Re: Help with calculating an interesting probability
- From: "Michael" <mchlgibs@xxxxxxx>
- Date: 22 Mar 2007 13:46:54 -0700
What's the probability that from this conversion the 6 characters
would spell a 6 letter word?
So we have an alphabet consisting of 36 characters. And there is a random
process (this is me assuming the number being converted has no restrictions)
that picks 6 letters and strings them together.
First, what is the probability of the process randomly selecting six
characters which could form an offensive word?
Let's just pretend the word is "NUMBER".
p(#) - probability of attaining characteristic "#"
p(n)=1/36
p(u)=1/36
p(m)=1/36
p(b)=1/36
p(e)=1/36
p(r)=1/36
So, the probability of attaining these six letters is:
(1/ (36^6))
Now there is the problem of getting them in the right order. How many ways
can {n, u. m, b, e, r} be ordered? i,e: {nmbuer, rebmun, ...}.
6*5*4*3*2*1=720
So the probability of ordering them "offensively" is:
1 / 720
Actually, the 720 is wrong. The probability of getting "NUMBER" is 1/
(36^6). Period. Because, as you've stated above, the probability of
getting N as the first character is 1/36, U as the second is 1/36,
etc.
You could do the calculation where you figure out the probability of
getting any of the letters as the first one, then as the second, etc.,
but that calculation would essentially give you 1/(36^6) * 720 / 720
(although probably not in so obvious a form).
1/(36^6) ~= 4 x 10^-10.
But now assume there are 1000 offensive words, then the probability of
getting one of them is 4 x 10^-7 (still less than one in a million).
There is one further wrinkle, though. Your original problem involved
a 10-digit number converted to base 36 - there are 10,000,000,000 10-
digit numbers, but only 2,176,782,336 6-base 36 numbers. Presumably
that means either 1) some of your numbers are 7-base 36 numbers, or 2)
numbers repeat, in which case your probability is about 5x greater.
To answer your question, though, it is very unlikely by chance.
Michael
.
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