http://mcadams.posc.mu.edu/factoid.htm
Likewide, it's considered an unlikely "coincidence" that, when Oswald put a phony "office address" on a pamphlet of his phony "Fair Play for Cuba Committee" it was the building (although not the address) where Guy Banister had an office. And the Banister connection also establishes a "connection" with David Ferrie, who hung around Banister's office in the summer of 1963.
Just how unlikely are such "coincidences?"
To simplify this question a bit, let's ignore leap years and assume that each year has 365 days.
If there are two people in a room, the odds of their having the same birthday are one in 365.* The probability is therefore 0.0027. That's pretty unlikely.
Suppose you have three people in the room: A, B and C.
There are three possibilities for two of them to have the same birthday. A and B might. A and C might. Or B and C might.
What is the probability that two of the three will have the same birthday? We have to start by noting that the probability of A and B not having the same birthday is 364/365 = 0.99726.
Thus the probability of A and B not having the same birthday and B and C also not having the same birthday is .99726 x .99726 = 0.994528.
And it follows that probability of A and B not having the same birthday and B and C also not having the same birthday and A and C not having the same birthday is .99726 x .99726 x .99726 = 0.9918.
Following this logic, when there are five people in the room there are ten possibilities for two of them to have the same birthday (4 + 3 + 2 + 1 = 10), and if there are six people in the room there are 15 possibilities (5 + 4 + 3 + 2 + 1 = 15).
In this latter case (six people in the room), the probablility of none of the six having the same birthday is:
(364/365)15 = 0.959683
Therefore there are about four chances in a hundred that at least two of the people will have the same birthday.
Since the number of possibilities of two people having the same birthday increases roughly as the square of the number of people, the probability of at least two having the same birthday rises rapidly as the number in the room increases. With 20 people in the room, there are 190 opportunities for two people having the same birthday. The probability that no two will is:
(364/365)190 = 0.59377
Repeating the analysis with 30 people, the probability is:
(364/365)435 = 0.30318
In other words, the odds are only three in ten that no two people will have the same birthday.
But the situation is even worse when we go beyond requiring solid face-to-face connections between two people who had a "connection" with Oswald. Thus, David Ferrie had a (very tenuous) "connection" with Oswald, and a tenuous "connection" with Jean Aase. Aase, in turn, had a real connection with Lawrence Meyers who had a real connection with Jack Ruby who had a real connection with Lee Oswald: he shot and killed him.
Given that each of these people had equally close "connections" with at least dozens, and sometimes hundreds of people, there had to be hundreds of thousands, and probably millions, of people with as close a connection to Oswald as Ruby had through Meyers and Aase and Ferrie.
Let's assume, for example, that there were a hundred people with a "connection" to Oswald as close as Ferrie's, and each of them (during the months before the assassination) called a hundred different buildings like the one Aase lived in, and there were 200 people on average in each of those buildings, and each of the 200 people in each building knew 50 people like Meyers, and each of those "persons like Meyers" knew 100 people, then the total number of people in the network is:
100
plus . . .
100 x 100 = 10,000
plus . . .
10,000 x 200 = 2,000,000
plus . . .
2,000,000 x 50 = 100,000,000
plus . . .
100,000,000 x 100 = 10,000,000,000
totals:
10,102,010,100
This, of course, is several times the population of the entire earth in 1963.
This analysis is simplistic because of multiple paths leading to the same individual. This can be dealt with by specifying how many new people (not already in the population "connected with Oswald") are added at each step. The proportion of "redundant links" will be large early in the chain — since many of the people connected directly with Oswald will be connected with each other. It will then become smaller as the chain of connections moves out to different regions and different social circles. And it will probably become larger again as the number of people at the end of the chain becomes a substantial proportion of the entire population of the United States. At this point, any given person at the end of the chain is quite likely to also be at the end of another chain — or two or three or five.
Let's redo the above example, assuming that that there were a hundred people with a "connection" to Oswald as close as Ferrie's, and each of them (during the months before the assassination) called fifty different buildings like the one Aase lived in that had not already been called by someone among the hundred people "connected with Oswald." Further, let's assume that there were on average 100 people in each of those buildings who had no connection with Oswald nor with any of the first hundred people in the chain. Assume that each of the 100 people in each building knew 25 people like Meyers (not already "connected"), and each of those "persons like Meyers" knew 25 people not already connected by a chain of "connection" with Oswald. The total number of people in the network is:
100
plus . . .
100 x 50 = 5,000
plus . . .
5,000 x 100 = 500,000
plus . . .
500,000 x 25 = 12,500,000
plus . . .
12,500,000 x 25 = 312,500,000
totals:
325,505,100
If there are 325 million people in the network, it's not surprising that one of them is Jack Ruby. The implication is obvious: literally millions of people were "connected" to Oswald with a chain of "connections" no more tenuous than the one that linked Oswald to Ruby through Ferrie, Aase, and Meyers.
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