Question

Suppose I have tossed a fair coin 13 times and one of the following sequences was my outcome: (1) HHHHHHHHHHHHH; (2) HHTTHTTHTHHTH; (3) TTTTTTTTTTTTT. Demonstrate that prior to actually tossing the coins, the three sequences are equally likely to occur (what is this probability?). What is the probability that the 13 coin tosses result in all heads or all tails? Now compute the probability that the 13 coin tosses result in a mix of heads and tails. Although the chances of the three specific sequences occurring randomly are equal, which sequence would you reason to be the genuine result, and why?

Answer 1

Solution

Since the coin is given to be fair, probability of getting a Head = probability of getting a Tail = ½. …............................… (1)

Further, result of any toss does not depend on the result of any other toss, i.e., the tosses are independent. Thus,

P(HHHHHHHHHHHHH)

= P(H) x P(H) x P(H) x P(H) x P(H) x P(H) x P(H) x P(H) x P(H) x P(H) x P(H) x P(H) x P(H)

= (½)(½)(½)(½)(½)(½)(½)(½)(½)(½)(½)(½)(½)

= (½)¹³ ………………………………………………………………………………………...................................……………… (2)

P(HHTTHTTHTHHTH)

= P(H) x P(H) x P(T) x P(T) x P(H) x P(T) x P(T) x P(H) x P(T) x P(H) x P(H) x P(T) x P(H)

= (½)(½)(½)(½)(½)(½)(½)(½)(½)(½)(½)(½)(½)

= (½)¹³ …………………………………………………………………………………………..................................…………… (3)

P(TTTTTTTTTTTTT)

= P(T) x P(T) x P(T) x P(T) x P(T) x P(T) x P(T) x P(T) x P(T) x P(T) x P(T) x P(T) x P(T)

= (½)(½)(½)(½)(½)(½)(½)(½)(½)(½)(½)(½)(½)

= (½)¹³ ………………………………………………………………………..................................……………………………… (4)

(2), (3) and (4) demonstrate that the probabilities of the three sequences are equally likely. Answer 1

Although the probabilities of the these three specific sequences are equally likely, a sequence of mix of H and T is more genuine that a sequence of all H or all T. This is so, because all H and all T have only possible chance, but a mix of H and T have more number of chances. For example, suppose we are only concerned with the number of H and T and not in the order in which they occur, the given sequence is only one possibility. One other sequence could be HHHHHHHTTTTTT; yet another possibility is TTTTTHHHHHHH – in fact there are ¹³C₇ = 1716 possibilities. Answer 2    

DONE