What are the odds of the Sagan Signal being a random event? Local college math professor, Sean Rule, in exchange for a gift certificate at a local restaurant, agreed to crunch the numbers for me. To avoid bringing in the Bible or religion, I framed the question as a thought experiment that involved a bingo ball machine, the kind they use in selecting winning lottery numbers, where numbered bingo balls bounce around in a Plexiglas container. At the magic hour a chute opens up and bingo balls are randomly drawn, one-by-one, into a tube. The winning numbers are then recorded in the order in which they were sucked into the tube.

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In our hypothetical bingo machine there are only three balls. One red, symbolizing grain, one white, symbolizing wine, and one blue, symbolizing oil. The machine is turned on and the balls are allowed to bounce around for a few minutes. The chute is then opened and the balls are randomly drawn into the tube and the results recorded: which color comes up first, which second, and which third. The machine is then turned off and the balls cleaned up to remove any oil from the palms of the operator that might prejudice the results. The balls are then dropped back into the container, the blower turned on, and the process starts over again.

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To make Sean’s job a little easier I rounded off the number of total sequences from 54 to 50, and the number of red, white, and blue sequences I was looking for to 40 rather than 46. The percentage of “hits” I was asking for was 80%, slightly less than the 85% rate found in the Sagan Signal.

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The question I posed to Sean was this: In this experiment, what are the mathematical odds of a red, white, and blue sequence randomly coming up 40 times in 50 repetitions? Following is his response: 

 

Hey there, Don!

 

OK...I've got it now.  What you have in this bingo game is a textbook

binomial distribution problem.  Here's how we can proceed:

 

There are 6 possible sequences that the balls can take on any one trial

(of the 50 of which you spoke): RWB, RBW, WRB, WBR, BWR, BRW.   

The chance of getting a sequence of RWB on any one of the 50 trials is 1/6.

 

Now, you need the chance of a set number of successes (i.e., RWB

sequences) out of 50; in particular, the chance that you get at least 40

out of the 50.  We can find this by summing the following:

 

Chance of getting at least 40 out of 50 = chance of getting 40 + chance

of getting 41 + chance of getting 42 + ... + chance of getting 50.

 

Each one of those terms that we have to add is a binomial expression.

Let's sum them all up:

 

(50 nCr 40)(1/6)^40(5/6)^10 + (50 nCr 41)(1/6)^41(5/6)^9 + ... + (50 nCr

50)(1/6)^50(5/6)^0 =...

 

...a very, very, very small number.  It's basically zero.  If you need a

number, it's in the neighborhood of 10^(-21).  Wow.  Very unlikely.

 

(makes sense; you would expect to get about 8 such runs; nowhere near

40)

 

Let me know if you need anything else, Don!

 

The Rule Equation establishes the mathematical odds of the sequences being the result chance at less than one in ten billion trillion, a ratio not generated by an algorithm - which is why skeptics haven’t used probability theory to explain the symmetry.