Microsoft running out of XP product keys

  WhiteTruckMan 23:47 13 Aug 07
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Apologies to anyone who thinks this is old news, but its new to me. I wonder how many (legit) editions are actually out there? I suppose while the possible range of keys in the current set of characters is very large indeed, only a relatively small number will actually be valid keys. Otherwise when it came to entering the key on installation you could put in any old rubbish.

WTM

  Stuartli 09:56 14 Aug 07

Must be a vast number of Product Keys in circulation.

The maximum number available is easily worked out (with a calculator!)

Numbers represent 10 choices and letters 26. Starting with, for instance, a letter and then a number would be 26x10 and so on for the first block of five letters and numbers.

This carries on with the subsequent blocks, multiplying by 10 or 26 as appropriate. The calculator may not be able to handle the full range...:-)

  johndrew 11:59 14 Aug 07

All a bit like telephone numbers really. BT never realised they would be so popular and needed to add the odd digit or two!!!!!!

  interzone55 13:41 14 Aug 07

Assuming a 25 digit code consisting of numbers & letters, the total available licence keys is somewhere in the region of 808281277464764000000000000000000000000.

Which is, give or take the odd billion 124350965763810000000000000000 for every man, woman & child on earth.

  DieSse 13:51 14 Aug 07

Except that licence keys are not exactly like that.

Normally the last group is a mathematical calculation based on the other groups - I don't know how the XP licences are done, but one original method was a "half-add" - an addition but ignoring the carries.

These days they're commonly called Check Digits - and used as a self-verification for validity.

Also some parts of the licence keys are used to identify versions - Language, OEM or End-User, Home or Pro, Which revision level, etc. Once again I don't know exactly how they are used.

So this cuts down the actual numbers available for purely the serial number.

  bjh 15:31 14 Aug 07

There is, indeed, a surprisingly small number of VALID codes that are possible. The last set of five characters is a checksum of the previous four sets, making them redundant as far as the calculation goes. (I believe that the last digit of each previous set also acts as a checksum, but that seems a little pointless to me). The vast majority of codes for the four remaining sets are invalid, as the characters are not independent of each other. If you change one character, the code is invalid, (excluding the checksum element that would invalidate that anyway).

My guess, and it is only an educated guess is that about 4/36ths of each character combination are likely to be valid. That needs to be multiplied (downsized) as each additional group is considered. I think it comes out lower than the 1 billion mark, maybe considerably lower.

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