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# Are these bills special? Answered

I got 2 bills back in change today at Dunkin Donuts. Here are the serial numbers:

22077552

and

22077551

They are 1 digit away.

Also, there are 3 doubles in each (22, 77, 55). Are these worth anything or are they special in any way? **PLEASE REPLY*

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## Discussions

Oh lol those kinds of things are sometimes very entertaining. There was this one guy that called my friends phone so I called him and was speaking with some weird accent and we had a couple of very entertaining conversations. He told me he really didn't like kids because they got their sticky hands all over stuff.... It was hilarious

I'm not sure but what denomination were the bills? \$1, \$2, fives, tens, etc. A \$3 bill would be worthless. They were probably just were consecutive bills printed in the same batch that was just gotten from the bank. Now if the serial numbers were 13371337, that would be, you know...

Here's a lesson in [probability]. The serial numbers on U.S. dollar bills have eight digits. That means there's a total of 100 million (108) possible serial numbers. How many of those have three doubles (adjacent identical digits)?

Let split this problem in two halves. First, suppose you already have one SN with three doubles. How many ways can you shuffle around the numbers and still have three doubles? You've got five "entities" (one, two, three pairs, plus the two extra digits), so the answer is 5*4*3*2*1 = 120 different arrangements.
NOTE We assume here that each entity is distinguishable; otherwise some of those arrangements would result in 3-of-a-kind or 4-of-a-kind blocks, which I think that you don't want to count.

So now we can restrict ourselves to asking how many three-doubles can we make for just one specific arrangement. We multiply that answer by 120 to get the total. As above, there are five slots to be filled with digits. The first one can be any digit from 0 to 9; the second any of the nine remaining digits; the third any of the eight that are neither the first nor second; and so on. The total is 10*9*8*7*6 = 30240.

So the final total of "triple-double" serial numbers, all with different digits, is 30240*120 = 3628800, or 3.6% of the total number of dollar bills.

Now, keep in mind that the "actual" number is even higher. After all, there are combinations where the digit groups don't have to be unique, for example "55855855" is a triple-double, with repeated digit groups. I leave it as an exercise to the reader to go through those special cases and work out their rates.

Not that special after all, are they?

Pretend the digit groups are playing cards. You've already been dealt five cards. How many different ways can arrange them in your hand? Well, for the first card you can pick any one of the five; for the second, you have to pick one of the four remaining; for the third, you have to pick one of the three that are left; for the fourth card, you only have two to choose from; and the last card is forced on you 'cause it's the only one left.

So if you have five "things", the number of ways you can arrange them is 5 * 4 * 3 * 2 * 1 = 120.

This kind of probability stuff is just counting and multiplication. Do you play RPG's (like D&D, Chill, Deadlands, etc.)? What we're doing here is exactly what you do to work out your odds when rolling 4d6 or 2d10 or 3d8 or whatever.

If you require the four pairs to be in sequence, then yeah, that's extremely rare -- just seven possible SN's out of 100 million:
`   00112233    11223344    22334455    33445566   44556677    55667788    66778899`
(or 14 possible if you accept them in reverse sequence :-).

If you don't require sequential pairs, then you can compute the number the same way I did before: (4*3*2*1) * (10*9*8*7) = 120,960 or 0.12%. Rarer than the triple-doubles, but still a large number :-)