The last red candy

A jar holds 10 red, 20 blue, and 30 green candies — 60 total. You draw them one at a time, randomly. What's the probability that when you've drawn all 10 red candies, at least 1 blue and 1 green candy remain in the jar?

Running many trials

Now let's run many trials at once and see how often red is the first color to be completely drawn:

After enough trials, the fraction converges to about 58% — surprisingly high given that red is the smallest group.

The symmetry insight

Here's a key observation. Instead of tracking all 60 draws, focus on just three special candies: the last red, the last blue, and the last green. Their relative order determines everything:

If the last candy overall is green or blue (not red), then red was fully emptied while at least some of that last color remained. This reframing turns a 60-candy problem into a 3-candy problem.

Reframing the problem

Let $T_R$, $T_B$, $T_G$ be the positions (1 through 60) at which the last red, blue, and green candies are drawn, respectively.

This is equivalent to asking: among the three "last candy" positions $T_R$, $T_B$, $T_G$, is the red one the smallest?

Not quite — we need to be more careful. There are two mutually exclusive scenarios that satisfy $T_R < T_B$ and $T_R < T_G$:

$P(T_R < T_B \text{ and } T_R < T_G) = P(T_R < T_B < T_G) + P(T_R < T_G < T_B)$

Computing each piece

Case 1: $T_R < T_B < T_G$ (green candy is drawn last overall)

Step 1. What's the probability the last candy drawn (position 60) is green?

Each of the 60 candies is equally likely to be in position 60. There are 30 green candies:

$P(T_G = 60) = \frac{30}{60} = \frac{1}{2}$

Step 2. Given the last candy is green, what's the probability the last red is drawn before the last blue?

Remove the last green candy from consideration. We now have 59 candies: 10 red, 20 blue, 29 green. Among these 59, which is drawn last — the last red or the last blue?

Equivalently: among the 30 red-and-blue candies (ignoring the 29 green ones that are irrelevant to this ordering), which one is last? Each of these 30 candies is equally likely to be the very last red-or-blue candy drawn, and 20 of them are blue:

$P(T_R < T_B \mid T_G = 60) = \frac{20}{30} = \frac{2}{3}$

Case 1 total:

$P(T_R < T_B < T_G) = \frac{30}{60} \times \frac{20}{30} = \frac{1}{2} \times \frac{2}{3} = \frac{1}{3}$

Case 2: $T_R < T_G < T_B$ (blue candy is drawn last overall)

Step 1. The last candy is blue:

$P(T_B = 60) = \frac{20}{60} = \frac{1}{3}$

Step 2. Given the last candy is blue, the last red is drawn before the last green? Among the 40 remaining red-and-green candies (10 red, 30 green), 30 are green, so:

$P(T_R < T_G \mid T_B = 60) = \frac{30}{40} = \frac{3}{4}$

Case 2 total:

$P(T_R < T_G < T_B) = \frac{20}{60} \times \frac{30}{40} = \frac{1}{3} \times \frac{3}{4} = \frac{1}{4}$

The answer

$P(T_R < T_B \text{ and } T_R < T_G) = \frac{1}{3} + \frac{1}{4} = \frac{7}{12} \approx 58.33\%$

The elegant shortcut

There's an even cleaner way to see this. Focus only on the last red candy, the last blue candy, and the last green candy — three specific candies in the sequence of 60. The question is: what's the probability the red one comes first among these three?

This isn't quite a symmetry argument (the three "last" candies aren't equally likely to be in any position), but the conditional approach above gives us the exact answer.

Competition appearances

• Quant interviews: This exact problem with 10/20/30 candies is a classic Jane Street question
• Putnam training: Conditional probability on orderings of "last elements" is a powerful technique
• Programming contests: Efficient simulation of draw-without-replacement processes
• AMC/AIME: Problems involving drawing colored balls until one color is exhausted

Takeaway

The candy jar problem looks combinatorial but yields to conditional probability. Instead of counting arrangements of 60 candies, we focus on the relative order of three special candies (the last of each color). The technique of conditioning on "which color finishes last" and then asking about the remaining two is a general tool for any multi-color depletion problem.