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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)
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Last edited by Stanley_Marsh (2007-04-05 14:29:37)
Numbers are the essence of the Universe
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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? :-)
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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)
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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)
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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 findLast edited by Stanley_Marsh (2007-04-06 02:38:12)
Numbers are the essence of the Universe
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You will find for n=odd number.
Last edited by Stanley_Marsh (2007-04-05 21:29:24)
Numbers are the essence of the Universe
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Actually when n=even number, the equality above still holds, but I haven't proven it yet,
Then the probability will be 1/2.
Numbers are the essence of the Universe
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