Problems in the Probability of Matching Envelopes
Problems with matching envelopes are frequently used when teaching basic probability theory. For educators, these problems offer a simple but effective illustration of some of the more important rules of probability. The problem can be simply stated and easily understood, but is not necessarily easy to solve unless students are familiar with concepts such as the probability of multiple events.
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Basic Envelope Problems
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The basic envelope problem is usually some variation of the following scenario: an absent-minded person writes six letters and addresses six matching envelopes but forgets to put the right letter in the corresponding envelope. Instead, the person simply randomly picks letters and puts them in envelopes without trying to match them. The student is then asked to answer some basic questions concerning the probability of the placement of the letters.
Solving for All Envelopes Matching
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One question frequently asked is: what is the probability that all the envelopes receive the correct matching letter? If the picking of each letter for each envelope is considered a statistical "event," this can be answered by following the rule that the overall probability of a sequence of events is simply the product of their individual probabilities. The probability of the first envelope receiving the correct letter is 1/6, since there are six letters to choose from, but only one matching one. If the first envelope does get the correct letter, the odds of the next envelope also getting the correct letter are 1/5, then 1/4 and so on. The combined probability is then 1/6 x 1/5 x 1/4 x 1/3 x 1/2 x 1/1 = 0.00139.
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Solving for at Least One Match
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Another common question is: what is the probability that at least one of the envelopes will have the correct letter in it? This is more complex and requires the concept of the union of probabilities. The union of two events A and B is the chance of either A or B occurring and its probability: P(A or B) = P(A) + P(B) - P(A and B). By using a bit of math and expanding this concept to six envelopes, you can show that the P(at least 1 correct) = 1 - (1/2!) + (1/3!) - (1/4!) + (1/5!) - (1/6!), where n! = (n)(n-1)(n-2) ... (2)(1). This calculation gives a probability of 0.632.
Variations on the Envelope Problem
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There are a number of variations on this problem which offer different scenarios, but all of which can be solved in similar fashion. One variant posits a group of people (the particular number of people varies) and students are asked to calculate the probability that no two people have the same birthday, or that all have the same birthday. Another variant is a dance at which there are a number of married couples who pair off randomly to dance, and the probability that each woman will dance with her own husband is calculated.
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