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Let $X$ be the random variable which denotes the number of times a die has been rolled till each side has appeared. The order does not matter. We are trying to find $E[X]$.

Let $X_i$ be a random variable which denotes how many times a die has to be rolled till side i has appeared.

So,

$$E[X]= E[X1+X2+X3+X4+X5+X6] = E[X1]+E[X2]+E[X3]+E[X4]+E[X5]+E[X6]$$

$$E[X1]=E[X2]=E[X3]=E[X4]=E[X5]=E[X6]=6$$

$$E[X]=36$$?

Why is this solution wrong?

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  • $\begingroup$ As mentioned by @Did, I saw the solution at math.stackexchange.com/questions/28905/…. I know my expected value is wrong. I am just trying to figure out why. $\endgroup$ Commented Aug 4, 2013 at 10:17
  • $\begingroup$ What do you fail to understand in this solution? (And why do you post a question here instead of a comment there?) $\endgroup$ Commented Aug 4, 2013 at 10:21
  • $\begingroup$ Your $X_i$ are not defined appropriately. You should let $X_1$ be the number of rolls made to obtain one of the numbers (it's always $1$). Let $X_2$ be the number of additional rolls made to obtain a number distinct from the first number collected. Then, let $X_3$ be the number of additional rolls made to obtain a third number, distinct from the first two numbers collected ... $\endgroup$ Commented Aug 4, 2013 at 10:34

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That's the expected value, which means it will be around the value, however it might very well take different values too.

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Why do you think $X = X_1 + X_2 + \dots + X_6$?

Think of a coin. Is the number of trials needed for both heads and tails to appear the sum of number of trials for heads to appear and number of trials for tails to appear?

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  • $\begingroup$ I assumed that the no. of times all 6 sides appear will be the sum of number of times side 1 appears + no. of times side 2 appears and so on. $\endgroup$ Commented Aug 4, 2013 at 10:20

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