I'm surprised no one got interested in my post about
L03160387B 1963A
L03166965B 1963A
Think about it: two of the small number of bills that went to auction, end up being successive serial numbers in the FBI sorted list. I wasn't joking about that.
EDIT: I've expanded this to show that 6 of the auction bills are from the same single FBI ransom list page! i.e. 6 out of 300 on one of the 34 pages of the ransom list)
links to auction photos for closer examination: (create login and sign in to get highest res photo. Can't post hi res here because of size limitations) ....
Even crazier is the # of auction bills that are present from that page of the FBI ransom list: 6 in total. Out of 300 bills on that page. That strongly suggests the original bill distribution was not random. Right?
the two bills above, which are successive:
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Login The other 4 auction bills that are on the same page 168 of fbi file 55
L01781113A 1969
L01842041A 1969
L02882111B 1963A
L06832736A 1969
links for these 4
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1) Bills weren't as randomized as we believe. It can be very hard to determine if a distribution is random or not. Dilbert covered this well. See attached.
2) An error in the ha.com determination of the fragment serial number? I've not re-examined the pic I posted closely
3) The calculation of the probability of this happening, is complicated to determine. First: we don't know if the original bills actually were randomly distributed in packets among all the 9998 bills. I have a suspicion that certain groups of bills were more likely to be in a single packet. Like maybe star notes were in a packet? Not sure.
Too bad we don't have the original 1 can of microfilm referred to in the FBI files, that had the $230,000 worth of bills. That had some order. I wonder if the order in that microfilm matched the order within each bill stack (strapped or otherwise). Since start/end pairs (serials) were recorded, and that was used to "lookup" groups of bills in the microfilm, that would mean the start/end pairs were ordered amongst themselves (to determine the bills between a start/end, the start/end had delineate a "group" of bills that were likely to be ordered? It's hard to imagine how a start/end would delineate an unordered group, int erms of how the microfilm was likely created with restrapping post-recording?...which raises the possibilty that all the bills in the microfilm order, were ordered in the same order in the packets Cooper received
Although FBI had difficulty figuring out which $30k to remove from the original list of $230k, they originally "thought" just 15 packet start/end pairs, for the extra $30k, (15 packets of 100 bills) where sufficient to determinate what entire set of 1500 bills to remove from the original list.
Which makes me think microfilm order may have matched "delivery to Cooper" order.
There's no way to know for sure. It is tantalizing to ponder if the order was recorded and preserved on delivery to Cooper, Then it would have been possible to see if any of the Ingram bills in a corroded "packet" were out of order relative to the order cooper received packets.
4) I've been pondering how to calculate the probabilty of the auctioned bills having a pair of bills that are successive in the sorted FBI list, assuming random distribution of bills in packets originally
I think it's similar to "birthday paradox" calcs, except using 9998 bills rather than 365 days. And calculating probably of occurence assuming 300 random bill choices. That would be a close approximation.
I've looked and one interesting thing is that if you remove the leading and trailing letters from all the ransom serials, the numbers are unique.
I'm guessing though, that the probability is low, of the sequence occurence. Hard to explain. A non-random original distribution of bills to packets, would make it more likely, and is probably what happened, contrary to the "fully random" assumption that's currently told about the bills.
So I think this is a new train of thought on what the bills can tell us.
Because of the number of unique serials (9998) compared to 365 days in the year, I think the probability of this event is much less than the birthday paradox, for the set of bills that were auctioned.
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LoginIn probability theory, the birthday problem or birthday paradox concerns the probability that, in a set of n randomly chosen people, some pair of them will have the same birthday. In a group of 23 people, the probability of a shared birthday exceeds 50%, while a group of 70 has a 99.9% chance of a shared birthday. (By the pigeonhole principle, the probability reaches 100% when the number of people reaches 367, since there are only 366 possible birthdays, including February 29.)
These conclusions are based on the assumption that each day of the year is equally probable for a birthday. Actual birth records show that different numbers of people are born on different days. In this case, it can be shown that the number of people required to reach the 50% threshold is 23 or fewer.[1]
The birthday problem is a veridical paradox: a proposition that at first appears counterintuitive, but is in fact true. While it may seem surprising that only 23 individuals are required to reach a 50% probability of a shared birthday, this result is made more intuitive by considering that the comparisons of birthdays will be made between every possible pair of individuals. With 23 individuals, there are (23 × 22) / 2 = 253 pairs to consider, which is well over half the number of days in a year (182.5 or 183).
Real-world applications for the birthday problem include a cryptographic attack called the birthday attack, which uses this probabilistic model to reduce the complexity of finding a collision for a hash function, as well as calculating the approximate risk of a hash collision existing within the hashes of a given size of population.