Math Is Fun Forum

clooneyisagenius

Member

Registered: 2007-03-25

Posts: 56

Summation help.

Problem: When rolling n dice, what is the probability that the sum of the numbers rolled is even.

To answer this I found that:   
1. There always has to be an even number of odd dice.

For example, if n=10, there has to be 0,2,4,6,8, or 10 odds for the outcome to be even.
0 odds = (10 choose 0)*3^10
2 odds = (10 choose 2)*3^2*3^8
4 odds = (10 choose 4)*3^4*3^6
6 odds = (10 choose 6)*3^6*3^4
8 odds = (10 choose 8)*3^8*3^2
10 odds = (10 choose 10)*3^10*3^0
Therefore, for any number there must be (n choose k)*(3^n)*(3^(n-k)). but it only works when k=even. Then the number must be divided by 6^n since that's the total number of possibilities there were. And the answer WILL BE 50%.

My question:
How do I find the sum of all of the possible outcomes?  I know that if it were for all k (not just evens) it would be:
   
∑(from k=0 to n) (n choose k)*(3^n)*(3^(n-k)) but I'm not sure how to do it for only k's being even.
Any ideas?
THANKS!

Last edited by clooneyisagenius (2007-04-05 14:59:39)

Stanley_Marsh

Member

Registered: 2006-12-13

Posts: 345

Re: Summation help.

If u can prove that the set of the Sum is equal to the set of all positive integer.  Then the probability of getting a random even number is 50%.  I am not sure tho, it's just my opinion.

Last edited by Stanley_Marsh (2007-04-05 14:29:37)

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Numbers are the essence of the Universe

clooneyisagenius

Member

Registered: 2007-03-25

Posts: 56

Re: Summation help.

I found that if I did:

∑(n choose k)*(3^n)*(3^(n-k)) :: (sum from k=0 to n)

Then divided that by 2.
Then divided that by 6^n.

And it gets 50% for all values of n. But not sure why it would work to divide by 2?
Any ideas about that? or the question above? :-)

pi man

Member

Registered: 2006-07-06

Posts: 251

Re: Summation help.

I'm not exactly following what you're trying to do but shouldn't it be:

∑(n choose k)*(3^k)*(3^(n-k)) :: (sum from k=0 to n)

I changed the 3^n to 3^k.

And then you can simplify:

∑(n choose k)*(3^n) :: (sum from k=0 to n)

clooneyisagenius

Member

Registered: 2007-03-25

Posts: 56

Re: Summation help.

Yes you are right pi man... I made a mistake... In my notes i did write 3^k... Made an accident here.  But I see what you are saying with:

∑(n choose k)*(3^n) :: (sum from k=0 to n)

But I think that it will be
∑(n choose 2k)*(3^n) :: (sum from k=0 to n/2)

and then divide that by 6^n to get .5?

Last edited by clooneyisagenius (2007-04-05 17:20:36)

Stanley_Marsh

Member

Registered: 2006-12-13

Posts: 345

Re: Summation help.

I just thought of something , every time you throw a dice , the probability of an even number is 1/2 which is equal to getting an odd number.  If the sum is even , then 

is the probability of getting an even sum . And

you will find

Last edited by Stanley_Marsh (2007-04-06 02:38:12)

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Numbers are the essence of the Universe

Stanley_Marsh

Member

Registered: 2006-12-13

Posts: 345

Re: Summation help.

You will find for n=odd number.

Last edited by Stanley_Marsh (2007-04-05 21:29:24)

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Numbers are the essence of the Universe

Stanley_Marsh

Member

Registered: 2006-12-13

Posts: 345

Re: Summation help.

Actually when n=even number, the equality above still holds, but I haven't proven it yet,
Then the probability will be 1/2.

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Numbers are the essence of the Universe