Average Number of Throwing a Dice

Consider a null sequence $A=\{\}$. Each time I will include the output in $A$ after throwing a dice. I will stop when $\{1,2,3\} \subseteq A$. On average how many number of throws is required?

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If you successively toss a biased coin with probability of heads $p$, the expected number of throws before you see the first success is $\frac{1}{p}$. Your question can be decomposed into three such games. Game 1: Keep throwing a dice until one of the three numbers $\{1,2,3\}$ occurs. This has a success probability of $1/2$ and hence the expected number of trials is $2$.

Game 2: Assume without loss of generality that you obtained $1$ when you succeeded in Game 1. Then you keep throwing dice until one of $\{2,3\}$ occurs. This has success probability $1/3$ and hence expected number of trials is $3$.

Game 3: Assume without loss of generality that you obtained $2$ when you succeeded in Game 2. Then you keep throwing dice until $3$ occurs. This has success probability $1/6$ and hence requires $6$ trials on average.

Thus the average number of throws required is (by linearity of expectation) $2+3+6=11$.

Note: To answer your inconsistency with experiments, let's compute the variance. Note that the 3 games are independent of each other. Therefore, variance is also linear. Variance of a geometric distribution is $\frac{1-p}{p^2}$. Hence variance of your game is $$\frac{1/2}{1/4}+\frac{2/3}{1/9}+\frac{5/6}{1/36}=2+6+30=38.$$ Hence the standard deviation is $\sqrt{38}\approx 6.1644$. Thus if you were to repeat the experiment $n$ times, the mean is $9$ and the standard deviation is $6.1644/\sqrt{n}$. Thus as long as your answer is within 3 standard deviations, your experiment is correct. To be more accurate, increase the number of trials $n$.

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The expected value of the waiting time for an event occuring with probability $p$ is $\frac1p$. The event "found a new one" occurs with probability $\frac36=\frac12$ initially, with $\frac26=\frac13$ after I have collected one, with $\frac16$ after I have collected two of the target numbers. Therefore the total expected waiting time is $2+3+6=11$.

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But experimentally I get 9.97. –  user89867 Aug 11 '13 at 22:13