Pessimist Singularitarian

#4
January 10th, 2013,
16:52
I see we aren't meant to either use technology, or use an approximation.

Intuition tells us that since there are 26 favorable outcomes and 25 unfavorable that the probability will be slightly greater than 1/2. But how much greater?

Perhaps if we rewrite the sum:

$\displaystyle P(X)=\frac{1}{3^{50}}\sum_{k=0}^{25}\left[{50 \choose 2k}2^{2k} \right]$

Now, let's let Y be the event of getting an odd number of heads, and write:

$\displaystyle P(X)-P(Y)=\delta$

$\displaystyle \frac{1}{3^{50}}\sum_{k=0}^{25}\left[{50 \choose 2k}2^{2k} \right]-\frac{1}{3^{50}}\sum_{k=0}^{24}\left[{50 \choose 2k+1}2^{2k+1} \right]=\delta$

$\displaystyle 3^{50}\delta=\sum_{k=0}^{50}\left[{50 \choose k}2^{k}(-1)^{50-k}\right]$

By the binomial theorem, we have:

$\displaystyle 3^{50}\delta=(2-1)^{50}=1$

and so:

$\displaystyle \delta=\frac{1}{3^{50}}$

Hence:

$\displaystyle P(X)-P(Y)=\frac{1}{3^{50}}$

$\displaystyle P(X)+P(Y)=1$

Adding, we find:

$\displaystyle 2P(X)=1+\frac{1}{3^{50}}$

and so finally, we have:

$\displaystyle P(X)=\frac{1}{2}\left(1+\frac{1}{3^{50}} \right)$