Pirate's Game: What if proposer can't vote I had read the Pirate's game 
With other number of pirates and coins we can think the same way
But I supposed maximum number of coins that pirate A might get if plan still accepted when get a half of votes but the proposer can't vote for his plan ?
The problem will be  :

There are 5 pirates, they must decide how to distribute 100 gold coins
  among them. The pirates have seniority levels, the senior-most is A,
  then B, then C, then D, and finally the junior-most is E.
Rules of distribution are:
  
  
*
  
*The most senior pirate proposes a distribution of coins.
  
*All pirates, who not proposes the distribution, will vote on whether to accept the distribution.
  
*If the distribution is accepted, the coins are disbursed and the game ends.
  
*If not, the proposer is thrown and dies, and the next most senior pirate makes a new proposal to begin the system again.
  
*In case of a tie vote the distribution will be accepted
  
  
  Rules every pirates follows.
  
  
*
  
*Every pirate wants to survive and doesn't believe in anyone else
  
*Given survival, each pirate wants to maximize the number of gold coins he receives.
  

 A: As an example, let's consider the case with $5$ pirates ($A$, $B$, $C$, $D$, and $E$) and $100$ coins:


*

*If only $D$ and $E$ are left, then the distribution is $(0,100)$.  If $D$ offers anything other than all the coins to $E$, then $E$ votes against the proposal and takes all $100$ for himself/herself.  Even with the distribution $(0,100)$, pirate $E$ can vote against the plan if he/she chooses (perhaps due to personality conflicts, etc).

*If $C$, $D$, and $E$ are left, then the distribution is $(99,1,0)$.  $D$ will vote for this plan as it's better than getting $0$.

*If $B$, $C$, $D$, and $E$ are left, then the distribution is $(97,0,2,1)$.  In this case, $B$ needs two votes to survive.  $B$ won't get $C$'s vote unless $B$ offers $C$ more than $99$, which isn't going to happen.  To guarantee the other two votes, $B$ offers one more to $D$ and $E$ than what they would get without $B$.

*If all pirates are alive, then the distribution is $(97,0,1,0,2)$.  In this case, $A$ needs two votes and isn't going to get $B$'s vote without offering more than $97$, which isn't going to happen.  Therefore, $A$ offers a little extra gold to two who would lose out (and get the smallest amounts) if $A$ were killed.
Just for fun, here are the next few cases:


*

*With $6$ pirates, $(96,0,1,2,1,0)$

*With $7$ pirates, $(96,0,1,2,0,0,1)$ or $(96,0,1,0,0,2,1)$

*With $8$ pirates, things get a little interesting.  Since there are two options for $7$, $(96,0,1,0,1,1,1,0)$ might be enough because the pirates who get $2$ in the case of $7$ might not want to risk getting $0$.

*With $9$ pirates, any of $(95,0,1,2,1,0,0,0,1)$, $(95,0,1,0,1,2,0,0,1)$, $(95,0,1,0,1,0,2,0,1)$, or $(95,0,1,0,1,0,0,2,1)$ should get enough votes.

*With $10$ pirates, $(95,0,1,0,1,0,1,1,1,0)$ should be a good distribution.
The number of coins doesn't really matter, just as long as there are enough coins to distribute nicely.  If there are more pirates than half the coins, things might get quite interesting.  Likely there will be a lot of killing in those cases to bring the number of pirates down.
In these types problems, sometimes the pirates are defined to be "bloodthirsty."  In this case, the pirates choose to kill whenever their two votes would give the same result.  This would mean that in the first case (only $D$ and $E$ left, $E$ would always vote to kill $D$).
