Bertrand’s tartine

A riddle from The Riddler on cutting a square (toast) into two parts and keeping at least 25% of the surface on each part while avoiding Bertrand’s paradox. By defining the random cut as generated by two uniform draws over the periphery of the square. Meaning that ¼ of the draws are on the same side, ½ on adjacent sides and again ¼ on opposite sides. Meaning one has to compute

P(UV>½)= ½(1-log(2))

and

P(½(U+V)∈(¼,¾))= ¾

Resulting in a probability of 0.2642 (checked by simulation)

Leave a comment

This site uses Akismet to reduce spam. Learn how your comment data is processed.