# Re: probability; check me, someone!

[ Follow Ups ] [ Post Followup ] [ Main Message Board ] [ FAQ ]

Posted by Denis Borris on October 21, 2002 at 21:59:11:

In Reply to: Re: probability; check me, someone! posted by MathBard on October 20, 2002 at 17:28:42:

: : I do not have the answer but I think your denominator should be
: : 6^12 = 2176782336

yep; assume one per second 6^12/(365*24*60*60) = 69 years !

: : 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
: : 6^2 = 36 possibilities for the die.

agree; and the leader is "7" with 6 possibilities; agree?

: : 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.)

repeating: there are 197 "number combinations" that add up to 35.

: TECHNICALLY there are 12 ways to roll a sum of 13.
: any ONE of the twelve dice could roll a two. Here are three...
: 1 1 1 1 1 1 1 1 1 1 1 2
: 1 1 1 1 1 1 1 1 1 1 2 1
: 1 1 1 1 1 1 1 1 1 2 1 1
: etc.
: Here is where i begin to lose it. With THIS in mind there is ONE combo for 12,
: 12 combos for 13, and I THINK 122 combos for 14.

agree with 12 and 13; but for 14, methinks it's 78:
1 1 1 1 1 1 1 1 1 1 1 3 : 12
1 1 1 1 1 1 1 1 1 1 2 2 : 66 ???????? You think it's 110 ?
The way I get 66 is:
the 2nd last 2 can move left spot by spot: 11
the last 2 then moves 1+2+......+10
sum 1 to 11 = 66.

Testing the above "gem" on a shorter one:
1 1 2 2
1 2 1 2
1 2 2 1
2 1 1 2
2 1 2 1
2 2 1 1
That's definitely 6; and 1+2+3 = 6

BUT that'll only work with cases like x x x x ... x x x y y
So that's not nuff for me to get a job at the crap table:)

Name:
E-Mail:

Subject: