The red candies run out first
A jar holds 10 red, 20 blue and 30 green candies. They are taken out one at a time, in a uniformly random order.
Find the probability that when the last red candy is taken out, at least one blue and at least one green candy are still in the jar. In other words: red is the first colour to run out.
Fractions, decimals with at least 3 significant digits, and expressions such as sqrt(2)/2, e^-1 or pi/4.
Read the order backwards, from the last candy taken out. The colour that runs out first is the colour seen last when reading backwards.
Which colour is the very last candy? Then, among the other two colours, which one has its last candy later?
h
s Shown after a correct answer. Revealing it first marks the problem as seen, not solved.
Idea. Read the sequence from the end. Red runs out first exactly when red is the last colour to be seen from the end: the last candy overall is blue or green, and among the other two colours, the later-finishing one is not red.
Step 1: the last candy. In a uniform random order, the last candy is green with probability , blue with probability , red with probability .
Step 2: if the last candy is green. Then green runs out last. Red runs out first exactly when, among the red and blue candies only, the last one is blue. The red and blue candies are in uniform random order among themselves, so this has probability .
Step 3: if the last candy is blue. Similarly, red runs out first when, among the red and green candies, the last one is green: .
Step 4.
Why it works. Conditioning on the last candy being green says nothing about the relative order of the red and blue candies, which is still uniformly random. Reading backwards turns “first to run out” into “last to appear”, which is easy to compute.