Comments on: The 2 Cake Cutting Problem http://cmalabs.com/blog/2009/09/the-2-cake-cutting-problem/#utm_source=rss&utm_medium=rss&utm_campaign=the-2-cake-cutting-problem Technology and ideas Wed, 19 Jan 2011 05:18:20 +0000 hourly 1 http://wordpress.org/?v=3.2.1 By: cma http://cmalabs.com/blog/2009/09/the-2-cake-cutting-problem/#comment-400 cma Tue, 22 Sep 2009 12:38:31 +0000 http://cmalabs.com/blog/?p=563#comment-400 Yep, you're absolutely right! In fact, if Kim decides to play this game, Ted will cut it at exactly 25%|75%. Here is why. Suppose Kim picks the larger slice of the first cake, then Ted would make the smaller slice for the second cake infinitesimally small. So he would effectively get 1/n + 1 cakes where n > 1 If Kim decides to let Ted have the larger slice of the first cake then he would cut the second cake almost in half. So he would effectively get 1 - 1/n + 1/2 cakes. Depending on what n is, Kim could always try and maximize the amount of cake she gets by choosing the option with min(1/n+1, 1 - 1/n + 1/2) for Ted. Thus, to maximize the amount of cake Ted gets. Ted would have to find n such that 1/n + 1 = 1 - 1/n + 1/2 Solve for n and we find that n = 4. So Ted ends up getting 1.25 cakes regardless of what Kim gets. If n > 4, Kim will pick the larger slice of the first cake and Ted would get < 1.25 cakes. If n < 4, Kim would let Ted have the larger slice of the first cake and again, Ted would have < 1.25 of a cake. So n = 4 is the equilibrium. :) Yep, you’re absolutely right! In fact, if Kim decides to play this game, Ted will cut it at exactly 25%|75%. Here is why.

Suppose Kim picks the larger slice of the first cake, then Ted would make the smaller slice for the second cake infinitesimally small.
So he would effectively get 1/n + 1 cakes where n > 1

If Kim decides to let Ted have the larger slice of the first cake then he would cut the second cake almost in half. So he would effectively get 1 – 1/n + 1/2 cakes.

Depending on what n is, Kim could always try and maximize the amount of cake she gets by choosing the option with min(1/n+1, 1 – 1/n + 1/2) for Ted.

Thus, to maximize the amount of cake Ted gets. Ted would have to find n such that 1/n + 1 = 1 – 1/n + 1/2

Solve for n and we find that n = 4. So Ted ends up getting 1.25 cakes regardless of what Kim gets. If n > 4, Kim will pick the larger slice of the first cake and Ted would get < 1.25 cakes. If n < 4, Kim would let Ted have the larger slice of the first cake and again, Ted would have < 1.25 of a cake. So n = 4 is the equilibrium. :)

]]> By: StudentLeader http://cmalabs.com/blog/2009/09/the-2-cake-cutting-problem/#comment-398 StudentLeader Mon, 21 Sep 2009 07:30:13 +0000 http://cmalabs.com/blog/?p=563#comment-398 If Ted wants the most cake he should offer a reasonable cut for the first slice (say 75%|25%) - it is important that Kim accepts this so he can make the second slice 99%|1% If Ted wants the most cake he should offer a reasonable cut for the first slice (say 75%|25%) – it is important that Kim accepts this so he can make the second slice 99%|1%

]]> By: StudentLeader http://cmalabs.com/blog/2009/09/the-2-cake-cutting-problem/#comment-397 StudentLeader Mon, 21 Sep 2009 07:22:30 +0000 http://cmalabs.com/blog/?p=563#comment-397 Not a good deal for Kim. Assume first cut is 90%/10% If Kim accepts 90% then Ted should cut an extremely large portion of the second cake (say 99%|1%) This means Ted = 109% while Kim = 91% If Kim declines the first cake then Ted should cut the second slice as equal as possible (say 51%|49%) This means Ted = 141% while Kim = 59% Kim should thus allways accept the first slice. Not a good deal for Kim.

Assume first cut is 90%/10%
If Kim accepts 90% then Ted should cut an extremely large portion of the second cake (say 99%|1%)
This means Ted = 109% while Kim = 91%
If Kim declines the first cake then Ted should cut the second slice as equal as possible (say 51%|49%)
This means Ted = 141% while Kim = 59%

Kim should thus allways accept the first slice.

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