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P48 of Intro to Pr, 2nd Ed by Bertsekas and Tsitsiklis describes a simplification to the binomial formula when the probability is $\dfrac12$. I don't understand how it is derived.

http://imgur.com/LZ3T2l1

If the probability is $p = \dfrac12$, shouldn't the formula be $$\sum_{k=0}^n {{n} \choose {k}}p^n = \sum_{k=0}^n {{n} \choose {k}}(1/2)^n $$

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Yes, where $p=\frac{1}{2}$, and your sum is equal to 1. If you multiply both sides by $2^n$, you get the given formula.

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  • $\begingroup$ Can you elaborate? I don't understand this still. I do understand how the sum of all n choose k equals 2^n. That just ends up being the set of all subsets which has 2^n elements. What I don't understand is what the 1/2 probability has to do with deriving that formula. $\endgroup$
    – mr real lyfe
    Oct 13, 2013 at 7:20
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    $\begingroup$ Using your above formula $\sum_{k=0}^n \binom{n}{k}p^n=1$ for $p=\frac{1}{2}$, so $\sum_{k=0}^n \binom{n}{k}\left(\frac{1}{2}\right)^n=1$, so $\sum_{k=0}^n \binom{n}{k}=2^n$. $\endgroup$
    – peterwhy
    Oct 13, 2013 at 15:55
  • $\begingroup$ ahh i see. Thanks! $\endgroup$
    – mr real lyfe
    Oct 15, 2013 at 6:30

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