## Final Problem B10 |
edelapena |
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A coin has a probability of landing on heads equal to 1/10 expected number of trials for the coin to land on heads on two consecutive rolls? What is the expected number of trials for the coin to land on heads on two consecutive rolls? |

Lilyhui: March 19, 2015, 9:10 p.m. |
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The answer I got was 110 using the recursive formula |

arpanshah: March 20, 2015, 12:48 a.m. |
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what exactly is the recursive formula that you mentioned above? |

weisbart: March 20, 2015, 1:37 a.m. |
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You can do it recursively or more directly. Here is a more direct approach. You could have T (tails first), HT (heads then tails), or HH (heads followed by heads--meaning you're done). Let $X$ count the number of trials until you get the repeated heads. Therefore you get that \[E[X] = E[X|T]P(T) + E[X|HT]P(HT) + E[X|HH] P(HH)\] \[= (1 + E[X])\frac{9}{10} + (2 + E[X])\frac{9}{100} + 2\times \frac{1}{100}\] \[ = \frac{99}{100}E[X] + \frac{9}{10} + \frac{18}{100} + \frac{2}{100}.\] Solving for $E[X]$, we get that \[E[X] = 90 + 18 + 2 = 110.\] Of course, this is the same answer give by the recursive formula. |