Jellybean Party Favors
Probability puzzles require you to weigh all the possibilities and pick the most likely outcome.
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?
HintThe 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.
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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*(N1)*(N2) ways to choose three jelly beans from the bag for a total of
(N+1)*N*(N1)*(N2)
ways to draw three jellybeans.
Just considering one flavor, there are (X) * (X1) * (X2) 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) * (X1) * (X2)}
This summation is equivalent to:
(N+1)*N*(N1)*(N2)/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*(N1)*(N2)/2
Divide this by the total number of ways to draw three jellybeans from a bag:
((N+1)*N*(N1)*(N2)/2) / ((N+1)*N*(N1)*(N2))
Canceling out the common terms leaves:
(1/2)/1 = 1/2
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Comments
Zag24
Mar 02, 2009
 javaguru, you have made a lot of interesting puzzles. I love these ones that, on first blush, seem like way too much computation to figure out, but suddenly simplify if you just look at it the right way. Great puzzle. 
(user deleted)
Mar 07, 2009
 wow you must be really smart but this 2 hard try and make it more easy k thanks 
EntangledQuark
Mar 11, 2009
 Great teaser! I was a good ways into solving it the hard way when I realized there was a pattern I was missing. Took me a while to figure out what it was and then to see that the solution was a constant. I did it without the hint!
Oh, and PLEASE don't dumb down your teasersthere are already enough easy and dumbeddown teasers here. Your teasers are real diamonds amongst the coal. I'm working my way through themI'm determined to solve all of them without using the hints. Please DO make more of the same. 
javaguru
Mar 11, 2009
 Thanks! Glad you like them. They're a lot of work to create, but it's a real blast when I get responses like these from Zag24 and you! If you like hard, you'll love my next one: "Rex and Ralph: Conglomerate Valuation". 
WildArabian
Mar 17, 2009
 Wow I blow a few brain circuits every time I try 2 solve 1 of these 
RebelAir
Mar 21, 2009
 2 flavors, 50% chance.
All that math confused me. 
bradon182001
Apr 30, 2009
 I must admit I was overwhelmed at first, but after settling down and reasoning it out, I got it. Thanks for posting. Great teaser. 
(user deleted)
May 17, 2009
 Great teaser! 
Paladin
Aug 07, 2009
 The hint helped tremendously. I never took any probability, but I was a math major. Once I knew 42 & 41 had the same probability as 26 & 25, I automatically went to the simplest form (4 & 3) and wrote out the permutations:
CCC
CCL
CLL
LLL
And from there I could see these were also the 4 possible draws one could make when drawing out 3 beans, and of the 4 possibilities, 2 of them yielded the required results, so there was a 50% chance.
I love the math teasers, and yours are the best. Keep them coming! 
Rainbownerd
Dec 03, 2011
 i got freaked by the answer its so long 
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