2

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?

al-Hwarizmi
  • 4,370
  • 2
  • 21
  • 36
lostboy_19
  • 143
  • As mentioned by @Did, I saw the solution at http://math.stackexchange.com/questions/28905/expected-time-to-roll-all-1-through-6-on-a-die. I know my expected value is wrong. I am just trying to figure out why. – lostboy_19 Aug 04 '13 at 10:17
  • What do you fail to understand in this solution? (And why do you post a question here instead of a comment there?) – Did Aug 04 '13 at 10:21
  • 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 ... – David Mitra Aug 04 '13 at 10:34

2 Answers2

1

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

gen
  • 1,608
1

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?

ronno
  • 12,914
  • 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. – lostboy_19 Aug 04 '13 at 10:20