# How do I go about finding how many different number/letter combinations are possible with the 10-digit codes on the iTunes Pepsi bottles?

March 1, 2004 11:07 PM Subscribe

How do I go about finding how many different number/letter combinations are possible with the 10-digit codes on the iTunes Pepsi bottles? [more]

Is it 10^36? The code is made up of ten letters or number which, in my experience, don't have to only appear once in the code. 0-9 and A-Z makes that 36 possible letters or numbers in each of the ten spaces.

If I haven't just answered my own question, should I instead be thinking about factorials? Or am I simply remembering something from Algebra II that's useless in this case?

There could be as many as 100 million winning codes, assuming Pepsi and Apple planned for that many winners. I just wanted to figure out in the grand scheme of things how many codes were theoretically possible.

Is it 10^36? The code is made up of ten letters or number which, in my experience, don't have to only appear once in the code. 0-9 and A-Z makes that 36 possible letters or numbers in each of the ten spaces.

If I haven't just answered my own question, should I instead be thinking about factorials? Or am I simply remembering something from Algebra II that's useless in this case?

There could be as many as 100 million winning codes, assuming Pepsi and Apple planned for that many winners. I just wanted to figure out in the grand scheme of things how many codes were theoretically possible.

36 options in 10 different slots means: 36^10 = 3,656,158,440,062,976 (call it 3.6x10^15).

posted by brool at 11:23 PM on March 1, 2004

posted by brool at 11:23 PM on March 1, 2004

Factorial (!) would be used if a number or letter should not be repeated, so you would get 36! possible Combinations (36 X 35 X 34 ...) But since in this case apparently there can be repetition 10^36 would be the correct answer. And if there are really 100 million winning codes (10^8) the chances of getting a winner would be 10^8 / 36^10, approximately one in 36 millions.

posted by golo at 12:38 AM on March 2, 2004

posted by golo at 12:38 AM on March 2, 2004

(assuming that all codes are printed - if there were only 100 million bottles made, each with a winning code, then you'd be guaranteed a winner)

also, 36! is only correct if there's no repitition

posted by andrew cooke at 6:20 AM on March 2, 2004

also, 36! is only correct if there's no repitition

*and*the code has 36 digits. in fact it is has 10, so if there were no repetition the number of different codes would be 36!/26! (or 36x35x34x33x32x31x30x29x28x27 - note that there are 10 terms for the 10 digits).posted by andrew cooke at 6:20 AM on March 2, 2004

A minor point: there are probably slightly less. I don't have any caps in front of me, but chances are, they use either the zero or the letter O, but not both, for clarity's sake. probably the same with one and the letter I.

posted by GeekAnimator at 7:02 AM on March 2, 2004

posted by GeekAnimator at 7:02 AM on March 2, 2004

I haven't seen the codes, so I can only speak from unrelated experience, but sometimes when working with codes (or license numbers) companies will exclude certain letters or numbers, such as 0 and O or I,l and 1 to prevent confusion. and reducing the total number of possible combinations.

Pepsi may have made one or more of the numbers a check digit (reducing further the number of possible combinations) to confirm the winning number. That way they only need to keep a record of 1 (or a few) algorithms for calculating the check digit based on the other digits in the code, instead of keeping 100 million records in the database.

As prizes are clamed, the used winning numbers could then be added to the db to provide a positive denial file, where prize codes that pass the algorithms are checked, and if there are in the database, deny a second claim on that number.

posted by Mahogne at 7:02 AM on March 2, 2004

Pepsi may have made one or more of the numbers a check digit (reducing further the number of possible combinations) to confirm the winning number. That way they only need to keep a record of 1 (or a few) algorithms for calculating the check digit based on the other digits in the code, instead of keeping 100 million records in the database.

As prizes are clamed, the used winning numbers could then be added to the db to provide a positive denial file, where prize codes that pass the algorithms are checked, and if there are in the database, deny a second claim on that number.

posted by Mahogne at 7:02 AM on March 2, 2004

just for reference, if you don't win, there is no code.

you only receive a code on a winning cap.

odds are supposed to be 1-in-3 on marked 20-oz soda bottles.

posted by fishfucker at 7:18 AM on March 2, 2004

you only receive a code on a winning cap.

odds are supposed to be 1-in-3 on marked 20-oz soda bottles.

posted by fishfucker at 7:18 AM on March 2, 2004

I seem to get more than 1:3, but that might be confirmation (observation?) bias talking.

posted by callmejay at 8:54 AM on March 2, 2004

posted by callmejay at 8:54 AM on March 2, 2004

Thanks, shepd and brool. I was pretty sure x^y or y^x was the trick to use, but I had it backwards.

posted by emelenjr at 11:58 AM on March 2, 2004

posted by emelenjr at 11:58 AM on March 2, 2004

And GeekAnimator, you're probably right. I haven't seen a code with a 0 and O both in it. Actually, I don't think I've seen a zero at all. I wonder whether the code input page will accept either 0 or O in those cases. I've always used the letter instead of the number because I expected the number to have a line through it so as to differentiate between the two.

posted by emelenjr at 12:01 PM on March 2, 2004

posted by emelenjr at 12:01 PM on March 2, 2004

I am sorry to write off topic, but HEY MAHOGNE! I have always wanted to thank you for your riffing in the Cat-Scan thread, but I thought you were long gone. Anyway, thanks!

posted by onlyconnect at 5:29 PM on March 2, 2004

posted by onlyconnect at 5:29 PM on March 2, 2004

This thread is closed to new comments.

We know that for three digits the lowest number is 000 and the highest is 999, and we know for certain that means there's 1000 combinations. Which is... 10 ^ 3.

So, the answer is (options) ^ (numbers), or 36 ^ 10.

This can also be shown for binary numbers. It is well known an 8 bit number has 256 combinations, which is 2 ^ 8.

(And I wonder how I passed math, too)

Factorials are for figuring out things where you remove previous possibilities, like playing lotteries, IIRC.

posted by shepd at 11:22 PM on March 1, 2004