Expected value of prize winners Consider a group consisting of $w$ women, $c$ children and $r$ rabbits. Assume that no two individuals
have same height. Moreover, each rabbit is shorter in height than each child, who in turn is shorter
than each women. Consider two types of games. Each game starts with a set of participants
arranged in a line from left to right in a uniformly random order. There is a pile of prizes at the left
end of the line. Assume that a participant at place $i$ would see the prize if none of the participants
preceding him/her is taller than him/her.
(a) Game A : all rabbits, women and children participate in the game. What would be the
expected number of children who can see the prize.
(b) Game B : all rabbits, women and children participate in the game. A woman can receive
the prize only if she can see the prize and there is at least one child preceding her in the line.
What is the expected number of women that would win the prize ?
My try I think I should find the probability for a child of a given height to see prize, then sum it for all children, which would give total expected value by linearity of expectation, but I can't find a closed expression. I am also not sure if it's correct.
 A: Solution on Game A.
It is evident that the rabbits play no part in this.
Let's number the human participants as $P_1,\dots,P_{c+w}$ and $P_i$ has larger length than $P_{i+1}$ for $i=1,\dots,c+w-1$. So $P_1,\dots, P_w$ are women and $P_{w+1},\dots,P_{c+w}$ are children.
Further let $B_i\sim\mathsf{Bern}(p_i)$ taking value $1$ if $P_i$ sees the price.
The probability that person $P_1$ sees the price is $p_1=1$. Further $P_{2}$ will see it if she is on the correct side of $P_1$, so this event has probability $p_2=\frac12$. The probability that $P_{3}$ sees the price is $p_3=\frac13$ (i.e. the probability that she is the utmost left (if I understand well, else right) of $P_1,P_{2}$ and $P_{3}$) et cetera.
That gives expectation: $$\mathsf E(B_{c+w}+\dots+B_{1+w})=\sum_{i=1+w}^{c+w}\frac1{i}$$ for the number of children that see the price.
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Solution on Game B.
Here we work with $C_i\sim\mathsf{Bern}(q_i)$ for $i=1,\dots,w$ where $q_i$ denotes the probability that woman $P_i$ can receive the price. Here we have:$$q_i=1-\left[\frac{i-1}{i}+\frac1{c+i}\right]=\frac1{i}-\frac1{c+i}$$
This because $\frac{i-1}i=1-\frac{1}{i}$ denotes the probability that woman $i$ will not stand on the right place when it concerns the women (a taller woman keeps the price out of her sight) and $\frac1{c+i}$ is the probability that she will stand on the right place when it concerns the women but all children are standing on the side of the taller women as well, so that no child is between her and the price, as is requested. This involves the location of $c+i$ persons.
So then we end up with:$$\mathsf E(C_1+\cdots+C_w)=\sum_{i=1}^w\left[\frac1i-\frac1{c+i}\right]$$
