**1.**Again, suppose Jeremy cuts the first cake into the fractions

**f**and

**1-f**, where

**f**is at least 1/2. Then if Marie chooses first, Jeremy will get 1 1/4 cakes from the last two. If Marie chooses to go second, then Jeremy will get

**f**of the first cake and then will divide the last two cakes in half.

So, **f** **+ 1/2 + 1/2 = (1 - f) + 1 1/4.**

That is, **f**** + 1 = 2 1/4 - f, or 2f = 1 1/4.**

That is **f** **= 5/8.** Therefore Jeremy will receive 1 5/8 cakes and Marie will receive 1 3/8 pieces.

**2.** There is an inductive pattern to these problems. If for **k** cakes where Marie gets the first choice in all but one, Jeremy still has an advantage **A** (that is, Jeremy would receive **A** more cake than Marie among those **k**), then Jeremy will still have an advantage for **k+1** cakes.

Here is why. Jeremy, knowing that he will have an advantage of **A** if Marie chooses first for the first cake, cuts the first piece into two pieces, **1/2 + A/4** and **1/2 - A/4**. If Marie, chooses the piece having **1/2 + A/4**, then Jeremy will suffer a disadvantage of **A/2** for that first cake, but will gain an advantage of A for the remaining k cakes, so have a net advantage of **A/2**. By contrast, if Marie chooses to go second for the first cake, then Jeremy will cut the remaining cakes in half. Again he will have an overall advantage of **A/2**. Thus, Jeremy's advantages for from 1/2 for two cakes, to 1/4 for three cakes, to 1/8 for four cakes, to ... 1/64 for seven cakes.

**3.** Yes, simply make it so that Marie gets the first choice every time. Jeremy will then be forced to cut evenly every time.