Say Alice divides pie as A, 1-A, and A>=0.5. I split it into 4 cases:

1) A>=0.8 which means A/4 >= 1-A – the best Alice can do is 1-A/2, so this gives best case 60% for A=0.8.

2) 2/3<=A= 1-A but A/4 < 1-A, Bob can split into A/2, A/2 and 1-A ensuring that Alice cannot get more than 3A/4 < 60%.

3) 0.6<=A<2/3. Here A/2 < 1-A. Alice can't get more than 55%.

4) A<0.6 – Bob can split this into (A, 1-A, 0), now Alice cannot get more than A+0 < 0.6, and we already know Alice can get this much with A=0.8. Bob can probably do better here…

@R If Alice divides into equal pieces, why does Bob have to do the same? Bob can split into (0.5, 0.5, 0), and now Alice can't get more than 0.5.

]]>ðŸŸ¡ 40% , 40% , 20%

ðŸŸ¢ 40% , 20% , 20% , 20%

ðŸ”° A: 60% , B: 40% The Best

ðŸŸ¢ 50% , 50%

ðŸŸ¡ 50% , 49% , 1%

ðŸŸ¢ 50% , 24.5% , 24.5% , 1%

ðŸ”° A: 51% , B: 49% The Worst

A Will Always divides the middle piece in two in the last step. If she is greedy in the First step, b Will divide the biggelt part in equal pieces

]]>Alice cuts the pie into 80-20. Bob has to cut the piece of 80 otherwise Alice takes more than 80.

Assume the 2 pieces are x and 80-x, where x x/2 and 80-x > 20 when x <= 40. 80-x is the biggest piece.

Because x <= 40, x/2 = 60.

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