# Probability Dice Game! Help me! [closed]

"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=$4; and so on. What is the expected value of her winnings, in dollars?" If anyone can help me figure this out that would be amazing! Obviously there is a 1/6 chance that she rolls a six, but I'm not sure where to go from there! ## closed as off-topic by Austin Mohr, user1551, user175968, Em., Edward JiangMay 16 '16 at 1:32 This question appears to be off-topic. The users who voted to close gave this specific reason: • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Austin Mohr, user1551, Community, Em., Edward Jiang If this question can be reworded to fit the rules in the help center, please edit the question. • This is the mean of a geometric distribution. – Masacroso May 15 '16 at 20:43 ## 2 Answers If she wins n games, then she gets n^2 dollars. The probability of winning 0 games is 1/6, as this is the chance her first roll is 6. The probability of winning 1 game is 1/6*5/6, as she needs to win the first game and lose the second. So the probability of winning n games is (5/6)^n*1/6. and the amount she wins is n^2, so the expectation is: Sum from 0 to infinity of n^2*(5/6)^n*1/6 which comes out to be 55 HINT: we have a geometric random variable$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 $$\Bbb E[(X-1)^2]=\sum_{x\in\Bbb N} (x-1)^2\Pr[X=x]=\frac16\sum_{x\ge 0}x^2\left(\frac56\right)^x$$ • But$\mathbb{E}[X]^2 \neq \mathbb{E}[X^2]\$. – Peter Shor May 15 '16 at 22:23
• Yes @PeterShor ... I dont get what you try to say. – Masacroso May 15 '16 at 22:30
• I'm saying the last line of your answer is wrong. The correct solution is 55. – Peter Shor May 15 '16 at 22:32
• @PeterShor oh, I understand, you are right, I will fix it. – Masacroso May 15 '16 at 22:42