Numbers question

[Cobra]

Throwing out a numbers question to anyone that can help verify if I have come up with the right answer.

There is a contest running on a local radio station where you have to have certain letters/numbers on the serial number of a $20 bill in order to win. I'm trying to figure out the odds of having a bill with the correct order.

Here's the facts:

Our bills here have 3 letters followed by 7 numbers. The contest specifies that you need the letters KL and the numbers 939 (in that order).

My calculation was as follows:

1 in 26 * 1 in 26 * 1 in 10 * 1 in 10 * 1 in 10 = 1 in 676,000 BUT there are 2 different ways of getting the letters portion right (if the first two letters are not K and L, then the 2nd and 3rd letters could be K and L) Also, there are 5 different ways of getting the numbers in the right order (the first 3 spots, the 2nd, 3rd and fourth spot, etc...until you get to 5th, 6th and 7th spot)

Therefore, you need to divide the 676,000 by 10 (2*5) which gives you the odds of 1 in 67,600 of having a $20 bill with these numbers.

Anyone agree or disagree with what I came up with ? It's been a few years that I've been out of school so not totally sure I'm right on this one.:shrug:

 

[fletcher]

can't help you there,hell i have trouble spelling

i know there are some numbers guys on here but i would look for kotysdad for help he would be the man for this, right in his ball park,i am sure he will see this and try to give you a hand if he has time.

 

[yyz]

Hold the phone........

Of course, I have no cash (perpetually poor!) so i can't look, but are you sure that the bills have 3 letters in the front?

I am almost certain that there are only two. If that is the case, I think you are being had.

The second letter on the serial number represents which Fed Res Dist it came from. (These go from A-L)

The first letter is indicative of what series the note is from. "A" indicates the first series. A "series" occures whenever there is a change in the plate used to mint the bill. (The signitures on the bottom, etc.)

I seriously doubt that we have had enough changes to get that second letter up to the "K-L" mark. I could be wrong, but since the new bills have only been around for about 4 years, it seems that many changes would be unlikely.

I think the radio station is having a good time with the listeners.

 

[Cobra]

I live in Canada so I'm guessing that our bills have a different format of serial numbers than you guys in the States. The contest is definitely legit; I am just trying to prove to the girlfriend that the chances of having such a bill are very slim before she empties out the whole bank account in $20 bills - HA HA.

By the way, the contest is paying out $2,000 for such a bill.....

 

[yyz]

WTF do I know about "funny money"?

 

[KotysDad]

can't help you there,hell i have trouble spelling

i know there are some numbers guys on here but i would look for kotysdad for help he would be the man for this, right in his ball park,i am sure he will see this and try to give you a hand if he has time.

Fletcher,

Thanks for the referral I love these kinds of problems. Right up my alley.

Cobra,

This is one of those problems that will get you pulling your hair out by the end of the problem and if my answer is right it only differs from yours by 329 extra serial numbers. My answer is 1 chance in 68,047. Here is how I did it.

You have the following sequence LLL-NNNNNNN

You are correct in that the KL sequence can occur in 2(26) = 52 ways in the first 3 letter slots. But the hard part is the number sequence.

Lets say you set the 939 in the first 3 number slots and allow the remaining 4 slots to be anything. One of those combinations will give you the following sequence 939939N. Now if you let the 939 occur in slots 4-6 and allow slots 1,2,3,7 to be anything, you can also get the same 939939N.....an overcount...and this is just one scenario where this overcount can occur. So here is what I did:

Forget the letters, got them taken care of. Whatever number we come up with when counting the numbers, we will just multiply it by 52 to get the total number of sequences that you are looking for.

Consider the following 5 sequences.

939NNNN
N939NNN
NN939NN
NNN939N
NNNN939

I am going to start with the top sequence, count how many sequences can occur with that pattern, then record that number. Next, I am going to jump down to the next sequence, count those, but making sure that none of the new sequences already occurred in any of the above sequences. Then I will move down to the next sequence until I am at the last sequence, each time counting the number of sequences with that pattern while making sure they didnt already occur above. Believe it or not, you only overcounted by 319 (assuming I am thinking straight at 2am).

Ok, so start with 939NNNN..........10000 possibilites for the remaining 4 numbered slots.

939NNNN
N939NNN *

Cant be any duplicate sequences here so again the remaining 4 slots can be anything.......10000

939NNNN
N939NNN
NN939NN *

As long as the first two slots arent 93, then we cant get any duplicates with the first sequence. No need to worry about duplicates with the second sequence. So making sure the first two slots are not 93, and the last two can be anything, we have 99*100 = 9900 possibilites for the third sequence.

939NNNN
N939NNN
NN939NN
NNN939N *

Now, the first 3 slots can be any one of 1000 possibilites but we need to remove duplicates. If the first 3 slots are 939 we have a duplicate with the first sequence, so eliminate that one. If slots 2 and 3 are "93" then you have 10 duplicates with the second sequence. Cant have any duplicates with the third sequence. Therefore, we need to remove 11 sequences from the original 1000 leaving 989 sequences. Since the last digit can be anything, you have 989*10 = 9890 sequences of the form NNN939N.

939NNNN
N939NNN
NN939NN
NNN939N
NNNN939 *

Finally, we have the last sequence to consider. The first 4 slots can be any one of 10000 combos to start. Cant have any duplicates with the 4th sequence right above it. Looking at the 3rd sequence, slots 3 and 4 cant be "93". The first two slots can be anything so far, so eliminate the 100 sequences that start NN93. Move up to the 2nd sequence. The last sequence cant begin with N939 or else you have duplicates with the 2nd sequence. So eliminate those 10 sequences since the first digit can be anything. Moving up to the top sequence, the first 3 digits cant start with 939 or else we have duplicates with the top sequence. But dont eliminate all 10 sequences that start 939N because one of those is 9393 which makes the last sequence 9393939 and we already removed that one from the 3rd sequence where the 939 is in the middle, so only remove 9 of the sequences that start 939N. So, all total we are removing 100+10+9 = 119 from the original 10,000...leaving 9881 sequences of the form NNNN939.

So, the total sequences of 7 digits is 10000+10000+9900+9890+9881 = 49671.

Like I said, not far off from the 50000 you got. Only overcounted by 329.

So if you divide the total number of LLL-NNNNNNN sequences by 49671*52, you get odds of 1 in 68,047.

 

[taoist]

...now I know why I got that "C" in statistics class... :shrug: ...and I also know why I keep coming back to MJ's day after day.

...when a man need some serious answers in life, this is the place to be.

p.s. ...very nice explanation from a "numbers man." :toast:

 

[yyz]

All I know? I will be getting 3 doughnuts for a dollar at the Kwik Trip this morning!