Thursday, January 27, 2005
Is the violence in Iraq enough to defeat the January 30 elections? How does it compare to violence in the US?
The answer to the first question is, of course: I don't know, we'll find out -- but in a moment I'll take a guess.
The second question is interesting. One of the most common of human failings is to make a judgment without first gaining relevant perspective. And when danger is involved people are notorious at this -- both greatly over- and underestimating risk in various circumstances. (In particular, we overestimate risks imposed on us by others while underestimating risks we incur voluntarily, right?)
A couple weeks ago the New York Times reported this...
Now if you want to talk about a US city where there's a high risk of being murdered then of course the first that comes to mind is the seat of the our US government, Washington DC.
The first decent reference for murder rates in DC that I hit upon presents these numbers: a peak murder count of 482 in 1991 when the population was 598,000, falling to a count of 239 in 2000 with a population of 572,000. (I'm not going to bother looking up post-2000 data -- this isn't about measuring things to decimal points, only looking for scale and obtaining perspective.) OK, so these give murder rates for DC ranging from about 0.0008 to 0.0004 per resident.
As to the killings caused by the Iraq insurgency, I have no idea how accurate the number cited by the Times is, but let's trust the paper in this case and take it at face value. We see that 202 killings in two weeks amount to an annual rate of 5,252 in a country with a population of 25.4 million, for a rate of about 0.0002 per resident.
So that puts the rate of killings caused by the insurgency at one-half to one-quarter the recent murder rate that people have lived with in Washington DC, as it has served as our nation's capital and seat of government over the last 15 years.
Of course this rate is for all of Iraq -- the rate of killings is far worse in some areas, Baghdad and the Sunni triangle, than in others (just as it's worse in some parts of DC than the whole).
But that's pretty much the point. Subtract the concentrated violence that is occurring in a small portion of the country -- in parts of four of 18 provinces -- and the average level everywhere else throughout the other 14 provinces is much less.
Lets guesstimate that outside of Baghdad and the Sunni triangle the rate of killing caused by the insurgency is only half of the total rate including those areas. Then we are talking about a rate of 0.0001 per resident -- which, as it happens, in addition to being only one-quarter of the murder rate in DC in 2000, and one-eighth of the DC murder rate of 1991, is almost exactly the same as the total nationwide US murder rate of 1991, (9.8 per 100,000) -- and for that matter, pretty much the same as the average total US rate for the 15 years until 1991.
Should a level of violence experienced by most people across the nation, aimed at disrupting the election, equal to the average murder rate that we all experienced as Americans throughout 1991, be enough to defeat democratic elections among a populace that by all accounts on the whole -- in those 14 provinces at least -- wants them?
I wouldn't think so, so I'll guess "no". I'll guess that the elections will be a success in most of Iraq, in the 14 of 18 provinces that are Kurdish, Shiite, and away from Bahgdad and the Sunni triangle.
As to how the election will work out in those troubled, insurgent-infested areas, I have no idea.
But if the elections work in 14 of 18 provinces to bring into being the mid-east's first functioning democracy (apart from Israel) they won't only be a success but will risk becoming an historic success, I should think.
But we will see.
The answer to the first question is, of course: I don't know, we'll find out -- but in a moment I'll take a guess.
The second question is interesting. One of the most common of human failings is to make a judgment without first gaining relevant perspective. And when danger is involved people are notorious at this -- both greatly over- and underestimating risk in various circumstances. (In particular, we overestimate risks imposed on us by others while underestimating risks we incur voluntarily, right?)
A couple weeks ago the New York Times reported this...
"In the first two weeks of January, at least 202 people died as a result of the insurgency in Iraq. The killings have been indiscriminate. The dead include Iraqi officials, police officers, civilians..."... which is certainly bad enough on its face. But how bad is it compared to, say, the risk of being murdered in a US city that its urban residents take on voluntarily by living in it every day? Or even the risk of being murdered incurred by the average American living in the average place? I meant to look it up then, when I saw that 202 number in the paper, but just got around to it today.
Now if you want to talk about a US city where there's a high risk of being murdered then of course the first that comes to mind is the seat of the our US government, Washington DC.
The first decent reference for murder rates in DC that I hit upon presents these numbers: a peak murder count of 482 in 1991 when the population was 598,000, falling to a count of 239 in 2000 with a population of 572,000. (I'm not going to bother looking up post-2000 data -- this isn't about measuring things to decimal points, only looking for scale and obtaining perspective.) OK, so these give murder rates for DC ranging from about 0.0008 to 0.0004 per resident.
As to the killings caused by the Iraq insurgency, I have no idea how accurate the number cited by the Times is, but let's trust the paper in this case and take it at face value. We see that 202 killings in two weeks amount to an annual rate of 5,252 in a country with a population of 25.4 million, for a rate of about 0.0002 per resident.
So that puts the rate of killings caused by the insurgency at one-half to one-quarter the recent murder rate that people have lived with in Washington DC, as it has served as our nation's capital and seat of government over the last 15 years.
Of course this rate is for all of Iraq -- the rate of killings is far worse in some areas, Baghdad and the Sunni triangle, than in others (just as it's worse in some parts of DC than the whole).
But that's pretty much the point. Subtract the concentrated violence that is occurring in a small portion of the country -- in parts of four of 18 provinces -- and the average level everywhere else throughout the other 14 provinces is much less.
Lets guesstimate that outside of Baghdad and the Sunni triangle the rate of killing caused by the insurgency is only half of the total rate including those areas. Then we are talking about a rate of 0.0001 per resident -- which, as it happens, in addition to being only one-quarter of the murder rate in DC in 2000, and one-eighth of the DC murder rate of 1991, is almost exactly the same as the total nationwide US murder rate of 1991, (9.8 per 100,000) -- and for that matter, pretty much the same as the average total US rate for the 15 years until 1991.
Should a level of violence experienced by most people across the nation, aimed at disrupting the election, equal to the average murder rate that we all experienced as Americans throughout 1991, be enough to defeat democratic elections among a populace that by all accounts on the whole -- in those 14 provinces at least -- wants them?
I wouldn't think so, so I'll guess "no". I'll guess that the elections will be a success in most of Iraq, in the 14 of 18 provinces that are Kurdish, Shiite, and away from Bahgdad and the Sunni triangle.
As to how the election will work out in those troubled, insurgent-infested areas, I have no idea.
But if the elections work in 14 of 18 provinces to bring into being the mid-east's first functioning democracy (apart from Israel) they won't only be a success but will risk becoming an historic success, I should think.
But we will see.
