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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.

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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.

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    $\begingroup$ This is a bit late, but I believe the two answers by Ron Gordon and user136194 are the correct ones. The OP's question asked for the probability that more heads would show up than tails, and therefore casework is necessary, with the cases that there are 5, 6, 7, and 8 heads. I don't see how finding the number of ways to obtain 4 heads would do anything -- it doesn't even fit in with casework, as it doesn't even fulfill the requirements. $\endgroup$ – Junlin Yi Feb 10 '16 at 23:02
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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

$$\frac{\binom{8}{5}+\binom{8}{6}+\binom{8}{7}+\binom{8}{8}}{2^8} = \frac{56+28+8+1}{2^8} = \frac{93}{256}$$

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P(getting more Heads in 8 tosses)=(8C5+8C6+8C7+8C8)/(2^8)

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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.

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Use Pascal's Triangle to find the numerator. For this particular problem, the numerator is 70.

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