### Brain Teasers

# Jellybean Party Favors

Probability
Probability puzzles require you to weigh all the possibilities and pick the most likely outcome.Probability

Rex is having a party and has prepared party favors for his guest. As each guest arrives they are given a small paper bag of licorice and cherry jellybeans. Each bag has the same number of jellybeans, but no two bags have the same combination of cherry and licorice jellybeans.

Altogether there are 42 bags of jellybeans and 41 jellybeans in each bag. What is the probability that the first three jellybeans a guest randomly removes from her bag are the same flavor?

Altogether there are 42 bags of jellybeans and 41 jellybeans in each bag. What is the probability that the first three jellybeans a guest randomly removes from her bag are the same flavor?

### Hint

The number of (licorice, cherry) in each bag are (41, 0), (40, 1), (39, 2), ... (1, 40), (0, 41).The answer would be the same if there were 26 bags of 25 jellybeans each.

### Answer

50%The answer is the same for any value of N jellybeans per bag in N+1 bags where N >= 3.

In the simplest case (4 bags of 3 jellybeans each), it's easy to see that the two guests that got a bag with only a single flavor of jellybean will always remove three of the same flavor and that the two guests that have two of one flavor and one of the other flavor will never remove three of the same flavor. 2/4 = 50%

With 5 bags of 4, the two bags with all one flavor always produce three of the same flavor. The two bags with three of one flavor and one of another flavor have a 25% change of leaving the single flavor jellybean in the bag. The bag with two of each flavor never has three of the same. So the probability is:

2/5 * 1 + 2/5 * 1/4 = 2/5 + 2/20 = 4/10 + 1/10 = 5/10 = 1/2

For N jellybeans in a bag, choose one of (N+1) bags and then N*(N-1)*(N-2) ways to choose three jelly beans from the bag for a total of

(N+1)*N*(N-1)*(N-2)

ways to draw three jellybeans.

Just considering one flavor, there are (X) * (X-1) * (X-2) ways to draw three of the flavor from a bag containing X jellybeans of that flavor. The total number of ways to choose three of that flavor of jellybean is then the summation:

sum{X=0..N: (X) * (X-1) * (X-2)}

This summation is equivalent to:

(N+1)*N*(N-1)*(N-2)/4

There are two flavors, so the total number of ways to draw the first three jellybeans of the same flavor are:

(N+1)*N*(N-1)*(N-2)/2

Divide this by the total number of ways to draw three jellybeans from a bag:

((N+1)*N*(N-1)*(N-2)/2) / ((N+1)*N*(N-1)*(N-2))

Canceling out the common terms leaves:

(1/2)/1 = 1/2

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