# 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.

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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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Thank you Andrew! – user81136 Jun 6 '13 at 1:00
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. – Junlin Yi Feb 10 at 23:02

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

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

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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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