Question

Three pirates must divide 100 gold doubloons. The doubloons cannot be cut into pieces. Pirate A is the strongest, followed by Pirate B, followed by Pirate C. Because of ancient pirate tradition, the coins are divided in the following manner. First Pirate A proposes a division of the coins. The three pirates then vote on whether to accept the proposed division. If the proposal gets a majority vote, it is accepted, and the game is over. If the proposal fails to get a majority vote, Pirate A is executed. It is then Pirate B's turn to propose a division of the coins between the two remaining pirates. The same rules apply, with one exception: if the vote is a tie, Pirate B, being the strongest remaining pirate, gets an additional vote to break the tie.
If we assume that in any proposal there are no doubloons left over, how many different proposals could Pirate A make?

Answer 1

Answer:

$C^{102}_{100}=5151$

Step-by-step explanation:

Let's imagine the following situation, if we want to distribute 100 coins between three pirates we could represent this situation with a line arrangement. For example if we had7 coins and 3 pirates one possible distribution of coins would be given by  CC|CCCC|C, the C's represent coins and the bars the boundaries between two pirates, for the particular line arrangement shown, we have that pirate A has 2 coins, B has 4 coins and C has a single coin. Another possible arrangement is,

|CCC|CCCC, where pirate A has no coin, pirate B has 3 coins and C has 7 coins. If we take notice of the fact that the arrangement representing a distribution is composed of 9 elements, that is 7 C's and 2 | (bars), then a way to make an arrangement would be to fill 9 empty boxes with our available coins and bars in all the possible ways. This means that if we first choose to fill 7 out of 9 boxes with  coins then the number of possible combinations is $C^{9}_7=\frac{9!}{7!(9-7)!}36$. In general if we want to distribute n elements in k boxes, where the boxes can either be filled with any number of elements (including 0 number of elements), we have that the number of possible distributions will be $C^{n+k-1}_{n}=\frac{(n+k-1)!}{n!(k-1)!}[/[tex], where we used the fact that we need k-q bars to represent k boxes. Thus pirate A can choose from [tex]C^{102}_{100}=5151$ possible divisions.

Bonus:

If every pirate wants to have the maximum number of coins possible without being executed, here's how pirate A has to divide the coins in order to keep the largest amount of coins.

We have to think backwards to figure this out. Imagine pirate A was executed and there are only two remaining players. Pirate B should propose to keep all the coins, pirate C could oppose but pirate B's vote would break the vote and keep all the loot. Pirate A, B and C are all aware of this, so pirate A should propose to keep 99 coins and give the remaining gold piece to pirate C, Pirate B will of course oppose the division, but pirate C should accept because if not he would get no coins. Thus the division would be.

A: 99 coins

B: 0 coins

C: 1 coins