Brain Teasers
Cereal Gift
Fun: (2.23)
Difficulty: (2.68)
Puzzle ID: #8806
Submitted By: lesternoronha1 Corrected By: Winner4600
Submitted By: lesternoronha1 Corrected By: Winner4600
Probability
Probability puzzles require you to weigh all the possibilities and pick the most likely outcome.Probability
There is a free gift in my breakfast cereal. The manufacturers say that the gift comes in four different colors, and encourage one to collect all four (thus eating lots of cereal). Assuming there is an equal chance of getting any one of the colors, what is the expected number of boxes I must consume to get all four?
Answer
25/3 trials are required.First I'll show that if the probability of an event happening is p then the mean number of trials to obtain a success is 1/p.
Number of Trials -- Probability of Success
----------------------------------------------
1 -- p
2 -- p*q
3 -- p*q^2
4 -- p*q^3
. -- .
. -- .
. -- .
Since there must eventually be a success, the sum of probabilities is:
p + p*q + p*q^2 + p*q^3 + ... = 1.
The mean number of trials (m) is: m = p + 2p*q + 3p*q^2 + 4p*q^3 + ...
m*q = p*q + 2p*q^2 + 3p*q^3 + ...
m-q*m= p + p*q + p*q^2 + p*q^3 + ...
m(1-q) = 1.
m = 1/(1-q) = 1/p.
Secondly, the answer to the problem can be express as the sum of the following:
Number of trials to get a first toy
Number of trials to get a second toy once you have one toy
Number of trials to get a third toy once you have two toys
Number of trials to get the final toy once you have three toys
The number of trials to get one toy is obviously one.
Once you have one the probability of getting a different toy in the next box is 3/4, thus the expected number of trials is 1/(3/4) = 4/3 to get the second toy.
By the same logic the number of trials to get the third is 1/(1/2)=2 and 1/(1/4)=4 for the final toy.
Summing these yields 1 + 4/3 + 2 + 4 = 25/3.
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Comments
huh
I think this is one of the best Probability teasers on here. I have been working on this for a month to convince myself of an easier way to understand it. The last couple of paragraphs of your explanations are the key but I was too bogged down at first in the inroductory part. Thanks for this! Expected value and conditional probability are not my strengths but problems like this help me to sharpen my skills.
Excellent teaser and double tough. Thank you.
Here, this number is correct if you are seeking the mean. There are other important measures of central tendency that are also pretty interesting, though. The median number of boxes (i.e., half the time it will take you this number or less) is 7, and the mode (i.e., the most common outcome) is 6. The probability you can do it in exactly 4 is 4!/4^4 = 3/32, just under 10%. Thanks for the question!
Nice puzzle.
I knew there was a simple way to calculate it, but had forgotten what the expected number of attempts was given the probability of success p. I "rediscovered" the 1/p formula by working the problem out exactly as you describe by solving for the sum of the series n = 1 to infinity of
np * (1-p)^(n-1) = 1
Really though, the formula 1/p is intuitive as it produces a result that moves propotionally from 1 to infinity as p moves from 1 to 0. Shortly into solving the above equation it occurred to me that this was the formula, but I finished solving the problem just to (re)prove it to myself.
I knew there was a simple way to calculate it, but had forgotten what the expected number of attempts was given the probability of success p. I "rediscovered" the 1/p formula by working the problem out exactly as you describe by solving for the sum of the series n = 1 to infinity of
np * (1-p)^(n-1) = 1
Really though, the formula 1/p is intuitive as it produces a result that moves propotionally from 1 to infinity as p moves from 1 to 0. Shortly into solving the above equation it occurred to me that this was the formula, but I finished solving the problem just to (re)prove it to myself.
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