# Re: probability; check me, someone!

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Posted by Brad Paul on October 20, 2002 at 16:35:53:

In Reply to: Re: probability; check me, someone! posted by Denis Borris on October 20, 2002 at 16:13:44:

: : In a throw of 12 dice,what is the probability of obtaining a sum of 35 ?

: I'll be as brief as possible, yet try to be clear;
: (hate typing!); there is:
: 1 way of getting 12: all 1's
: 1 way of getting 13: 11 1's and a 2
: 2 ways of getting 14: 11 1's and a 3, 10 1's and 2 2's
: ....
: 184 ways of getting 34
: 197 ways of getting 35**
: 212 ways of getting 36
: ....
: 252 ways of getting 42 (that's tops: so bet 42!!)
: ....
: 1 way of getting 72: all 6's

: These "ways" total 6188.

: So the probability of getting 35 is 197/6188.

: I did printout all of them (12 to 72) but am too
: lazy to type 'em all...

I do not have the answer but I think your denominator should be

612=2176782336

Think of just two die. If you work out a table of all the possibilities
you have a square with 6 entries on each side. Which gives
62=36 possibilities for the die. Here we just have a 12
dimensional cube.

The hard part of the question is: How many ways can one add 12 integers
that can range from 1 to 6 to get a value of 35 where order does make
a difference (because of how the denominator has been calculated.)

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