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Suppose we have one $12$-sided dice and one $20$-sided dice. Now, we keep rolling the dice together until we see the same number of both dice. What is the expectation of the number of times we roll the dice?

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    Any thoughts? What's the probability that they match on a single trial? – 2017-02-09
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    If you have calculated this probability (Let us denote it with $p$), the expected number of trials is simply $\frac{1}{p}$ – 2017-02-09
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    @lulu $ 12 \times \frac{1}{12} \times \frac{1}{20} = \frac{1}{20}$ ? – 2017-02-09
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    That's correct. Do you see why this means that the answer to your question is $20$? – 2017-02-09
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    I'll post something below. – 2017-02-09

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More generally, suppose we have a Bernoulli process in which the probability of success is $p$. Here, $p=\frac 1{20}$. We want to argue that the expected number of trials it takes to see the first success is $\frac 1p$.

Let $E$ be this expected number. Consider what happens on the first trial. We either succeed (with probability $p$) or we fail (probability $1-p$). If we succeed, it took $1$ trial. If we fail, we now expect it to take $E+1$ trials (as we are back at the start having already tried once). Thus $$E=p\times 1 +(1-p)\times (E+1)\implies pE=1\implies E=\frac 1p$$

Note: it is a good exercise to do this via geometric sums. After all, the probability that it takes exactly $n$ trials is $(1-p)^{n-1}p$ so $E=\sum n(1-p)^{n-1}p$.

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You wish to conduct an indefinite sequence of trials, counting until you stop when an event happens.

The probability of this event, that the die show the same result, is $\binom {12}1\frac 1{12}\frac 1{20}$, that is $1/{20}$.

So the probability of taking $n$ trials until this event is $(\frac{19}{20})^{n-1}\frac 1{20}$ or ${19^{n-1}}/{20^n}$.   That is the probability of $n-1$ "failures" then one "success".

This should perhaps be a familiar type of distribution, so you can obtain the expectation directly from your knowledge base.   Otheriwise use your knowledge of series:

$$\mathsf E(X) =\sum_{n=1}^\infty \dfrac{19^{n-1}n}{20^n} = \ldots$$