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#1 2007-04-09 14:58:54

Stanley_Marsh
Member
Registered: 2006-12-13
Posts: 345

Is it true


I've already prove the case when n is odd, then I used induction to prove this , the calculation is tiresome , not sure it's right tho.

Last edited by Stanley_Marsh (2007-04-10 10:42:08)


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#2 2007-04-09 15:12:23

Stanley_Marsh
Member
Registered: 2006-12-13
Posts: 345

Re: Is it true

K=odd , this holds, if k= even ,this holds too , then throw n dices , the probability of getting an even sum will be 0.5

Last edited by Stanley_Marsh (2007-04-10 10:42:38)


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#3 2007-04-11 12:12:34

whatismath
Member
Registered: 2007-04-10
Posts: 19

Re: Is it true

You could see it this way:
denote nCk = (n choose k)
The Binomial Theorem says
(1+x)^n  = summation (k=0 to k=n) nCk(x^k)
Put x=-1: 0 =  summation (k=0 to k=n) (-1)^k(nCk)
So sum of odd terms = sum of even terms.
Put x=1 to see both sums are 2^(n-1).

With this we could see the probability questionthis way:
Prob(sum of n dice is odd)
= 1/(2^n) (# of combinations in which an odd number of dice is odd)
= 1/(2^n) (nC1 + nC3 + nC5 + ...)
= 1/(2^n) (2^(n-1))
= 1/2
Similar for the even case.
Thanks.

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#4 2007-04-11 13:45:21

Stanley_Marsh
Member
Registered: 2006-12-13
Posts: 345

Re: Is it true

My proof works either , I just need to prove the case when n=even


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