I'm trying to get $E(Y)$ where $Y$ is the number of trials to get two consecutive successes.

I'm trying to do this by introducing a random variable $Z$ that represents the number of trials until the first success, where successes happen with probability $p$.

So I did the following:

$$ E(Y) = [1+E(Z)]p+[2+E(Y)](1-p) $$

The logic is that, if you get a success, then you will need just an extra trial with probability $p$. However, you may need to start all over again, with probability $1-p$. When I compute this formula I get:

$$ E(Y)=\frac{3-p}{p} $$

Which doesn't seem correct. Can someone help me identify where my logic is wrong? I know there are other ways to compute the expected value, but I need to use these two random variables for the purpose of this exercise.

  • $\begingroup$ Let $q = 1 - p.$ Then you have $Z = 1$ with probability (wp) $p,$ $Z = 2$ wp $qp,$ $Z = 3$ wp $q^2p,$ and so on. You will get a geometric series which is easy to calculate. $\endgroup$
    – William M.
    Commented Feb 2, 2023 at 17:38
  • $\begingroup$ I know $E(Z)=1/p$, but I’m having trouble with $E(Y)$. $\endgroup$
    – Melanie
    Commented Feb 2, 2023 at 18:44

2 Answers 2


You first carry out $Z$ trials, the first $Z-1$ of which are failures, and the last of which is a success. (This is the definition of the random variable $Z$.) Then you do one more trial. If it succeeds (with probability $p$), you've found the value of $Y$. Otherwise (with probability $1-p$), you start fresh and expect to need $E[Y]$ more trials. In short: $$ E[Y] = E[Z] + 1 + (1-p)E[Y], $$ or $$ E[Y]=\frac{1}{p}\left(E[Z] + 1\right). $$ Note that this logic extends to longer runs as well. If you want $n$ consecutive successes, you find the first run of $n-1$ successes, and then do one more trial, possibly starting over afterwards. Generally, $$ E[R_n]=\frac{1}{p}\left(E[R_{n-1}] + 1\right). $$

  • $\begingroup$ Thanks. This was so helpful. This is part of a bigger problem where I'm asked to find the number of trials, X, till I see the pattern of two successes and one failure. Could you tell me if my formulation is correct? $$ E(X) = E(Y) + 1 * (1-p) + [1+E(X)]p $$ $\endgroup$
    – Melanie
    Commented Feb 2, 2023 at 19:43
  • $\begingroup$ $$E(X) = (E(Y) + 1) (1 - p) + (E(Y) + 1 + E(X)) p = E(Y) + 1 + E(X) p$$ $\endgroup$
    – Essaidi
    Commented Feb 2, 2023 at 22:27
  • $\begingroup$ Does that mean two successes and then one failure, in that order? Then @Essaidi is correct, it should be $E[X]=E[Y] + 1 + pE[X]$, or $E[X]=(E[Y]+1)/(1-p)$, by the exact same reasoning as the answer I gave. $\endgroup$
    – mjqxxxx
    Commented Feb 3, 2023 at 0:06

Your equations were incorrect. Go step by step

$E(Z) = 0.5\times1 + 0.5*[E(Z)+1)] \Longrightarrow E(Z) = 2$

$E(Y) = E(Z) + 0.5*1 + 0.5[E(Y)+1] \Longrightarrow E(Y)=6$

Oh, if $p$ is not 0.5. substitute $p$ and $(1-p)$ appropriately

$E(Z) = p + (1-p)[E(Z)+1]$
$E(Y) = E(Z) + p + (1-p)[E(Y)+1]$

You should now solve

You can simplify the last equation to
$E(Y) = E(Z) +1 + (1-p)*E(Y)$, or
$E(Y) = \frac{E(Z)+1}{p}$


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