Jordan has five green gumdrops and five orange gumdrops left from Halloween. His mother tells him
that he must let his younger brother have one of the gumdrops. Jordan prefers the orange gumdrops over
the green gumdrops, but his mother tells him that he must let his younger brother choose.

To make things interesting, the mother gives Jordan two plates and tells him that he can distribute the
gumdrops any way he likes between the two plates so long as all ten gumdrops are there. The younger
brother will then choose a plate at random and then take a gumdrop from that plate. How can Jordan
distribute the gumdrops to minimize the chances that his brother takes an orange gumdrop?

Note: neither of the plates can be empty and all 10 candies must be placed (no hiding the orange
candies somewhere else).

**ANSWER**:

*Place a single green gumdrop on one plate and the remaining 9 gumdrops on the other plate.*
**EXPLANATION**: The probability that his younger brother will choose a orange gumdrop is as follows.
There is a 50% chance the younger brother will choose the first plate (where the probability of him
taking a orange gumdrop is 0%) and a 50% chance the younger brother will choose the second plate (where
the probability of him taking an orange gumdrop is 5/9 or 55.6%). Added together, the combined probability
that the younger brother will take an orange gumdrop is (50% × 0%) + (50% × 55.6%) = 27.8%

Do you have a

suggestion for this puzzle (e.g. something that should
be mentioned/clarified in the question or solution, bug, typo, etc.)?