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A fair 6-sided die is thrown repeatedly until two different numbers appear. What is the expected number of rolls?


My intuition tells that this is a geometric distribution with parameter $(\dfrac{5}{6})$ . So the expected value is $\dfrac{6}{5}$. But I am not sure. The place where I am confused is a general geometric distribution random variable is defined by its pdf $P(X=k) = (1-p)^{k-1}p, k = \{1,2,3,...\}$.

However, in this question, it seems it is off by 1 since $P(X=1)=0$.

Can someone explain this to me?

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2 Answers 2

up vote 4 down vote accepted

Note that there should be a minimum of two throws. Hence, your expected value of $6/5$ is incorrect.

Probability that you stop after $n$ throws is $\displaystyle \left(\frac16 \right)^{n-2} \frac56$.

This is so since given what you get on the first throw for you to stop at $n^{th}$ throw you need to repeatedly throw the same number for the next $n-2$ throws (i.e. from the second throw to the $(n-1)^{th}$ throw) and throw a different number of the $n^{th}$ throw.

The expected value now is hence given by $$\sum_{n = 2}^{\infty} \frac{5n}{6^{n-1}} = 5 \times \frac{11}{25} = \frac{11}{5}$$

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I really like your explanation. –  geraldgreen Dec 9 '11 at 23:26

It certainly can't be $\frac{6}{5}$ as you need at least two rolls. After the first roll, on each roll you have $\frac{5}{6}$ chance of getting a different number, so it is a geometric distribution. The expected number is then $1+\frac{6}{5}=\frac{11}{5}$. You might be interested in the coupon collector's problem.

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