Question

You roll a pair of fair dice repeatedly. Let X denote the number of rolls until you get two consecutive sums of 8(roll two 8 in a row). Find E[X]

Answer #1

The probability of getting an 8 in two dice throws is first computed here as:

Total outcomes = 6*6 = 36

Sum of 8 can be obtained as: 2 + 6, 3 + 5, 4 + 4, 5 + 3 or 6 + 2
that is 5 cases. Therefore the probability of getting a sum of 8 is
computed here as: 5/36

The expected number of rolls until we get two consecutive sums of 8 be X. Also when one 8 has already occurred let the same probability be Y.

Then, we have here:

X = (5/36)*(Y+ 1) + (31/36)*(1 + X)

This is because there is a 5/36 probability that an 8 will come in
which case there would be Y expected more rolls to get two
consecutive 8s.

(5/36) X = 1 + (5/36)Y

Also, for Y we have here:

Y = (5/36)*1 + (31/36)*(1 + X)

Y = 1 + (31/36)X

Putting this in the previous equation, we get here:

(5/36) X = 1 + (5/36)(1 + (31/36)X )

(5/36)X = 1 + (5/36) + (155/36^{2})X

X = 59

**Therefore 59 is the required expected number of tosses
required here.**

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c) Find the probability that the sum of the dice is greater than
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d) Given that X less than or equal to 4 find the probability
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I roll a fair die until I get my first ace. Let X be the number
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Please follow the comment.
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(c) What is the probability that the first time you win is
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consider the following experiment:
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