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What is the expected number of rolls until outcomes 1 and 2 are both observed. They can be in any order. I assume this is something related to a geometric distribution.

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2 Answers 2

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This is indeed related to the geometric distribution. The first number (either $1$ or $2$) will come up with probability $\frac26$ on each roll, and the number of rolls needed follows a geometric distribution, so $3$ rolls are expected. Then the remaining number comes up with probability $\frac16$, and similar to the first number, the reciprocal number of rolls are expected, or $6$ rolls.

Thus $9$ rolls are expected on average.

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  • $\begingroup$ do we need to add the probability of the remaining numbers? because we are only looking for the expected number of rolls until 1 and 2 both are observed. $\endgroup$ Apr 10, 2020 at 7:53
  • $\begingroup$ @shadowwalker The remaining numbers may be ignored. It's either a one or two or it's not. $\endgroup$ Apr 10, 2020 at 7:53
  • $\begingroup$ Why do we need to add the expected number of not getting 1 or 2? $\endgroup$ Apr 10, 2020 at 17:48
  • $\begingroup$ @ForumWhiner One follows the other. $\endgroup$ Apr 10, 2020 at 17:52
  • $\begingroup$ @Parcly Taxel, I don't understand what do you mean. Could you pls explain? $\endgroup$ Apr 10, 2020 at 17:57
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Let $X$ the random variable "first moment when both $1$ and $2$ appeared". (The space is the space of all sequence of possible tosses. ) We'll find $$P(X\ge k)$$ Let $Y_{1,k-1}$ the event that $i$ did not appear in the first $k-1$ tosses. Then $$P(X\ge k) = P(Y_{1,k-1}\cup Y_{2,k-1})= P(Y_{1,k-1})+P(Y_{2,k-1})-P(Y_{1,k-1}\cap Y_{2,k-1}) =2(5/6)^{k-1}-(4/6)^{k-1}$$ We have now $$E(X)=\sum_{k=1}^{\infty} k P(X=k)= \sum_{k=1}^{\infty}P(X\ge k) = 2\cdot 6 - 3 = 9$$

Note that the probability $$p_k =P(X\le k) = 1 - P(X\ge k+1) = 1 -(2 (5/6)^k -(4/6)^k)$$

Some values: $p_1=0$, $p_2=0.055\ldots$, $p_7=0.50036\ldots$ (just breaking even), $p_9=0.6383\ldots$, $p_{20}=0.948\ldots$

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