**Scenario -2:**

You toss a fair coin thrice. Since there are two outcomes at each toss, namely “Heads” and “Tails” the total number of possible outcomes is

Let us list down all the outcomes for clarity’s sake; represents “Heads ” and represents “Tails”:

Since the coin is fair, we have sufficient reason to believe that each of these outcomes is equally likely, i.e. equi-probable. Also, since these are the only outcomes possible and one of them must occur, we can say that the probability of any outcome is . For example,

Now, let us pose the following question: in the sequence of these three throws, what is the probability of occuring before ?

This is easy; we count all cases satisfying the stated constraint:

Since there are “favorable” cases to our constraint out of the total cases possible, it should be correct to say that the possibility of occuring before should be . Similarly, the possibility of occuring before should also be .

Now let us try to answer the same question from a slightly different perspective. Let us define the events and as follows for this sequence of three tosses:

: occurs before | |

: occurs before |

The task is to find and , which we earlier did by explicitly counting the favorable cases. Here we’ll try to calculate and differently. Observe that one of the events or must occur; there is no other possibility. If we denote the outcomes of our “experiment” (involving the sequence of three tosses), by a rectangle, we can divide the rectangle into exactly two halves, one representing the event and one the event .

Now, it should be obvious that

Also, since the coin is fair, and must be equi-probable. This is because and are equi-probable, i.e, equivalent outcomes in terms of probability. Thus, occuring before must be as likely as occuring before . Hence, we must have

From and , we must have

Let us now pose a further question for this sequence of three tosses. Let the events and be defined as follows:

: There are more than | |

: There are more than |

Our task is to calculate and . These should be immediately apparent. Since there are an odd number of tosses, one of the events or must occur. Also, and have to be equi-probable.

Thus,

and | |

Your are urged to arrive at the same result by the explicit counting of out comes.