What sequence is most likely to result from flipping a fair coin 5 times?

Answer 1

Let x be the aleatory variable representing the number of heads appearing when

tossing a fair coin 5 times.

Using the 'Binomial Distribution': P(x) = 5Cx (1/2)5 , where 5Cx = 5!/[x!(5-x)!]

Then:

P(x=0) = (1/2)5 = 1/32

P(x=1) = 5(1/2)5 = 5/32

P(x=2) = 10(1/2)5 = 10/32

P(x=3) = 10(1/2)5 = 10/32

P(x=4) = 5(1/2)5 = 5/32

P(x=5) = (1/2)5 = 1/32

So the most likely combination of heads-tails to come out are; (3H,2T) and (2H,3T), with a probability of 10/32 (=5/16=0.3125=31.25%) either one.

For any one of these outcomes there are 10 different sequences, each of which has

a probability of happening of 1/32.

If one observes the list of results, it can be inferred that any head-tails sequence

that can be thought of in 5 tosses has the probability of turning out of 1/32.