't say which one. We could say equivalently that at least one is showing heads. The probability is the expected 1/2 once we specify ahead of time which coin it is.
Given this preliminary, consider the following problem, which has been making its way around the Web of late:
I will post the answer in a few days.
The answer most people jump to is 1/2, since we've had it drilled into us that coins are independent, so the outcome of one shouldn't affect the other. But the phrasing of the question contains a subtle trick. By stating that one of the coins is showing heads, I have eliminated the possibility that both coins are showing tails. But there are actually three remaining possibilities: heads/tails, tails/heads, or heads/heads. In only one of these three are both coins heads, so the answer is in fact 1/3.
The trickiness of the phrasing comes from the fact that when we say that one of the coins is heads, we deliberately don't say which one. We could say equivalently that at least one is showing heads. The probability is the expected 1/2 once we specify ahead of time which coin it is.
Given this preliminary, consider the following problem, which has been making its way around the Web of late:
I have two children, one of whom is a boy born on a Tuesday. What is the probability the other one is a boy? (You can make all the expected simplifying assumptions: no twins, boys and girls born in equal proportions, each day of the week is equally likely, etc. It's a math problem, not an obstetrics exercise.)
I will post the answer in a few days.
Thursday, July 1, 2010
The Bargain
Back in the days of the Clinton White House,
It was said we had two Presidents when we'd only elected one.
Hillary being the other, you see.
Quite a bargain, one might say.
But the Obama administration has given us an even
More attractive package. All in one man:
Ideological consistency
And pragmatism,
Bipartisanship
And combativeness,
Populism
And elitism,
Fiscal discipline
And big-government liberalism,
Taciturnity
And logorrhea.
He is extremely moderate about being moderately extreme.
Come to think of it, he is a contradiction.
He contains multitudes.
Ten Presidents for the price of one!
It was said we had two Presidents when we'd only elected one.
Hillary being the other, you see.
Quite a bargain, one might say.
But the Obama administration has given us an even
More attractive package. All in one man:
Ideological consistency
And pragmatism,
Bipartisanship
And combativeness,
Populism
And elitism,
Fiscal discipline
And big-government liberalism,
Taciturnity
And logorrhea.
He is extremely moderate about being moderately extreme.
Come to think of it, he is a contradiction.
He contains multitudes.
Ten Presidents for the price of one!
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