Fairly Simple Puzzle

[jds30]

I was given a puzzle. I read the solution. I don't understand the solution. Given that the puzzle was a just a warm-up, I'm slightly embarrassed and exasperated.

Here it is:

There are 2 rectangular identical cakes. Jeremy is the cutter. His friend Marie is there hoping to get as much cake as possible (along with Jeremy). Jeremy will cut the first cake into two pieces, perhaps evenly, perhaps not. After seeing the cut Marie will decide to choose first (and take the bigger piece...both parties want as much cake as possible). If she goes first, she will take the larger piece. If she goes second, she can assume Jeremy will take the larger piece.

Next, Jeremy will cut the second cake into two pieces (remember that one of the pieces can be vanishingly small if he so chooses). If Marie had chosen first for the first cake, then Jeremy gets the larger piece of the second cake. If Marie had chosen second for the first cake, then she gets the larger piece of the second cake.

Assuming each person will strive to get the most total cake possible, what is an optimal strategy for Jeremy?

Hint: Assume that Jeremy divides the first cake into fractions f and 1-f where f is at least 1/2.

While this isn't the whole solution, here's what is relevant:

If Marie takes the fraction f piece, then Jeremy will take the entire second cake. So, Marie will get exactly f and Jeremy will get (1-f) +1. If Marie takes the smaller piece of the first cake (fraction 1-f), Jeremy will do best if he divides the second cake in half. This gives Marie (1-f) +1/2. Jeremy follows this reasoning, so realizes that the best he can do is to make f = (1-f) +1/2. That is, 2f = 1.5 or f = 3/4.

My Question:

What I don't understand are the steps in Jeremy's reasoning that lead him to conclude the best he can do is when f = (1-f) +.5 ? WHY is the best Jeremy can do when the big piece of the first cake (f) is the same amount as the small piece of the first cake (1-f) plus half of the second cake?

 

[lionelhill]

I'll try to explain how I see it, another way.

Way 1
=====

There are two situations: (a) Marie takes small half first; (b) Marie takes large half first.

For each, you can plot what she gets versus how the first cake was cut. If she takes the big half first, you can assume she gets virtually nothing of the second cake, so her "winnings" start at 1 if Jeremy cuts the first cake right at the end, and fall to 0.5 by the time Jeremy decides to cut it exactly in half. Straight line graph from 1.0 to 0.5 as the cut changes from 0 to 0.5.

For scenario (b), Marie gets what she wants of the second cake, and you can assume Jeremy makes sure it is barely detectably more than 0.5. Therefore her winnings vary from 0 to 0.5 on the first cake, plus 0.5 of the second, i.e. a straight line from 0.5 to 1.0, as the cut changes from 0 to 0.5.

Straight lines from 1.0 to 0.5 and 0.5 to 1.0 intersect in the middle, i.e. winnings of 0.75 for a cake cut one quarter of the way from the end.

Way 2
=====

The original solution does the algebra for situation (a) and situation (b), calculating Marie's winnings. It then puts winnings (a) equal to winnings (b) and solves for f. This assumes situations (a) and (b) are in conflict, and that the optimum happens where the two give the same result, which is true, as seen above.

Comment
=======

But personally I have a strong leaning towards puzzles that are simply stated and turn out hairier than they first appear. I understand your exasperation. There's a lot less to this puzzle than meets the eye.

 

[bibulous]

Cutting the first cake into 1/4 & 3/4 means that Marie will only ever get 3/4 of one cake & Jeremy will always get 1 whole cake & a quarter (since both cakes are equal in size).

Case 1 - If Marie picks the 1/4 of the first cake then Jeremy will cut the second cake in half exactly, so it doesn't matter which section Marie chooses giving her only 3/4 of one cake.

Case 2 - If Marie Picks the 3/4 of the first cake, Jeremy will cut an infinitely small sliver off the second cake & take the big piece for himself. Again this leaves Marie with only 3/4 of one cake.

Hope this makes sense.

 

[CajunCenturion]

==> What I don't understand are the steps in Jeremy's reasoning that lead him to conclude the best he can do is when f = (1-f) +.5 ? WHY is the best Jeremy can do when the big piece of the first cake (f) is the same amount as the small piece of the first cake (1-f) plus half of the second cake?
Jeremy should reason that if he cuts the first cake such that the larger piece is greater than 3/4 of the cake, then Marie will actually take the larger piece, not the smaller. That will leave Jeremy with the entire 2nd cake, plus less than 1/4 of the first cake, since Marie took over 3/4 of the first cake. Jeremy's take < 1.25.

If Jeremy cuts the first cake so that the larger piece is between 1/2 and 3/4, then Marie will take the smaller piece, leaving Jeremy with 1/2 of the second cake plus somewhere between 1/2 and 3/4 of the first cake. That means Jeremy's take is between 1/2 + 1/2 and 1/2 + 3/4, or between 1.00 and 1.25.

If Jeremy cuts the first cake into exactly 3/4 and 1/4, then it doesn't matter which piece Marie takes. If she takes the 1/4, the second cake is 50/50 so Jeremy gets 3/4 + 1/2 or exactly 1.25. If Marie takes the larger piece (3/4), again, Jeremy gets exactly 1.25 with 1/4 of the first cake plus the entire second cake.

So to maximize his take, Jeremy should cut the first cake into a 3/4 and a 1/4 piece.

--------------
Good Luck
To get the most from your Tek-Tips experience, please read
FAQ181-2886
As a circle of light increases so does the circumference of darkness around it. - Albert Einstein