### 538 Riddler: Puzzle of the Robot Pizza Cutter

#### 26 June ’16

The robot's first cut will create 2 pizza slices. The robot's second cut will create either 3 or 4 pizza slices. The robot's third cut will create 4, 5, or 6 slices if starting with 3 slices or 5, 6, or 7 slices if starting with 4 slices. I began by thinking about the probability of these outcomes in a specific state. So that the geometry mimics the probability distribution, I'm setting the circumference of the pizza equal to 1.

Second Cut | Third Cut (Starting w/ 3 Slices) | Third Cut (Starting w/ 4 Slices) |
---|---|---|

Next, I integrated over the entire state space to get the probability of each of these outcomes. The state space for the second cut (after 1 cut has been made) is very easy to define; we only need 1 variable (x) to specify the size of each of the two slices. The state space for the third cut (after 2 cuts have been made) is more challenging; we need 3 variables to define the space (x, y, and z). Each integral is multipled by the number of circular permutations; in the case of the third cut, there are 4 points, so (4 - 1)! = 6.

For that second cut:

For that third cut:

Putting this all together: