http://en.wikipedia.org/wiki/Pirate_game
What happens if you remove the order of seniority? Whenever a pirate dies, you randomly pick the next pirate to propose a distribution.
Here's my solution for 5 pirates: A: 48 B: 26 C: 26 D: 0 E: 0
How about for $ n $ pirates and $ p $ gold coins?
Your solution is correct for $n=5$. I generalize and explicitly state the assumptions.
For ease of composition, I will assume we distribute 100 coins (just replace $100$ with $p$ in the following if you want). I also assume the probability of becoming the proposer is uniform (equal for all). Moreover, I assume the pirates are risk neutral (this was not necessary in the original game, where we have no risk). Finally, I assume the proposer can only distribute integer values (if the strategy space is the real numbers, then there might not be an equilibrium).
Denote the amount that the proposer allocates to himself by $x_{i=z}$ and the vector of allocations to the rest by $x_{-i,z}$, where $z$ is the number of pirates left. For example, if only 2 pirates are left, the allocation to non-proposers is just a nonnegative number: $0\le x_{-i,2}\le 100$.
We solve the game by backward induction. Consider first the case where only 2 pirates are left, and the proposer $i$ has been determined. The proposer then chooses $x_i=100,x_{-i,2}=0$. That's just because 1 out of 2 votes is enough to realize the proposed distribution.
Now consider what the proposer needs to offer if $z=3$. The two non-proposers (note their situation is symmetric) could throw the proposer over board and share the treasure. If they do that, they expect payoff $1/2*100+1/2*0=50$, because the probability to become the proposer is $1/2$. Thus, in order not to get thrown over board, the proposer needs to offer one of them at least $51$! Thus, $x_i=49,~x_{-i,3}=(51,0)$. (Note, by the rules of the game, an allocation of $50$ will not be enough, as the pirate would then rather throw the proposer over board than accept the offer.)
Now consider case $z=4$. The 3 nonproposers expect payoff $1/3*0+1/3*51+1/3*49=100/3$ when throwing the proposer over board. The proposer thus chooses $x_i=66,~x_{-i,4}=(34,0,0)$, so he has 2 out of 4 votes.
Case $z=5$. The 4 nonproposers expect payoff $1/4*0+1/4*0+1/4*66+1/4*34=100/4=25$. So the proposer chooses $x_i=48,~x_{-i,5}=(26,26,0,0)$. (Exactly what you have!)
By induction, if $z$ pirates are left, we have $z-1$ nonproposers, who expect $100/(z-1)$ from throwing the proposer over board. The proposer thus chooses $$x_{i=z}=\begin{cases} 100-\frac{z-1}{2}\lceil 100/(z-1)\rceil & \text{if } z \text{ odd} \\ 100-\left(\frac{z}{2}-1\right)\lceil 100/(z-1)\rceil & \text{if } z \text{ even} \\ \end{cases}$$ and $$x_{-i,z}=\begin{cases} (x_1=\lceil 100/(z-1)\rceil,\ldots,x_{(z-1)/2}=\lceil 100/(z-1)\rceil,x_{(z-1)/2+1}=0,\ldots,x_{z-1}=0) & \text{if } z \text{ odd} \\ (x_1=\lceil 100/(z-1)\rceil,\ldots,x_{z/2-1}=\lceil 100/(z-1)\rceil,x_{z/2}=0,\ldots,x_{z-1}=0) & \text{if } z \text{ even} \\ \end{cases}$$ where $\lceil x\rceil$ is the next highest integer from $x$ (thus, if $x=50$, then $\lceil x\rceil=51$). Informally, in equilibrium the proposer gives half the nonproposers $\lceil100/(z-1)\rceil$ if $z$ is odd, and if $z$ is even, he gives half minus 1 of the nonproposers $\lceil100/(z-1)\rceil$ (so that, with the own vote, he gets a tie).
Thus, as in the original equilibrium, nobody is thrown off board for sufficiently small $n$: the first proposal is accepted. But the proposer for large enough $n$ cannot buy enough votes to survive. For $n=103$, we get $100-\frac{z-1}{2}\lceil 100/(z-1)\rceil=100-51*1=49$ (so the proposer survives and keeps almost half of the treasure). For $n=1003$, we get $100-\frac{z-1}{2}\lceil 100/(z-1)\rceil=100-501*1=-400$ (proposer would have to add 400 in order to buy the votes to survive). This would not be a problem if the proposer were allowed to allocate non-integers, but as I wrote it, every nonproposer needs at least 1 in order to accept.