Nicholas and Peter share their nuts

[BMAD]

Nicholas and Peter are dividing (2n+1) nuts. Each wants to get more. Three ways for that were suggested. (Each consist of three stages.) First two stages are common. 

1 stage: Peter divides nuts onto 2 heaps, each contain not less than 2 nuts. 2 stage: Nicholas divides both heaps onto 2 heaps, each contain not less than 1 nut. 3 stage: The 3rd stage will be one of the following ways:   1 way: Nicholas takes the biggest and the least heaps. 2 way: Nicholas takes two middle size heaps. 3 way: Nicholas takes either the biggest and the least heaps or two middle size heaps, but gives one nut to the Peter for the right of choice.

Find the most and the least profitable method for the Nicholas.

[gavinksong]

With the first method, Nicholas can always take the larger pile after Peter splits the nuts. Thus, Peter will always split the nuts into n and n+1 nuts to minimize his losses, always leaving Nicholas with n+1 nuts.

With the second method, if the smaller pile is more than half the size of the bigger pile after Peter splits the nuts, then Nicholas can always take the bigger pile. If Peter instead splits the nuts such that the bigger pile is at least twice as large as the smaller pile, Nicholas can split the smaller pile into 1 and x nuts, and the larger pile evenly into y and y(+1) nuts. Since y >=x, this results in Nicholas winning x+y nuts, and Peter winning 1+y(+1) nuts. This is okay for Peter if x=1, which also wins him the extra nut because the larger pile is odd. Thus, Peter will always split the nuts into piles of 2 and 2n-1, leaving Nicholas with exactly n nuts every time.

With the third method, if Peter splits the nuts at all unevenly, Nicholas can take the larger pile. Thus, Perer will split the nuts evenly. The best that Nicholas can do now is take the slightly larger pile of n+1 nuts, but now since he has to pay Peter a nut, he ends up with only n nuts.

The first method is the most profitable for Nicholas. The second and third methods are tied for the least profitable.

[BMAD]

My spacing went away:

 

 

Nicholas and Peter are dividing (2n+1) nuts. Each wants to get more. Three ways for that were suggested. (Each consist of three stages.) First two stages are common. 

1 stage: Peter divides nuts onto 2 heaps, each contain not less than 2 nuts.

2 stage: Nicholas divides both heaps onto 2 heaps, each contain not less than 1 nut.

3 stage:

 

The 3rd stage will be one of the following ways:  

1 way: Nicholas takes the biggest and the least heaps.

2 way: Nicholas takes two middle size heaps.

3 way: Nicholas takes either the biggest and the least heaps or two middle size heaps, but gives one nut to the Peter for the right of choice.

 

Find the most and the least profitable method for the Nicholas.

[k-man]

First way is the most profitable for Nicholas. Nicholas can guarantee himself at least n+1 nuts by initially dividing 2n+1 into heaps of 2n-1 and 2. Peter is now going to try to make the largest heap as small as possible. which will be n nuts, so Nicholas gets n+1.

The second way is the least profitable for Nicholas. Regardless of how Nicholas divides initially, Peter can guarantee at least n+1 nuts for himself. To maximize his share Nicholas has to split into n and n+1 sized heaps.

In the third way, Peter can always split the 2 heaps created by Nicholas in such a way that the two middle size heaps differ from the other two heaps by exactly one nut, thereby winning with n+1 nuts every time.

[gavinksong]

First way is the most profitable for Nicholas. Nicholas can guarantee himself at least n+1 nuts by initially dividing 2n+1 into heaps of 2n-1 and 2. Peter is now going to try to make the largest heap as small as possible. which will be n nuts, so Nicholas gets n+1.

The second way is the least profitable for Nicholas. Regardless of how Nicholas divides initially, Peter can guarantee at least n+1 nuts for himself. To maximize his share Nicholas has to split into n and n+1 sized heaps.

In the third way, Peter can always split the 2 heaps created by Nicholas in such a way that the two middle size heaps differ from the other two heaps by exactly one nut, thereby winning with n+1 nuts every time.

Sorry, but Peter moves first.