I am considering the following process:
There are $m$ boxes in total and $n$ boxes are red and $0\leq n\leq m$. $m$ and $n$ are all very large. Now we have $K$ (very large) people, where $0\leq K\leq m$. Boxes are labeled from 1 to $m$.
In each round, everyone randomly submits a number form $1$ to $m$ that is completely independent of others. If the number a person submitted corresponds to a red box, then this person leave the system.
Because everyone's submission is completely random and is independent of others, hence, the number of submission each person makes is a geometric distribution with success probability $n/m$.
Now, consider this process as the base. We now put a treasure which has value $r$ in each red box. The whole process is the same as before. If it happens that a person submits a number which corresponds to a red box and that red box has a treasure, then the person will take that treasure (not the red box) with him. And hence, if later on this red box is called by other people, then there is no treasure in it. If in one round two people happen to call the same red box and this red box has never been called by people before, then the treasure is randomly assigned among the people who call it in this round.
My question is: what is the probability that when a person leaves the system, the red box he called has never been called by anyone else? Consider the situation when $n/m=p$ and $K/m=q$ and $m\rightarrow\infty$.
My attempt is the following:
Round $~~~$ People Finish the Game $~~~$ Red box with treasure left $~~~$ Prob. get treasure
1 $~~~~~~~~~~~~~~A_1=qp$ $~~~~~~~~~~~~~~~~~~~~~~~~~~~~~B_1=B_0-A_1B_0/p$ $~~~~~~~~~~~~~~1$
2 $~~~~~~~~~~~~~~A_2=A_1(1-p)$ $~~~~~~~~~~~~~~~~B_2=B_1-A_2B_1/p$ $~~~~~~~~~~~~~B_1/p$
$n$ $~~~~~~~~~~~~~A_n=A_{n-1}(1-p)$ $~~~~~~~~~~~~B_n=B_{n-1}-A_nB_{n-1}/p$ $~~~~B_{n-1}/p$
where $B_0=p$. Then we consider all the cases: if a man finishes the game at round 1, 2, 3,..., n,... So the corresponding probability is \begin{align*} &\sum_{i=1}^{\infty}(1-p)^{i-1}pB_{i-1}/p\\ &=\frac{1}{q}\sum_{i=1}^{\infty}A_iB_{i-1}/p\\ &=\frac{1}{q}(B_0-B_1+B_1-B_2+B_2-B_3+....)\\ &=\frac{1}{q}B_0=p/q \end{align*}
Is the last step correct? I think it may have some problem, as $B_i$ is bounded by $p$ for all $i$. Hence, the whole series should be bounded by 1 but now I get something that depends on $q$ and if $q$ is close to zero, then this $p/q$ can be greater than 1.
The above should be $\sum_{i=1}^{\infty}p(1-p)^{i-1}e^{-[1-(1-p)^i]q/p}$
The idea is that if a man finishes the game in round 1, then he must call a red box in round 1 and clearly all red boxes in round 1 has never been called by all the other people. If a man finishes the game in round 2, then he must call a non-red box in round 1, which occurs with probability $(1-p)p$, then as during the round 1, there are already $qp$ red boxes being tapped by people in round 1, the left red boxes are $p-qp$. Hence, in round 2, a man, conditioning on he calls a red box, will have chance $\frac{p-qp}{p}$ to call a red box that has not been called by previous people.
Assuming the treasure in each box is given to one person by some random process if more than one calls the number the first time it is called or the treasure is divided, just use symmetry. Everybody has an equal chance at a treasure, so the chance they get it is $\frac nK$. The value to each person is then $\frac {nr}K$. The details of how you select the winner don't matter as long as it is random. The only way this fails is if you give a similar treasure to all the people who call the right number the first time it is called so the total value of treasures is higher than $r$ in some cases.