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Independent trials resulting in outcomes $1,2,3$ with probabilities $1/2,1/4,1/4$ are performed. Let $N$ be the number of trials needed until the first outcome has occurred 3 times (not necessarily consecutively). I want to find $E(N)$. My thoughts were to condition on the first outcome. $$E(N)=E(N|1)\frac{1}{2}+E(N|2)\frac{1}{4}+E(N|3)\frac{1}{4}$$. But I do not know what to do with $E(N|i)$ which is the conditional expectation given the first outcome is $i$. Any help is much appreciated. Thank you.

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Let $N_1, N_2, N_3$ be the waiting times until the first, second, and third 1. Your $N$ is $N_3$. But $E(N_3) = 3 E(N_1)$ -- basically after each 1 the situation resets until you get the next 1.

So what's $E(N_1)$? You have

$$E(N_1) = E(N_1 | 1) (1/2) + E(N_1 | 2) (1/4) + E(N_1 | 3) (1/4).$$

Now clearly $E(N_1 | 1) = 1$ -- if you get a 1 on the first trial then the waiting time is 1. And $E(N_1 | 2) = 1 + E(N_1)$ -- if you get a 2 on the first trial then you have to start over, but with one trial already counted against you. Similarly $E(N_1 | 3) = 1 + E(N_1)$. So the equation above becomes

$$E(N_1) = (1)(1/2) + (1 + E(N_1)) (1/2)$$

which you can solve for $E(N_1)$.

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  • $\begingroup$ Awesome this is so simple! $\endgroup$ – Heisenberg Sep 16 '17 at 23:31
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You have a Markov chain with $3^3+1=28$ states. The first $27$ of them can be named like $012$, meaning you have thrown no $1$s, one $2$, and two $3$s. The last state is finish, where you have thrown three of something. You can start from the end to compute the expected number of throws. For example, $E(N|222)=7$ because you have made six throws and are guaranteed to finish next time. Knowing this, $E(N|122)=\frac 12 \cdot 7 + \frac 12 \cdot 6=6.5$ because you have half chance to throw a $1$ and go to $222$ and half chance to finish. The equality of probability between $2$ and $3$ makes the expected number of throws match in many pairs of states.

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$$\newcommand{\C}[2]{^{#1}\text{C}_{#2}}$$ Let, first outcome=$O_1$,

Suppose $N $ trials have been done,$$$$The last trial must be $O_1$ as you stop after the third trial, $$$$ So the other $O_1$ s are distributed in first $N-1$ trials, $$$$ hence probability that we get 2 $ O_1$ for $(N-1)$ trials is, $$=\C{(N-1)}{2} (\frac{1}{2})^2(\frac{1}{2})^2$$ $$=\C{(N-1)}{2} \frac{1}{16}$$ $$=\frac{\C{(N-1)}{2}}{16}$$ Now probability that 1 outcome occurs after N-1 th trial, $$=\frac{\C{(N-1)}{2}}{32}$$ Hence, $$P(N)=\frac{\C{(N-1)}{2}}{32}$$

The formula i used is, $$$$Let p be the probability an event will happen, $$$$q be it unit happening, $$$$Given n trials have taken place $$$$Probability that event occurred r times is $$P(n,r)=\C{n}{r} (p)^r (q)^{n-r}$$

PS:I don't understand what an expectation is , i can just guide you till here.

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