# 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?

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• Amit Y
Lv 5

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

• sd3r
Lv 7

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

So it's 1/9 chance.

I may be wrong, but somehow I don't really get the "1/2 no matter how many flips" theory.

the odds are the same for each flip.

even if you flipped the coin 100 times and it was tails every time, your next flip would still have a 1/2 chance

1/16

1/2

.