Re: probability; check me, someone!

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Posted by MathBard on October 20, 2002 at 17:28:42:

In Reply to: Re: probability; check me, someone! posted by Brad Paul on October 20, 2002 at 16:35:53:

: : : 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.)

In Mr. Boris' example he eliminated identical rolls, Mr. Paul Left them in.

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. I am not going to figure how many different ways to calculate a roll of 35 or a roll of 42 becasue i do nOt know the formula for that, but i bet one of you REAL Mathgeeks could. OR tell me i am completely off base. But that is the difference between Mr.B's and Mr.P's denominators.

MathBard
PS Remember I am an Englishgeek tring to fill the shoes of a Mathgeek 8-)

Follow Ups:

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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: : : : 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.) : In Mr. Boris' example he eliminated identical rolls, Mr. Paul Left them in. : 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. I am not going to figure how many different ways to calculate a roll of 35 or a roll of 42 becasue i do nOt know the formula for that, but i bet one of you REAL Mathgeeks could. OR tell me i am completely off base. But that is the difference between Mr.B's and Mr.P's denominators. : MathBard : PS Remember I am an Englishgeek tring to fill the shoes of a Mathgeek 8-)

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