James Deardorff, former Senior Scientist at the National Center for Atmospheric Research, in Boulder, Colorado, gives an outline of probabilities to calculate the Arecibo response. Deardorff poses that it is less than two chances, out of 10 billion, to be a hoax.
[What is] the probability that hoaxers could:
- a. be creative enough to construct a new type of glyph like that, involving rectangular "binary units" in the "Arecibo" response, and no circles,
- b. repeatedly practice making the Arecibo glyph first, in some field(s), without these practice attempts being spotted from the air and reported
- c. actually carry it out, producing all those right-angle corners in the Arecibo-like pattern, without making any mistakes
- d. do it all in just a few hours overnight
- e. do it without showing up on the security cameras there, one or more of which looked out towards the relevant direction …
- f. do it without leaving undesired trampled stalks or stake holes, etc., behind, from having accessed the location along some tram line and laying out the surveying lines, etc., which would be necessary
- g. not claim credit for it afterwards and not offer to show skeptics just how they did it by being willing to quickly reproduce the same designs within a pristine area of a wheat field while under the watchful eyes of veteran crop-circle researchers.
Concerning the probability of (a), we have, on a couple occasions, seen the handiwork of crop-circle hoaxers in a contest. Their patterns consisted of the same elements, and were of the same type, as in (genuine) preexisting crop-circle formations (circles, triangles, stars, and such). Very little creativity. Thus I would estimate the probability of (a) as being p(a) = 0.3 - possible, but not very likely. (Here, p=0 would mean no chance whatsoever it could be a hoax, and p=1 would mean absolute certainty it was a hoax.)
Concerning the probability of (b), since most of the crop-circle formations apparently do get noticed, including hoaxes, so would practice attempts be noticed and reported as either genuine or hoaxes. Surely several practice attempts would be needed in this case, and this would give away hoaxers' final version unless they trampled down each practice attempt right away after making it, without being noticed. However, such trampled areas would themselves likely be noticed from the air and/or the perpetrators reported. I estimate the probability of such going unnoticed and unreported as less than 50-50, say 0.3.
Concerning the probability of (c), I notice that there are some 700-1000 right-angle corners of standing stalk involved, on a relatively small scale, in all those binary units of the "returned" Arecibo message. It would be difficult to generate even 30 of them without making a mistake - and once a mistake is made, with the wrong stems bent over to stay, they can't be raised again. If the chance for error was only 0.5 (50-50) for each succession of 30 corner units, then the probability of making just one right-angle corner come out right is quite high, 0.9782. However, the probability of one or more persons continuing the process on 800 of them without botching any of the corners or trampling down the wrong spot would be this figure raised to the 800th power, which is only 2 x 10-8 = p(c).
- the time to attempt to accomplish this would be on the order of 20 seconds to correctly emplace each of some 2 x (23 + 73) = 192 tall stakes around the periphery (64 minutes in all)
- two minutes to string each of 23 parallel "grid" lines (cords) lengthwise and one minute each for 73 shorter parellel lines crosswise (119 minutes in all)
- 2 minutes each to flatten stalks around the roughly 800 "binary units" of wheat to be left standing (this includes the time necessary to identify where to move to next without trampling the wrong area in the dark, and ducking under the various criss-crossing lines to get there - 1600 minutes in all)
- some 20 minutes for a couple of rest breaks
- 45 minutes to remove all stakes and cords and carefully exit without leaving access tracks behind in the field.
This is some 31 hours, suggesting the need for a team of 5 or 6 people, each knowing what their specific tasks are. Since this seems possible, this consideration doesn't rule either against the hoax or against the "real thing," which means p(d) = 0.5.
Concerning (e): for a team of 5 or 6 people to do this at night would require a good deal of artificial lighting, along with walkie-talkies so that the head hoaxer could orchestrate the entire endeavor, directing each worker on where to step next or not to step. The odds are not good that such lighting would not have been detected when the security-camera video tapes were examined. Here I estimate p(e) = 0.2 that hoaxers could have done this without their night lights showing up.
Concerning (f), I believe that no stake holes were reported, but the probability that so many of them could have been filled in prior to the hypothetical hoaxing team's departure without the disturbed ground being noticed and reported later, and similarly for no disturbances along any tram lines showing up on the aerial photos, suggests a low probability of hoaxers getting away with this aspect, say p(f) = 0.1. (Bear in mind that if hoaxers get to a genuine formation prior to serious crop-circle researchers, such hoaxers could deceptively make stake holes, leave behind some string and cigarette butts, etc.)
Concerning (g), I believe the chances are less than even that, if hoaxers had made such unique crop glyphs, they wouldn't wish to claim credit for it (or them) within a couple weeks afterwards - after a goodly number of paranormal researchers had offered their opinions that the formations were not man-made. This hasn't happened. So I would estimate p(g) = 0.4, with this value decreasing somewhat as time rolls on without any viable confession forthcoming.
… it turns out that there's a mathematical way of combining individual probabilities on a yes-no type of hypothesis, in this case a hoax or no-hoax hypothesis, to arrive at an overall probability, P. That's because probabilities p(a)...p(g) involve independent elements all bearing on the same question of hoax or no-hoax.
- The formula is:
- P = M1/(M1 + M2)
where M1 = p(a)*p(b)*...*p(g) and M2 = [1 - p(a)]*[1-p(b)]*...*[1-p(g)], where the asterisks denote multiplication.
- Plugging in, we get:
- P = 7 x 10-11
That is - less than two chances out of 10 billion.
- Exopaedia, James Deardorff
- A Message Finally Received—In Answer to Carl Sagan?, by Richard Hoagland