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Suppose that we roll a fair die until a 6 comes up or we have rolled it 10 times. What is the expected number of times we roll the die?

My answer:

The ith roll of the dice may give a 6 where $i=1,2,3,…,10$

Let X be the random variable for the ith roll of the die.

All possible values of X={1,2,3,4,5,6,7,8,9,10}

So I am computing the expected value for the number of rolls of the fair dice until getting a 6 or stop with a max of 10 trials

$1(1/6)+2(5/6)(1/6)+3(5/6)^2 (1/6)+4(5/6)^3 (1/6)+5(5/6)^4 (1/6)+6(5/6)^5 (1/6)+7(5/6)^6 (1/6)+8(5/6)^7 (1/6)+9(5/6)^8 (1/6)+10(5/6)^9 (1/6)=3.415911$

However, the book answer is 5.03. How?

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The probability of rolling 10 times is not $(\frac{5}{6})^9\frac{1}{6}$, but $(\frac{5}{6})^9$, since once you roll the 10th time, whatever you get, you will stop

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  • $\begingroup$ oh.. The first 9 rolls would NOT be a 6 but the 10th roll could be anything. So probability of the 10th roll is 6/6=1. $\endgroup$ – Sakthi Aug 15 '14 at 19:04
  • $\begingroup$ Only the first 9 rolls play a part in calculating the probability. The 10th roll does NOT matter. $\endgroup$ – Sakthi Aug 15 '14 at 19:16

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