# a coin has two sides. when the coin is flipped, each side appears with probability 1/2.?

Supposed a coin is flipped 8 times. what is the chance that both heads and tails appear an equal number of times?

Relevance

The number of ways to flip a coin 8 times: 2^8 = 256

The number of ways to select 4 times out of 8, when tails appear:

8! / (4! * 4!) = (5 * 6 * 7 * 8) / (1 * 2 * 3 * 4) =

= 5 * 6 / (2 * 3) * 7 * (8 / 4) = 5 * 7 * 2 = 70

The chance is 70 / 256 = 35 / 128 = 0.2734375

• i agree with 1 in 16.

look at it this way, flipping a coin 8 times is 2^^8 flips or 256. if a 1 represents heads and a 0 represents tails, you need all the binary numbers between 0 and 255 with exactly four 1's and four 0's. now, whatever place the four 0's appear in the eight possible places, there is only 1 in 16 chance those will be all 0's in every combination of 8 bits.

so again, i'll go with 1:16.

people saying 1:2, read the problem, you're wrong.

• 1/16

What you are trying to find out is the probability of getting 4 heads (or 4 tails). Each time the coin is flipped, there is a 1/2 chance of its being a head. There is a 1/2 chance in each of the first 4 flips. Multiply them together.

• Anonymous

1 out of 16

• From what I see, there are 9 scenarios:

~ 4 heads, 4 tails <--- the one you want