A fair coin is flipped n times...?

Question

What is the probability that all the heads occur at the start of the sequence?

Answer 1

One question, does a sequence of all tails count as all the heads being at the beginning?

I'll assume it does.

n = 1
H
T
2 out of 2

n = 2
HH
HT
TT
3 out of 4

n = 3
HHH
HHT
HTT
TTT
4 out of 8

n = 4
HHHH
HHHT
HHTT
HTTT
TTTT
5 out of 16

General formula.
P(Coin flip n times, all heads occur at the start including sequence of all tails)
= (n+1)/2^n

Edit:
Or if you require there to be at least one head in the sequence, the answer is simply:
P(Coin flip n times, all heads occur at the start NOT including sequence of all tails)
= n/2^n

Answer 2

well the probability if you flipped it would be 1/2 then every time you flip it you multiply it by the probability

so 1 flip you get 1/2 or 50%
2 flips you get 1/2*1/2= 1/4 or 25%
3 flips you get 1/2*1/2*1/2= 1/8 or 12.5%

so your probability is (1/x)^n...x being the # of sides and n being how many times you flip it

i chose x for sides bc it can also go with a certain side on a role of dice also

Answer 3

1/2

EDIT - Trust me. These guys below are wrong.

You take every flip as a new flip of the coin. The previous flips have no effect on the next flip. It's a misconception that because you've flipped say 10 heads.. theres more chance of a tail next. It's always 1/2.

Answer 4

I don't know what you mean 'at the start of the sequence'.
However,
In one trial,the probability of getting a head is 1/2.
In two trials,the probability of getting two heads is 1/2 * 1/2=1/4
In three trials,the probability of getting three heads is 1/2*1/2*1/2=1/8
............................and so on.
In n trials,the probability of getting all heads is 1/2*1/2*1/2...=1/(2^n)

P.S.assume the experiment as an independent one.So,P(A and B)=P(A) * P(B)

Hope this help!