A spinner has 4 sectors of area 10%, 20%, 30% and 40%.
What is the expected # of spins for the spinner to stop on each sector at least once ?
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If areas were equal, it'd be 4/4 +4/3 +4/2 +4/1 = 8.33
and with the unequal probabilities above, it'd be > 10 as pointed out by @Hagen, but how do we get the exact value ?
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I have already provided some context in the first edit. I am not a mathematician, just a puzzle aficionado, so the simpler the explanation, the better for me.
See the coupon collector's problem formula for non uniform probabilities.
$\displaystyle E(T)=\int_0^\infty \big(1-\prod_{i=1}^n(1-e^{-p_it})\big)dt $
Substituting the probabilities for the problem and evaluating the integral gives
$\displaystyle E(T)=\frac{445}{36}\approx 12.3611$
Here's a completely elementary approach (although it's not as clean as gar's answer by more sophisticated methods).
First, let's analyze why you say that for equal sectors, the expected value is $\frac44+\frac43+\frac42+\frac41=\frac{25}3$. The expected number of spins to get at least one sector is $1$ (it's guaranteed to hit something). After this event, we collect all the sectors not yet hit, with total value $3/4$, and ask the expected number of tries to hit this region at least once: this is $4/3$. Repeating the process until there are no more regions, we get $25/3$.
For unequal sectors, we can use the same technique, although it gets more complicated. Let $S(\{k_1,\dots,k_n\})$ be the expected number of spins for the spinner to hit the each of the $P(k_i)$ sectors at least once. Then your goal is $S(\{.1,.2,.3,.4\})$, and $S(\emptyset)=0$. The relevant equation here is $S(A)=(1-\sum A)(S(A)+1)+\sum_{p\in A}p(S(A\setminus\{p\})+1)$, because the expected number of spins to get each of the sectors in $A$ is a weighted sum over the sectors. For each sector in $A$, we multiply the probability of hitting that sector times the cost of this try, $1$, plus the expected number of extra tries to hit the remaining sectors. If we hit a sector not in $A$, we try again, and this event has probability $1-\sum A$ and cost $1+S(A)$. This allows you to calculate $S(A)$ recursively, via the formula
$$S(A)=\frac{1+\sum_{p\in A}p\,S(A\setminus\{p\})}{\sum A}.$$
We can verify that $S(\{p\})=\frac{1+0}{p}=\frac1p$ meets our expectation. Working out the probabilities by explicit calculation (on Mathematica, 'cause I'm lazy) yields:
\begin{align} S(\emptyset)&=0\\ S(\{.1\})&=\frac{1+.1\cdot0}{.1}=10\\ S(\{.2\})&=\frac{1+.2\cdot0}{.2}=5\\ S(\{.3\})&=\frac{1+.3\cdot0}{.3}=\frac{10}{3}\\ S(\{.4\})&=\frac{1+.4\cdot0}{.4}=\frac{5}{2}\\ S(\{.1,.2\})&=\frac{1+.1\cdot5+.2\cdot10}{.1+.2}=\frac{35}{3}\\ S(\{.1,.3\})&=\frac{1+.1\cdot10/3+.3\cdot10}{.1+.3}=\frac{65}{6}\\ S(\{.1,.4\})&=\frac{1+.1\cdot5/2+.4\cdot10}{.1+.4}=\frac{21}{2}\\ S(\{.2,.3\})&=\frac{1+.2\cdot10/3+.3\cdot5}{.2+.3}=\frac{19}{3}\\ S(\{.2,.4\})&=\frac{1+.2\cdot5/2+.4\cdot5}{.2+.4}=\frac{35}{6}\\ S(\{.3,.4\})&=\frac{1+.3\cdot5/2+.4\cdot10/3}{.3+.4}=\frac{185}{42}\\ S(\{.1,.2,.3\})&=\frac{1+.1\cdot19/3+.2\cdot65/6+.3\cdot35/3}{.1+.2+.3}=\frac{73}{6}\\ S(\{.1,.2,.4\})&=\frac{1+.1\cdot35/6+.2\cdot21/2+.4\cdot35/3}{.1+.2+.4}=\frac{167}{14}\\ S(\{.1,.3,.4\})&=\frac{1+.1\cdot185/42+.3\cdot21/2+.4\cdot65/6}{.1+.3+.4}=\frac{937}{84}\\ S(\{.2,.3,.4\})&=\frac{1+.2\cdot185/42+.3\cdot35/6+.4\cdot19/3}{.2+.3+.4}=\frac{863}{126}\\ S(\{.1,.2,.3,.4\})&=\frac{1+.1\cdot863/126+.2\cdot937/84+.3\cdot167/14+.4\cdot73/6}{.1+.2+.3+.4}=\frac{445}{36},\\ \end{align}
which agrees with gar's answer by a different method.
This approach also leads to a proof of gar's integral answer $S(A)=\int_0^\infty\left(1-\prod_{p\in A}(1-e^{-pt})\right)\,dt$ (where $\sum A=1$). First, we need to generalize that result to $S(A)<1$. In this case, we can map each trial to a selection of the region of probability $\sum A$, followed by the same unequal-probabilities case where the probabilities are multiplied by $\frac1{\sum A}$, but a change of variables $t\mapsto t\sum A$ reveals that the formula is the same in this case. We can evaluate the integral:
\begin{align} \int_0^\infty\Big(1-\prod_{p\in A}(1-e^{-pt})\Big)\,dt&=\int_0^\infty\sum_{\emptyset\ne B\subseteq A}(-1)^{|B|+1}e^{-(\sum B)t}\,dt\\ &=\sum_{\emptyset\ne B\subseteq A}(-1)^{|B|+1}\frac1{\sum B}\\ \end{align}
It is this version of the formula that we will prove by induction. Obviously $S(A)=0$ is satisfied, and for the induction step:
\begin{align} S(A)\sum A&=1+\sum_{p\in A}p\,S(A\setminus\{p\})\\ &=1+\sum_{p\in A}\sum_{\emptyset\ne B\subseteq A\setminus\{p\}}(-1)^{|B|+1}\frac p{\sum B}\\ &\stackrel?=\sum A\cdot\sum_{\emptyset\ne B\subseteq A}(-1)^{|B|+1}\frac1{\sum B}\\ &=\sum_{p\in A}\sum_{\emptyset\ne B\subseteq A}(-1)^{|B|+1}\frac p{\sum B}\\ \end{align}
Canceling common terms, we want to prove $$\sum_{p\in A}\sum_{p\in B\subseteq A}(-1)^{|B|+1}\frac p{\sum B}=1.$$ Each $B\subseteq A$ appears $|B|$ times in the sum, once for each $p\in B$. Collecting these terms together, we have \begin{align} \sum_{p\in A}\sum_{p\in B\subseteq A}(-1)^{|B|+1}\frac p{\sum B}&=\sum_{\emptyset\ne B\subseteq A}\sum_{p\in B}(-1)^{|B|+1}\frac p{\sum B}\\ &=\sum_{\emptyset\ne B\subseteq A}(-1)^{|B|+1}\\ &=\sum_{n=1}^{|A|}\sum_{B\subseteq A,|B|=n}(-1)^{n+1}\\ &=\sum_{n=1}^{|A|}{|A|\choose n}(-1)^{n+1}\\ &=1-\sum_{n=0}^{|A|}(-1)^n{|A|\choose n}=1,\\ \end{align}
because of the identity $\sum_{n=0}^{|A|}(-1)^n{|A|\choose n}=0$.
