Coin Puzzle 4

July 31, 2020

Coin Puzzle 4

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 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 proposer can has the casting vote

Rules every pirates follows.

Every pirate wants to survive
Given survival, each pirate wants to maximize the number of gold coins he receives.

What is the maximum number of coins that pirate A might get?

Solution:

To know what amount coins A might get we need to see all those cases where A dies:

Case 1:

Let’s suppose a situation when A, B, and C died, such that D, and E are the only ones left. Now E knows that he will not get anything as D is senior and D will make a distribution of (100, 0). So E would be find with anything greater than 0.

Case 1: When A, B, and C dies and D, and E are left
Coins distribution in this case            
           A = 0 (as he is dead)
           B = 0 (as he is dead)
           C = 0 (as he is dead)
           D = 100 (as now he is only senior)
           E = 0 (as he is most junior)

In Case 1 distribution will be (100, 0) among D and E.

Case 2:

Let’s suppose a situation when A, and B dies and C, D, and E are only left. Now D knows he will not get anything as C is senior than D and C will make a distribution of (99, 0, 1). Now in earlier case E was not getting anything and now he is getting 1 coin so he will not oppose this situation and E will vote in favor of C.

Case 2: When A, and B dies and C, D, and E are left
Coins distribution in this case            
           A = 0 (as he is dead)
           B = 0 (as he is dead)
           C = 99 
           D = 0 
           E = 1

In Case 2 distribution will be (99, 0, 1) among C, D and E.

Case 3:

Let’s suppose a situation when only A dies and B, C, D, and E are left.

Now to survive B only needs to give 1 coin to D. So the distribution will be (99, 0, 1, 0).

Case 3: When A dies and B, C, D, and E are left
Coins distribution in this case            
           A = 0 (as he is dead)
           B = 99 
           C = 0 
           D = 1 
           E = 0

In Case 3 distribution will be (99, 0, 1, 0) among B, C, D and E.

Case 4:

Now, in similar manner when is alive he knows about Case 3, so he just need to give 1 coin to C and 1 coin to E to get them in favor. So, the distribution will be (98, 0, 1, 0, 1).

The idea is based on the fact that what B will distribute if A dies (B would always want A to die). If A gives more coins to 2 people than B would have given, A wins.

Case 4: When all A, B, C, D, and E are alive
Coins distribution in this case            
           A = 98 
           B = 0 
           C = 1
           D = 0 
           E = 1

In Case 4 distribution will be (98, 0, 1, 0, 1) among A, B, C, D and E.