If 8 fair coins are flipped, what are the odds that there will be more heads than tails?

Printable View

- Jun 19th 2010, 06:22 PMmathman88Coin Flip
If 8 fair coins are flipped, what are the odds that there will be more heads than tails?

- Jun 19th 2010, 06:40 PMSudharaka
Dear mathman88,

Since flipping coins in this case are independent events, we can use the Binomial distribution to solve the problem. That is if you define a success as getting a head,

$\displaystyle b(x;n,\theta)=~^{n}C_{x}\theta^{x}(1-\theta)^{n-x}$

where, $\displaystyle x-~number~ of ~succeses$

$\displaystyle n-~number~of~trials$

$\displaystyle \theta-~probability~of~a~success$

Probability of getting more heads than tails= $\displaystyle b(5;8,\frac{1}{2})+b(6;8,\frac{1}{2})+b(7;8,\frac{ 1}{2})+b(8;8,\frac{1}{2})$

Hope you can continue. - Jun 20th 2010, 06:59 AMawkward
I like

**Sudharaka**'s solution, but here is an alternative approach just to show there is more than one way. Let H be the number of heads, T the number of tails. Then

P(H < T) + P(H = T) + P(H > T) = 1

By symmetry, P(H < T) = P(H > T), so

2 P(H > T) + P(H = T) = 1

It's pretty easy to find P(H = T) because this is just the case H = T = 4. Then use the equation above to solve for P(H > T).