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"Megan is selected to play a Megamillions game in which she can win a potentially unlimited amount of money. In this game, she repeatedly rolls a fair six-sided die, and the game ends as soon as she rolls a six. If roll 0 is a six, then she wins nothing; if roll 1 is a six, then she wins a total of 1^2 = 1 dollar; if roll 2 is a six, then she wins a total of 2^2=$4; and so on. What is the expected value of her winnings, in dollars?"

If anyone can help me figure this out that would be amazing! Obviously there is a 1/6 chance that she rolls a six, but I'm not sure where to go from there!

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closed as off-topic by Austin Mohr, user1551, user175968, Em., Edward Jiang May 16 '16 at 1:32

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  • $\begingroup$ This is the mean of a geometric distribution. $\endgroup$ – Masacroso May 15 '16 at 20:43
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If she wins n games, then she gets n^2 dollars.

The probability of winning 0 games is 1/6, as this is the chance her first roll is 6.

The probability of winning 1 game is 1/6*5/6, as she needs to win the first game and lose the second.

So the probability of winning n games is (5/6)^n*1/6. and the amount she wins is n^2, so the expectation is:

Sum from 0 to infinity of n^2*(5/6)^n*1/6 which comes out to be 55

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HINT: we have a geometric random variable $X\sim G(1/6)$ where $X$ represent the total number of throws of a game (the game ends when a six appear) what means that

$$\Pr[X=x]=\frac16\left(\frac56\right)^{x-1}$$

and the expected value of a discrete random variable $X$ is defined as

$$\Bbb E[X]=\sum_x x\Pr[X=x]$$

However we want to know the expected value of the amount of dollars that you can win per game, this is

$$\Bbb E[(X-1)^2]=\sum_{x\in\Bbb N} (x-1)^2\Pr[X=x]=\frac16\sum_{x\ge 0}x^2\left(\frac56\right)^x$$

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  • $\begingroup$ But $\mathbb{E}[X]^2 \neq \mathbb{E}[X^2]$. $\endgroup$ – Peter Shor May 15 '16 at 22:23
  • $\begingroup$ Yes @PeterShor ... I dont get what you try to say. $\endgroup$ – Masacroso May 15 '16 at 22:30
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    $\begingroup$ I'm saying the last line of your answer is wrong. The correct solution is 55. $\endgroup$ – Peter Shor May 15 '16 at 22:32
  • $\begingroup$ @PeterShor oh, I understand, you are right, I will fix it. $\endgroup$ – Masacroso May 15 '16 at 22:42

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