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.)

Follow Ups:

Re: probability; check me, someone! MathBard 17:28:42 10/20/02 (7)
- Re: probability; check me, someone! Denis Borris 21:59:11 10/21/02 (6)
  - Re: probability; check me, someone! MathBard 22:43:41 10/21/02 (5)
    - 74,005,152 for "35" confirmed... Denis Borris 02:05:32 10/22/02 (4)
      - the formula.... Denis Borris 13:36:29 10/22/02 (3)
        
        AND HERE TIZZ: the full thing! Denis Borris 18:54:17 10/22/02 (2)
        
        IMPRESSIVE so what ARE the odds of rolling 35 on 12 dice? MathBArd 22:50:28 10/22/02 (1)
        
        not very high..... Denis Borris 00:29:09 10/23/02 (0)

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Comments:
: : : 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 : 6<sup>12</sup>=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 : 6<sup>2</sup>=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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