Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
up vote 2 down vote accepted

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.

share|cite|improve this answer
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

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

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

share|cite|improve this answer

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

share|cite|improve this answer

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

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.