The probability of getting more heads than tails in a coin toss A fair coin is to be tossed $8$ times. What is the probability that more of the tosses will result in heads than will result in tails? 

$\textbf{Guess:}$ I'm guessing that by symmetry, we can write down the probability $x$ of getting exactly $4$ heads and $4$ tails and then calculate 
  $\dfrac{1}{2}(1-x)$. 

So how does one calculate for $x$? I know that it should be a rational number (that is, $\dfrac{?}{2^8}$), but I am not sure how to get the numerator. 
 A: P(getting more Heads in 8 tosses)=(8C5+8C6+8C7+8C8)/(2^8)
A: Use the binomial distribution to get the probability of getting $k$ heads from $n$ flips:
$$p(n,k) = \binom{n}{k} \left ( \frac12 \right )^k \left ( \frac12 \right )^{n-k} = \binom{n}{k} \left ( \frac12 \right )^n$$
The probability you seek is $p(8,5)+p(8,6)+p(8,7)+p(8,8)$, or
$$\frac{\binom{8}{5}+\binom{8}{6}+\binom{8}{7}+\binom{8}{8}}{2^8} = \frac{56+28+8+1}{2^8} = \frac{93}{256}$$
A: The number in the numerator should be $\displaystyle \left( \begin{array}{c} 8 \\ 4 \end{array} \right) = \frac{8!}{4! ( 8-4)!} = 70$.
Why?  Because we have $8$ tosses, and out of these tosses, we have $4$ heads.
A: Since the number of outcomes with the 4H and 4T is C(8,4)=70, there is an equal possibility for the rest of outcomes to be more heads than tails, or more tails than heads.
Knowing that the total outcomes of flipping a coin 8 times are 256. the difference 256-70 will split equally.
so: (256-70)/2=93.
P(h>t)=93/256.
A: Use Pascal's Triangle to find the numerator.  For this particular problem, the numerator is 70.
