Suppose a die is rolled repeatedly until it comes up `1\' or is rolled n times,

Question

Suppose a die is rolled repeatedly until it comes up `1' or is rolled n times, whichever comes first.Let X be the random variable representing the number of rolls. Calculate E[X].

Explanation / Answer

The solution give in your link is quite clear (in the calculation at least), so I will just attempt to justify the calculations done in the pdf file. The idea is the following: X is the random variable as you defined it. Suppose the value of X is n: i.e., after n rolls of the dice, you end up with 4 consecutive 6s. Therefore you know that the last 4 rolls--the n'th, (n-1)th, (n-2)th, (n-3)th rolls--must all be 6. You also know that the (n-4)th roll cannot be a six: if it were, then at the (n-1)th roll, you would already have rolled 4 consecutive 6s, so X would be (n-1) and not n. So now you know what the last 5 rolls in the sequence must be like. And you reason: if the last four rolls are the first time four consecutive sixes appeared, then the first (n-5) rolls {since you already have some information about the last 5 rolls, you don't need to consider them} can be anything as long as there are no "four consecutive sixes". Hence: The probability that X is exactly n = The probability that { n, n-1, n-2, n-3 are all sixes} times The probability that { n-4 is anything but a six} times The probability that { in the first n-5 rolls, no 4 consecutive 6s} The last probability can be expressed 1 - prob { X = 1} - prob {X=2} - ... - prob {X = n-5} which is equal to probability { X > n-5 } and hence the formula given. So now you have a recursive formula for the probability. Prob { X = n } = (1/6)^4 (5/6) Prob { X > n-5 } Now, sum the identity over all n from 5 to infinity. The Expectation value of X is equal to E(X) = Sum(n from 0 to infinity) Prob {X>n} = Sum(n from 5 to infinity) Prob {X>n-5} So Sum(n from 5 to infinity) Prob{X=n} = 5/7776 E(X) Since X can only take positive integer values, the sum of probabilities that X = n over (n from 1 to infinity) must be 1. So the left hand side becomes 1 - Sum(n from 1 to 4) Prob{X=n} = 5/7776 E(X) Now: the four terms in the sum on the left hand side can be easily calculated: X cannot be 1, 2, or 3, since you need at least 4 rolls. If X is 4, it must be 4 consecutive rolls of 6, so the probability is (1/6)^4. So 1 - (1/6)^4 = 5/7776 E(X) Multiplying through by 7776 = 6^5, we get 7776 - 6 = 5 E(X) E(X) = 7770 / 5 = 1554

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