| Title: | Mathematics at DEC |
| Moderator: | RUSURE::EDP |
| Created: | Mon Feb 03 1986 |
| Last Modified: | Fri Jun 06 1997 |
| Last Successful Update: | Fri Jun 06 1997 |
| Number of topics: | 2083 |
| Total number of notes: | 14613 |
Put your mind to the test...
You want to find two US $1 bills with the same serial number.
Impossible you say!! Maybe...
Each $1 bill has an 8 digit numeric serial number with an alpha prefix and
suffix. If you were to have the first bill printed with the A letter prefix
you would have bill A 00000000 and if you were to have the first bill printed
with the B letter prefix, B 00000000, you have succeeded. Yet the chances of
getting these two bills would be astronomical. So you look for an easier
goal. If you were to look for bills whose individual digits inside the
numeric serial number added up to 1 you would have 8 bills under A series.
00000001 00000010 00000100 00001000 00010000 00100000 01000000 10000000
Your odds still aren't very good of getting an A and B series with the
serial number of 10000000. How about summing the serial number to 2?
00000011 00000101 00000110 00001001 00001010 00001100 00010001 00010010
00010100 00011000 00100001 00100010 00100100 00101000 00110000 01000001
01000010 01000100 01001000 01010000 01100000 10000001 10000010 10000100
10001000 10010000 10100000 11000000 00000002 00000020 00000200 00002000
00020000 00200000 02000000 20000000
Still pretty poor odds with all the $1 bills out there. Hey, there are
collectors, destroyed or lost bills, bill hoarding, misprints(*) and a host
of other reasons that could foil the attempt.
So which serial number sum should you look for. Since there are as many
bills with the number 00000000 as their are with the number 99999999, middle
ground would be nice. 9X9=81. � of 81 = 40.5 (middle ground). Bills with the
sum of 40 and 41 should have the same number of bills printed but how many and
what are the odds of holding ,simultaneously, two bills with the same serial
number, having only the alpha prefix different?
I believe there are 12 issued alpha prefixes - A through L. This betters
your odds by 12 (year of mint not considered a variable).
| T.R | Title | User | Personal Name | Date | Lines |
|---|---|---|---|---|---|
| 1408.1 | poor man's consolation | STATOS::BUCHANAN | Holdfast is the only dog, my duck. | Mon Apr 01 1991 16:07 | 14 |
>You want to find two US $1 bills with the same serial number. >Each $1 bill has an 8 digit numeric serial number with an alpha prefix and >suffix. If you were to have the first bill printed with the A letter prefix >you would have bill A 00000000 and if you were to have the first bill printed >with the B letter prefix, B 00000000, you have succeeded. Yet the chances of >getting these two bills would be astronomical. Not so fast. Surely the birthday paradox will apply on any billfold. How wealthy do you have to be so that with probability > � you've two bills with the same number, if you hold your wealth entirely in $1 bills? Do you have to be a millionaire? Regards, Andrew. | |||||
| 1408.2 | GUESS::DERAMO | duly noted | Mon Jun 24 1991 23:42 | 40 | |
> Not so fast. Surely the birthday paradox will apply on any billfold.
> How wealthy do you have to be so that with probability > � you've two bills with
> the same number, if you hold your wealth entirely in $1 bills? Do you have
> to be a millionaire?
If there are n equally likely "birthdays" and you
randomly select k of them (with replacement), then the
probability of them all being different is
P = (n/n) * ((n-1)/n) * ((n-2)/n) * ... * ((n-(k-1))/n)
= (1 - 0/n) * (1 - 1/n) * ... * (1 - (k-1)/n)
Taking logarithms and using the series
ln (1 - x) = -(x + (x^2)/2 + (x^3)/3 + ...)
and dropping terms in (1/n)^2 or beyond yields
- ln P = 0 + (1/n + (1/2)(1/n)^2 + ...) + ... + ((k-1)/n
+ (1/2)((k-1)/n)^2 + ...)
= (1 + 2 + ... + (k-1))/n + (1/2)(1^2 + 2^2 + ...
+ (k-1)^2)/(n^2) + ...
= k(k-1)/(2n) + terms in 1/(n^2)
If we want P <= 1/2, then we need - ln P >= ln 2 and so want
approximately k(k-1)/2n >= ln2. This gives
k^2 - k + 1/4 >= (k - 1/2)^2 = 2 n ln 2 + 1/4
or
k >= 1/2 + sqrt(2 n ln 2 + 1/4)
k >= approx sqrt( 2 n ln 2 ) = 1.1774100225... sqrt(n)
If the n "birthdays" are 100,000,000 eight digit serial
numbers, then you get probability P <= 1/2 of k of them
being different when k >= around 12K.
Dan
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