The task was to predict how the popular vote in the 2008 Presidential Election would break down, expressed as Obama’s fraction of the total votes that will go to Obama+McCain, and also to give a standard deviation. The winner will be the prediction whose Gaussian distribution function has the largest value at the real fraction, whatever it turns out to be.
Entries are now complete, and here they are, in handy graphical format:
Pretty, isn’t it? But a tad cluttered. Zooming in a bit:
And, in case you are one of those jumbled in the middle there, zooming in a bit more:
The mean (unweighted — sorry) prediction was that Obama would win 53.6 percent of the McCain/Obama popular vote, while the median was 53.2. The average standard deviation was 1.2%. Clearly, for predictions anywhere near the popular values, a fairly small standard deviation was required for one’s curve to poke up past the crowd; indeed, some predictions with large errors are already mathematically eliminated. Like — me. That’s what you get for going first. Here is an even closer zoom, vertically as well as horizontally, centered on my 55.5 +- 1.5 prediction:
See that aqua-colored bell curve, reaching a peak of about .27 at 55.5? That’s me, swamped by narrower neighbors. Confidence pays! I think there are about 20 entries out of 61 that have a nontrivial chance of winning.
(If we were serious and respectable, we would have kept the predictions [and, crucially, the total number of entries] secret until they were all announced. We are neither serious nor respectable.)
See you in November.
Brilliant! It looks like you plotted them in Mathematica–would you be willing to upload the .nb file for use to look at?
Looks like my turf goes roughly from 55.2 or so to 55.5. I had to plot my estimate to see compare in order to be sure of which Gaussian was mine.
Well, at least it isn’t a set of literal measure zero. 🙂
You averaged the standard deviations? ???
You should produce a summed data set and give it to your favorite neigborhood experimentalist to fit that spectrum with a single gaussian peak. That would make an interesting graph.
It makes perfect sense to average the standard deviations. These aren’t actually measurements of anything, they’re predictions. The average of sigma gives a feeling for how confident people are in their predictions, whatever they may be — or, more accurately, how much variance they thought would give them the best chance of winning the contest.
Yes, here is the Mathematica code — I had meant to include it and just forgot. Cut and paste. And let me know if I entered anyone’s prediction incorrectly!
—————
p[x_, f_, sigma_] = Exp[-(x – f)^2/(2sigma^2)]/(Sqrt[2 Pi] sigma)
predictions = {{55.5, 1.5}, {52, 2}, {48, 1}, {55, 1}, {54.5, .75}, {65, 4}, {54, .75}, {68, 3}, {51.9, .6}, {49.2, .7}, {51.5, 2}, {50.6, .5}, {52.2, 1}, {54.27, .75}, {53.8, 1}, {53.2, .15}, {50.5, .8}, {54.28, .85}, {47, 2}, {45, 3}, {49.999, .001}, {54.4, 1}, {61, 5}, {54.2, .35}, {53, .1}, {47, 1.5}, {53.45, .125}, {51.5, .5}, {57, .75}, {50.3, .05}, {44, 2}, {56.1, 1.8}, {60, 1}, {52.3, 0.3}, {51, 1.2}, {53.2, 0.9}, {94, 3}, {58.7, 0.7}, {57, 2}, {56.32, 0.7}, {53, 0.5}, {50.75, 0.25}, {62, 1.21347}, {56, 1.6}, {40, 0.8}, {48.632, 0.002}, {50.4, 0.2}, {51.8, 1.1}, {57.2, 0.3}, {55, 3.1}, {53.55, 0.1}, {52.35, 0.69}, {53.5, 2.5}, {51, 1}, {46, .8}, {59, .5}, {69, .69}, {53.5, 1.5}, {52, 1}, {53.7, .7}, {55, 1.5}, {52.67, .1}, {45, 1.2}, {30, 5}, {52.5,1}}
colors = Table[Random[], {n, Dimensions[predictions][[1]]}]
Plot[Evaluate[Table[p[x, predictions[[n, 1]], predictions[[n, 2]]], {n, Dimensions[predictions][[1]]}]], {x, 0, 100}, PlotStyle -> Table[Hue[colors[[n]], 1, .8], {n, Dimensions[predictions][[1]]}], PlotRange -> {0, 5}]
Here’s a possible way to deal with the bias towards later entries (not for this contest, as rules must be up front, but just as an idea): if a later entry y overlaps with x, then consider the area of overlap. Multiply x by a weight factor so that its integral increases by the (original) area of overlap.
Would that be fair, or would it overcompensate? Might the answer to that depend on the true PDF? Is there a statistician in the house?
This is a good contest, and let’s see how well the Delphi mean compares not only to the actual results, but to how well Intrade picks it (that has done really well over the past few years.) Once a sunken submarine was found to within a couple hundred yards by doing averaging of the guesses of experts, but apparently Intrade can do well with some approximation of “the public” pooling their so-called wisdom of the crowd. (How’s that for “mob rule”?)
http://www.intrade.com/
I’ve kept my prediction secret:
http://blogs.discovermagazine.com/cosmicvariance/2008/06/19/presidential-prediction-contest/#comment-318857
In case you want to include it, here it is: Obama’s fraction 51.2% avg., 0.07% stdv. (a sharp peak!). I’ve emailed you the original message, whose SHA-512 hash I announced.
Sean, it seems that you forgot to include Mike and homung’s predications (comments # 37 and 108).
There are 69 predications in total. 28 predications have been eliminated. 41 predications are still alive.
The following people have been eliminated: Alejandro, Asad, Brian Mingus, Erik, Eugene, Frege, Ian Paul Freeley, Jacob, Jennifer Ouellette, JoeB, Karl, Ktoaster, Lawrence B. Crowell, Matt, Matt (the real one), Mauro Guerra, Michael Bacon, Mike, Ryan, Sean, charly, cope, graviton383, greg, kmeson, lemuel pitkin, sadguy, and wds.
The following people still have a chance of winning. (Their names are following the winnig numbers.)
( 0.000, 38.003): Colin M
( 38.004, 41.479): Sili
( 41.480, 43.779): Vibi
( 43.780, 45.259): Ross Presser
( 45.260, 46.898): I can’t believe we voted no to Lisbon
( 46.899, 47.097): JCF
( 47.098, 48.509): Kurt
( 48.510, 48.624): zw
( 48.625, 48.639): Thomas W. Swidarski, CEO of Diebold
( 48.640, 49.863): zw
( 49.864, 49.995): Ellipsis
( 49.996, 50.002): mts
( 50.003, 50.034): Ellipsis
( 50.035, 50.051): AGray
( 50.052, 50.203): John Kemeny
( 50.204, 50.383): Nik
( 50.384, 50.586): John Kemeny
( 50.587, 51.058): Joshua
( 51.059, 51.340): cryptographically anonymous
( 51.341, 51.810): Ijon Tichy
( 51.811, 51.946): anonymous 37
( 51.947, 52.444): Randall Shane
( 52.445, 52.784): Joshua Zucker
( 52.785, 52.816): Fermi-Walker Public Transport
( 52.817, 53.108): Tim
( 53.109, 53.322): Elliot
( 53.323, 53.479): Anonymous Snowoboarder
( 53.480, 53.747): Gabe
( 53.748, 53.785): joulesm
( 53.786, 54.636): Evan
( 54.637, 55.071): Tad
( 55.072, 55.596): BlackGriffen
( 55.597, 56.177): Philip
( 56.178, 56.859): homung
( 56.860, 57.691): hegemon359
( 57.692, 57.858): A Curious Mind
( 57.859, 58.581): macho
( 58.582, 59.618): billyjohnson
( 59.619, 61.019): Zach
( 61.020, 63.903): Norman Costa
( 63.904, 65.669): Xenophage
( 65.670, 67.816): Count Iblis
( 67.817, 70.295): twaters
( 70.296, 78.044): Count Iblis
( 78.045, 81.452): Xenophage
( 81.453, 100.000): Trevor
Thanks for the catch, wqz. I didn’t forget homung’s prediction; sadly, it came after the deadline, so fairness dictates that it can’t be included.
I’m surprised that so many people are still theoretically in it, my eyeball estimate was significantly off.
Without homung,
There are 68 predications in total. 28 predications have been eliminated. 40 predications are still alive.
The following people have been eliminated: Alejandro, Asad, Brian Mingus, Erik, Eugene, Frege, Ian Paul Freeley, Jacob, Jennifer Ouellette, JoeB, Karl, Ktoaster, Lawrence B. Crowell, Matt, Matt (the real one), Mauro Guerra, Michael Bacon, Mike, Ryan, Sean, charly, cope, graviton383, greg, kmeson, lemuel pitkin, sadguy, and wds.
The following people still have a chance of winning. (Their names are following the winnig numbers.)
( 0.00000, 38.00361): Colin M
( 38.00361, 41.47961): Sili
( 41.47961, 43.77991): Vibi
( 43.77991, 45.25916): Ross Presser
( 45.25916, 46.89863): I can’t believe we voted no to Lisbon
( 46.89863, 47.09716): JCF
( 47.09716, 48.50953): Kurt
( 48.50953, 48.62496): zw
( 48.62496, 48.63903): Thomas W. Swidarski, CEO of Diebold
( 48.63903, 49.86385): zw
( 49.86385, 49.99529): Ellipsis
( 49.99529, 50.00271): mts
( 50.00271, 50.03469): Ellipsis
( 50.03469, 50.05152): AGray
( 50.05152, 50.20331): John Kemeny
( 50.20331, 50.38336): Nik
( 50.38336, 50.58681): John Kemeny
( 50.58681, 51.05876): Joshua
( 51.05876, 51.34059): cryptographically anonymous
( 51.34059, 51.81092): Ijon Tichy
( 51.81092, 51.94603): anonymous 37
( 51.94603, 52.44417): Randall Shane
( 52.44417, 52.78452): Joshua Zucker
( 52.78452, 52.81689): Fermi-Walker Public Transport
( 52.81689, 53.10869): Tim
( 53.10869, 53.32282): Elliot
( 53.32282, 53.47922): Anonymous Snowoboarder
( 53.47922, 53.74739): Gabe
( 53.74739, 53.78569): joulesm
( 53.78569, 54.63681): Evan
( 54.63681, 55.07141): Tad
( 55.07141, 55.59628): BlackGriffen
( 55.59628, 56.70126): Philip
( 56.70126, 56.78486): A Curious Mind
( 56.78486, 57.69133): hegemon359
( 57.69133, 57.85802): A Curious Mind
( 57.85802, 58.58114): macho
( 58.58114, 59.61879): billyjohnson
( 59.61879, 61.01964): Zach
( 61.01964, 63.90354): Norman Costa
( 63.90354, 65.66965): Xenophage
( 65.66965, 67.81627): Count Iblis
( 67.81627, 70.29544): twaters
( 70.29544, 78.04463): Count Iblis
( 78.04463, 81.45253): Xenophage
( 81.45253, 100.00000): Trevor
Given wqz’s analysis of the data, suppose the actual number lies on one of the boundary cases, such as the upper end of Xenophage and the lower end of Trevor’s guesses. Does that mean both win?
Typically, no — we just go to more significant figures.
I still call no fair.
Since humans, and especially humans in an election year could at best be typified as the ultimate analog random number generator, any permutations of purely mathematical could be possible. In other words, you are guessing with style.
When I saw at NPR that
Barack Obama
65,125,060
52.6%
I thought that meant I won.
But then I saw
John McCain
57,178,056
46.1%
and realized that f is not Obama’s vote percent, but Obama / (Obama + McCain), so I get .5325 for f, not .526, so my guess of 52.67% wasn’t so good after all.
Looks like Elliot is the winner.
Sean, could you make a post about who won?
I want to make sure the numbers have settled down, then we will certify a winner!
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