New Maryland License Plates

Pavel314 · science forum addict Joined: 29 Apr 2005 Posts: 78

Until recently, automobile license plates in Maryland were generally a group
of three letters and a group of three numbers. Recently, they've jumbled
them all up and now have the format [(number-two letters) and (letter-two
numbers)]. These are rather annoying as you could break the old three-digit
numbers into prime factors while stuck in rush hour traffic. Or play Semitic
words with the letters, inserting any vowels necessary to make an English
word.

I came up with a new game today; see if the two-digit number at the end of
the new format plate was divisible by the one-digit number at the beginning.
There are 9 one-digit numbers, as they don't start with zero, and 100
two-digit numbers, 00 through 99. To compute the odds, I first considered
the number of two-digit numbers divisible by a given one-digit number and
came up with the following:

First Digit ........ Number of Two Digits Divisible By It

1 100
2 50
3 33
4 25
5 20
6 16
7 14
8 12
9 11

Since each one-digit number is equally likely to occur, the odds of a
divisible two-digit number would be sum of 1/9 times the number of possible
divisible numbers. (100+50+33+....+11) / 9 = 281 / 9= 31.2222 out of 100 or
31.2%. Does anyone see an error in the logic thus far?

On the way home from work tonight, I only saw 3 out of 15, or 20%, divisible
new plates. Seems a bit low against my calculations. I'll try next week
after the New Moon on Sunday to see if the observed situation is closer to
the predicted results.

Paul

Pavel314 · science forum addict Joined: 29 Apr 2005 Posts: 78

Update: Saturnday morning's tally was 4 out of 21 divisible for 19%.

"Pavel314" <Pavel314@NOSPAM.comcast.net> wrote in message
news:4fOdnWp3-o20FKPeRVn-uw@comcast.com...

Dan Akers · science forum addict Joined: 19 Jul 2005 Posts: 56

Paul wrote;
"Until recently, automobile license plates in Maryland were generally a
group of three letters and a group of three numbers. Recently, they've
jumbled them all up and now have the format [(number-two letters) and
(letter-two numbers)]. These are rather annoying as you could break the
old three-digit numbers into prime factors while stuck in rush hour
traffic. Or play Semitic words with the letters, inserting any vowels
necessary to make an English word.
I came up with a new game today; see if the two-digit number at the end
of the new format plate was divisible by the one-digit number at the
beginning. There are 9 one-digit numbers, as they don't start with zero,
and 100 two-digit numbers, 00 through 99. To compute the odds, I first
considered the number of two-digit numbers divisible by a given
one-digit number and came up with the following:
First Digit ........ Number of Two Digits Divisible By It
1 100
2 50
3 33
4 25
5 20
6 16
7 14
8 12
9 11
Since each one-digit number is equally likely to occur, the odds of a
divisible two-digit number would be sum of 1/9 times the number of
possible divisible numbers. (100+50+33+....+11) / 9 = 281 / 9= 31.2222
out of 100 or 31.2%. Does anyone see an error in the logic thus far?
On the way home from work tonight, I only saw 3 out of 15, or 20%,
divisible new plates. Seems a bit low against my calculations. I'll try
next week after the New Moon on Sunday to see if the observed situation
is closer to the predicted results."
_____________________________________
Re;
There are 9 X 100 possible one digit/two digit combinations, with only
281 divisible by it's one digit number; so yes I would agree that the
probability of the two digit number being divisible by the single digit
number, without a remainder, is 281/900=31.2%.
Your total sample size, for the two outings is 36; the expectation is
that you would have found 11 divisibles in that sample. However, you
found only 7. The probability of deviating on either side of the
expected 11 by 4 or more, is about 14%. But the sample size is rather
small yet. I calculate you would need to sample 82 cars in order to
confirm your hypothesis to within 10% with 95% confidence.
One red herring may be that certain number/letter sequences are not
allowed given that some numbers such as "4" and "2" and "69" can be used
as implied words/situations in unsavory permutations that might arise in
the printing of the random number/letter sequences. I can't think of
any right off hand, but...

-Dan Akers

Pavel314 · science forum addict Joined: 29 Apr 2005 Posts: 78

"Dan Akers" <digikey@webtv.net> wrote in message
news:9587-433F144A-645@storefull-3134.bay.webtv.net...
Paul wrote;
"Until recently, automobile license plates in Maryland were generally a
group of three letters and a group of three numbers. Recently, they've
jumbled them all up and now have the format [(number-two letters) and
(letter-two numbers)]. These are rather annoying as you could break the
old three-digit numbers into prime factors while stuck in rush hour
traffic. Or play Semitic words with the letters, inserting any vowels
necessary to make an English word.
I came up with a new game today; see if the two-digit number at the end
of the new format plate was divisible by the one-digit number at the
beginning. There are 9 one-digit numbers, as they don't start with zero,
and 100 two-digit numbers, 00 through 99. To compute the odds, I first
considered the number of two-digit numbers divisible by a given
one-digit number and came up with the following:
First Digit ........ Number of Two Digits Divisible By It
1 100
2 50
3 33
4 25
5 20
6 16
7 14
8 12
9 11
Since each one-digit number is equally likely to occur, the odds of a
divisible two-digit number would be sum of 1/9 times the number of
possible divisible numbers. (100+50+33+....+11) / 9 = 281 / 9= 31.2222
out of 100 or 31.2%. Does anyone see an error in the logic thus far?
On the way home from work tonight, I only saw 3 out of 15, or 20%,
divisible new plates. Seems a bit low against my calculations. I'll try
next week after the New Moon on Sunday to see if the observed situation
is closer to the predicted results."
_____________________________________
Re;
There are 9 X 100 possible one digit/two digit combinations, with only
281 divisible by it's one digit number; so yes I would agree that the
probability of the two digit number being divisible by the single digit
number, without a remainder, is 281/900=31.2%.
Your total sample size, for the two outings is 36; the expectation is
that you would have found 11 divisibles in that sample. However, you
found only 7. The probability of deviating on either side of the
expected 11 by 4 or more, is about 14%. But the sample size is rather
small yet. I calculate you would need to sample 82 cars in order to
confirm your hypothesis to within 10% with 95% confidence.
One red herring may be that certain number/letter sequences are not
allowed given that some numbers such as "4" and "2" and "69" can be used
as implied words/situations in unsavory permutations that might arise in
the printing of the random number/letter sequences. I can't think of
any right off hand, but...

-Dan Akers

Dan,

I recorded 162 license plates by using a small tape recorder on the comute
to and from the office, and by walking around the parking lots at work with
a clipboard at lunch time for a couple of days. My wife was sure that
security would stop me but the guards already recognize me as a harmless
eccentric. This sample gave 29.6% divisibility. Some of the lead digits
yielded more or less divisibility than theory would indicate but they were
close enough for their sub-sample size. I am now assured that the Department
of Motor Vehicles is not engaged in some sinister plot via the new license
plate format.

Thanks for your guidance on this project.

Paul

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