We repeat independent trials when each time more than 2 events can happen:

  • event $A$ with probability $p$
  • event $B$ (in the complement of $A$) with probability $q$
  • something different with probability $1-p-q$

(For example we can consider rolling a die where $A=$"result is 6" and $B=$ "result is 5 or 4").

Problem: what is the expected waiting time for $A$ and $B$ to have both of them happened at least once in the trials so far?

Here is what I have done:

  1. I compute the probability that $A$ happens before $B$ in an infinite sequence of trials: $$ P(A \text{ happens before B})=\sum_{k=0}^\infty P(\text{A and B didn't happen in $k$ trials})p $$ $$ = p\sum_{k=0}^\infty (1-q-p)^k=\frac {p}{p+q} $$
  2. I assume event $A$ happens before $B$ and compute conditional expected waiting time, that is we compute how long we wait for $A$ to happen assuming $B$ is not going to happen: it is $T$ such that $$ T=1 \cdot P(\text{$A$ happens}|\text{$B$ doesn't happen})+(T+1)\cdot P(\text{$A$ doesn't happen}|\text{$B$ doesn't happen}) $$ that is $$ T=\frac p {1-q}+(T+1)\frac {1-p-q}{1-q} $$ solving fot $T$ we have $$ T=\frac {1-q}{p} $$
  3. I compute the expected time to have event $B$ (after $A$ has just happened): $T$ satisfies $$ T=1 \cdot q+ (T+1) \cdot (1-q) $$ that is $T=\frac 1 q$. This time must be added to the time computed in 2 in order to have the first time that both $A$ and $B$ have occurred that is $T=\frac 1 q$.
  4. I put all together to compute expected time to have $A$ and $B$ both happened. Call $T$ such expected time, then we must have: $$ T=P(\text{$A$ happens fist})\cdot \left(\frac {1-q}{p}+\frac 1 q\right)+P(\text{$B$ happens fist})\cdot \left(\frac {1-p}{q}+\frac 1 p\right) $$ (the first part follows from computation above, the second by symmetry) then we substitute the probabilities in point 1: $P(\text{$A$ happens fist})=\frac {p}{p+q}$ and $P(\text{$A$ happens fist})=\frac {q}{p+q}$ (by simmetry) and I get the result $$ T=\frac 1 p +\frac 1 q -1. $$ that seems quite implausible.

Questions: What is wrong with this computation? Are there quicker/better ways to do this computation?


It is easy to show that if neither $p$ nor $q$ is $0$, the expectation exists. Let $E$ be the desired expectation. Condition on the result of the first trial.

If $A$ occurs, then the expected additional time for $B$ to occur is $\frac{1}{q}$. If $B$ occurs, the expected additional time for $A$ to occur is $\frac{1}{p}$.

If neither occurs then the expected additional time is $E$. In all cases we have "used up" one trial. Thus $$E=1+p\cdot\frac{1}{q}+q\cdot\frac{1}{p}+(1-p-q)E.$$ Solve for $E$.

  • $\begingroup$ Superquick! The result seems to be $E=\frac 1 p + \frac 1 q - \frac 1 {p+q}$ (that could also suggest another way of reasoning). $\endgroup$ – Marco Disce Feb 24 '16 at 16:05
  • $\begingroup$ I'm a little bit confused now... The formula $E=\frac 1 p$ for the waiting time of $p$ comes from the equation $E=p+(1-p)(E+1)$, if instead we wanted to use $E-1=p+(1-p)E$ we would get $E=\frac 1 p +1$. They can both be ok, it depends on whether the last trial must count as 1 more unit of time or not. But shouldn't we be coherent when we concatenate formulas obtained in this way? $\endgroup$ – Marco Disce Feb 24 '16 at 16:32
  • $\begingroup$ For the expected additional time for $B$ to occur given that $A$ has occurred on the first trial, I simply quoted the standard result about the expectation of a geometric distribution. One can derive that expectation by using a simpler version of the reasoning I used in the answer, but one would not use the same letter $E$ for that calculation. Let the expectation of the additional time in that case be $F$. Then $F=q\cdot 1+(1-q)(1+F)$, which gives $F=1/q$. $\endgroup$ – André Nicolas Feb 24 '16 at 16:41

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