A big warning:
**always follow Murphy's laws to their full extent!**

I was working on a php application which uses log-ins.
The client wanted some friendly way to have users remember their password, maybe something with a combination of real words, -fine- said I.
As a prototype I prepared a simple script which used the /dict/words file, nothing over the top, the script would output a combination of two hyphenated word, so that the guy could play with it and tell me if he liked it.
Would you ever believe that the first two words the script produced where sh**-head?
Need I to say that the guy was NOT pleased at all?
Anybody fancies calculating the chances of that happening?

Well, wc says mine contains 45427 words. So, any two-word combination should be 1-in-(45427-squared), or 1-in-2-billion-ish. It's the same odds for "soda-pop" as it is for "dog-sh**".

According to the California Lottery page, your chance of winning the big pay-off are 1-in-41-million-ish. << http://www.calottery.com/games/super...lus/howtoplay/ >>

Go buy a lottery ticket. Great story, I just told everyone here at the office.

Got my Lotto ticket

Apologies for this, my math is a bit rusty, you used something like:
Dn,k=n.(n-1)....(n-k+1) where k=2 and n=45427 to calculate it? Dunno what is called in english.

No, I just used a simple squaring... If there were only five words [ A,B,C,D,E ], then all possible combinations would be...

A-A ,A-B,A-C,A-D,A-E
B-A,B-B ,B-C,B-D,B-E
C-A,C-B,C-C ,C-D,C-E
D-A,D-B,D-C,D-D ,D-E
E-A,E-B,E-C,E-D,E-E

If you are picking completely at random, it is possible to pick the same word twice, and thereby the number of combos is N-squared. I didn't bother to put it on the calculator, but it should be about 2.064-billion-ish.

If you are removing the first word picked, then you would need to subtract one from N for the second word, and thereby the number of combos would be N(N-1), or NN-N, or 2.064-billion-ish minus 45427, which is still 2.064-billion-ish. Remove the red combos above.

LOL, i cant seem to stop laughing at this,... just you luck eh?

/me is reminded of a Dilbert comic strip

Yes, the combination did not include repetition, the idea was to have unique values, althought the positioning of the words would have been random, that way A-B != B-A although A and B are the same words.
The D(n,k) would be D(n,k)=45427*(45427-1) I think, which would be fairly close to the square of the number.

Cheers!

I have removed all the "naughty"words from the word list so that it should give less "entertaining" results. Was thinking of setting up a cron to run the program and save the result to db, and then check to see what comes out of it. You know the thing about monkeys, typewriters and eternity