The Deadliest Day of the Year is the Safest · 4:51am Oct 5th, 2017
From Sperling's Best Places:
Kids have a greater chance of being fatally injured by a car on Halloween than any other day of the year, including the Fourth of July and New Year’s Day. State Farm®, the nation’s leading auto insurer, teamed up with research expert, Bert Sperling of Sperling’s BestPlaces, to better understand the risk kids face as they take to the streets in search of treats.
Sperling’s BestPlaces analyzed more than four million records in the Fatality Analysis Reporting System (FARS) from 1990 – 2010 for children 0-18 years of age on October 31. A description of the methodology follows the graphs below.
That detailed analysis revealed the following:
- Halloween Was Deadliest Day of the Year for Child Pedestrian Accidents
One hundred and fifteen child pedestrian fatalities occurred on Halloween over the 21 years of our analysis. That is an average of 5.5 fatalities each year on October 31, which is more than double the average number of 2.6 fatalities for other days.
Children are more than twice as likely to be killed by a car while walking on Halloween night than at any other time of the year, according to one safety group.
From ABC News:
Children are twice as likely to be hit by a car and killed on Halloween than on a typical night, a Centers for Disease Control and Prevention found.
Really; the CDC found that? Let's see what they actually said:
From chilling tales to creepy costumes, lots of things can be scary on Halloween night. But the real danger for children is walking in the dark. On average, children are more than twice as likely to be hit by a car and killed on Halloween than on any other day of the year.
Oh, so they "found" it on a website called SafeKids.org. Good job, CDC. SafeKids says:
Hard Facts
On average, children are more than twice as likely to be hit by a car and killed on Halloween than on any other day of the year.
(No citation.)
Please, someone, restore my faith in humanity by explaining in the comments why showing that twice as many children are hit by a car and killed on Halloween as on any other day, actually shows that Halloween is the day on which children are least likely to be hit by a car and killed. Please. It should be obvious if you just think for ten seconds.
(Someone other than Titanium Dragon. Sorry, TD. I know you can explain it, but you just don't count as representative of humanity. 
)
If there are 20 times as many children on the street and twice as many get hit then each individual child is ten times safer than normal.
Because on Halloween there's tons more kids out than any normal day, so if it was actually more dangerous that figure would be much higher?
People don't know how to play with proportions. As long as the number of kids outside is greater than twice as many as there usually are (which I assume to be true cause there's absolute swarms of those little fuckers around my neighborhood on Halloween), the percentage of kids getting hit by cars is lower. The absolute number of kids getting hit increases, but your chance as an individual decreases.
Also, herd mentality. Let the slow kids get hit by the cars! More candy for the rest of us!
P(Halloween|Kid died being hit by a car) = 2*P(<Random Day>|Kid died being hit by a car)
P(Kid dies being hit by a car|Halloween) < P(Kid dies being hit by a car|<Random Day>) because the ratio between # of kids dying from car accidents vs. # of kids walking around in places where they can get hit by cars is a lot smaller on Halloween. Because the number of kids walking around in places where they can get hit by cars on Halloween is more than twice the number that there would be normally.
Because candy.
4687579
does fimfic support typing math equations in LaTex
Ugh, I'm pretty ashamed at how long it took me to realize the answer, because it really is obvious and I've bitched about this exact sort of thing before. It's just like the whole airplane versus cars thing. For whatever reason, I wasn't processing that they didn't already account for the difference in the number of child pedestrians and misread that as you asking how an increase in vehicular manslaughter somehow makes kids safer, which sounds like something Steven E. Landsburg might write about
I am proud to say I could have restored your faith n humanity but others have beaten me to the punch. I’m still holding on to this one, though. It’s not often I get to restore someone’s faith in humanity!
I see the point you're trying to make, but I think you're making a logical error. Two children out on the street versus one child out on the street does not double the likelihood that a child will be struck and killed by a car. Try to think about why.
Hint: it has something to do with the drivers.
Math is hard.
4687707
It may or may not double it, but it certainly does increase it. Sure, there's a bit of crosstalk because child-killing accidents are not independent events, but out of all the factors influencing the lethality of a given night in absolute terms 'number of kids walking about in the dark, possibly wearing dark costumes, near where cars can get at them' is predictor number one, easy.
It's the same logical trap that people who dislike airlines or cars used when quoting statistics about auto vs airline travel. Per passenger mile, airlines are safer than hiding in your basement on a warm, summer day. Per accident, they're death traps. Or the same logic trap about 'people who own guns are more likely to get killed by them' (mostly from suicides, but that gets left out). How you measure statistics tells a lot about what result you're going to get, or as one of my high school teachers used to say, "Figures lie, but liars figure."
If you're an EMT, you are more likely to treat a child for being run over during Halloween than any other day of the year, due to the swarming multitudes of targets.
If you're a child walking one mile on Halloween, you are *less* likely to be run over than if you walk the same mile on any other night of the year, because drivers are alert.
Depending on what point of view you want to plug (i.e. Halloween is dangerous, no it's not), you use the appropriate statistic.
4687707
I don't need to think about why. I just need to divide the number of kids killed while out walking by the number of kids out walking.
4688006
I agree with your main point—let me get that out of the way first.
Still, I think there may a couple of minor flaws in your logic. The primary one is in thinking that kids being hit by cars are each independent events. They aren't. Drivers don't drive at random. More kids out on the street does not proportionately correspond to more fatalities because cars are not driven in random directions over streets and roads. If the factor were meteor strikes, then you would be correct: ten times as many kids outside means ten times the likelihood to have an event. But that isn't the only determining factor.
A smaller issue is that we don't have a good estimate for how often kids go walking on non-Halloween nights, especially in rural areas. We also don't know how short the period during which kids trick-or-treat is versus the fact they prepare costumes indoors and stay home the rest of the night to eat candy, which are both times they could be playing outside.
So the number you're dividing by may not be representative of the reality. Would it really be safer to let your kids trick-or-treat than having them walk around the neighborhood on another night? I don't have a strong enough basis for an estimation.
Regardless, don't send them out as the "black ghost".
Sadly, I'm not confident enough in humanity to say that, between trunk-or-treats and mall Halloweens and the other alternatives put in place to being a pedestrian with a pillowcase on a cool October night, there are still actually double the usual number of child pedestrians.
4688016
Well duh. That would be cultural appropriation.
It's good to be appreciated.
4688016
It doesn't matter whether they're independent events. The figure of interest is the probability that a randomly-chosen child out walking in America gets killed by a car. A number proportional to this can be computed by dividing number of children killed by cars by hours walked outside by children. It does not matter at all if the events cluster on Halloween; that doesn't change the resulting number. You could make an elaborate model of how likely kids are to walk together and where they are likely to walk on Halloween vs. on different days, or how drivers drive differently, but that doesn't matter, because all those things are wrapped into the question, which is how likely a randomly-chosen American child out walking is to get killed by a car on Halloween vs. on other nights. The question as I interpret it literally specifies that it is asking for one number divided by another number. So you just have to divide those numbers.
If you want to produce a probability for your child walking on your street, then all that would matter; but nobody has spoken of such numbers.
If you wanted to produce an actual number, you could do it by sampling. Have lots of observers count number of children observed walking at appropriately-selected intervals. I haven't done this formally, but I have lived through many Halloweens, and I'm confident the ratio of hours walked outside by children on Halloween to other nights is much larger than 2.
The last time I played Telephone was in the 8th grade. I, uh, guess I missed the bandwagon?
Because you've at least 50 times more children out and about, but your death rate only doubles, instead of jumping up by 50x or more.
Did I get it right?
*checks comments*
Woo! How much extra credit do I get?
This just goes to show: just because you can do math to it doesn't mean you're doing the right math/are interpreting it correctly. Case in point, never trust what's said of a study till you read it yourself. You shouldn't even trust the study's own synopses. Preaching to the choir here, but the choir are good listeners.
4688006
Whistling cheerfully all the while, if I know you.
This is like the 'most accidents occur at home' statistic. Well where else would they occur? You spend more time at home than any other single place. If you more time somewhere else, you'd just switch things around, not decrease accidents.
Here's another amusing little statistical anomaly: When icecream sales peak, so do boat accidents. Obviously, that's because of the warm weather, but it's a fun way to prove correlation is not causation.
4691799
And the tune is Hall of the Mountain King, I assume.
4697394 4691799 Mmm.
( Now you know what it's like to be esoterically referenced at, Ed.
)
4697788
4691799
Well, I hadn't thought of it as esoteric, since Peter Lorre and Fritz Lang are both pretty famous, but I suppose if anyone needs clarification (though I suspect Bad Horse doesn't from that Mmm):
Don't worry about copyright, M is public domain now.
4687707
Actually, it does - if all you care about is the statistic. It's the same reason why you can get fractions of a person with some event probabilities. If two children were killed by the same car, that's 2ch/day - or, twice as likely than 1ch/day. But in reality, the problem is more complex.
The entire issue stems from oversimplification and misinterpretation in the first place, and that was the entire point.