Probability - the Australian Mathematical Sciences Institute

{30} • ProbabilityIndependence for more than two eventsOur discussion of independence has so far been limited to two events. The extension toan arbitrary number of events is important.If the events A 1 , A 2 ,..., A n are mutually independent, thenPr(A 1 ∩ A 2 ∩ ··· ∩ A n ) = Pr(A 1 )Pr(A 2 )···Pr(A n ).(∗)This is a necessary condition for mutual independence, but it is not sufficient. As wemight reasonably expect, mutual independence is actually characterised by statementsabout conditional probabilities.Essentially, the idea is the natural extension of the case of two events: We say that theevents A 1 , A 2 ,..., A n are mutually independent if, for each event A i , all of the possibleconditional probabilities involving the other events are equal to the unconditional probabilityPr(A i ). Informally, this means that regardless of what happens among the otherevents, the probability of A i is unchanged; and this must be true for each event A i , wherei = 1,2,...,n.We can express this definition formally as follows: Events A 1 , A 2 ,..., A n are mutually independentifPr(A i ) = Pr(A i | A j1 ∩ A j2 ∩ ··· ∩ A jm ),for all i and for every possible combination j 1 , j 2 ,..., j m such that j k ≠ i .What happens, for example, when n = 3? This definition says that events A 1 , A 2 and A 3are mutually independent ifPr(A 1 ) = Pr(A 1 |A 2 ) = Pr(A 1 |A 3 ) = Pr(A 1 |A 2 ∩ A 3 )and similarly for A 2 and A 3 .The equation (∗) above follows from this definition. The reason that (∗) is not sufficientto indicate mutual independence, however, is that it is possible for (∗) to be satisfied,while at the same time A 1 and A 2 are not independent of each other.Example: Dice gamesA Flemish gentleman called Chevalier de Méré played games of chance using dice inaround 1650. He played one game where a fair die is rolled four times; it is assumed thatthe outcomes are independent. What is the chance of getting at least one six? De Méréreasoned as follows. On any single roll, the probability of getting a six is 1 6. There arefour rolls, so the probability of getting a six at some stage is 4 × 1 6 = 2 3. The flaw in thisreasoning should be obvious: What if there were seven rolls?