Similar Phone NumbersDate: 10/21/96 at 11:41:49 From: Don Opper Subject: Probability Dear Dr. Math, My brother and I have phone numbers with exactly the same numbers but in a different order. He thinks this is some sort of miracle. I think it's fairly common. What is the probability that two seven digit numbers will contain the same numbers but in a different order? How is this figured out? Thanks, D. K. Opper Date: 10/22/96 at 12:10:12 From: Doctor Tom Subject: Re: Probability Hi Don, This problem is probably a lot more complicated than you think. For example, the 3-digit prefixes are localized, so if you and your brother live near each other, the odds that you share a prefix are better, and that would improve the odds of a match. Similarly, if you live apart but in the same area code, it's likely the prefixes will not match. Some prefixes were illegal until recently - those with a "0" or "1" as the second digit were reserved for area codes. No prefix begins with "1" or "0", etc. I'll bet that given current religious superstitions, a lot of places will not use the "666" area code. I can state a mathematical problem that can be solved, but given the statements above, it's unlikely that it will represent the true situation. We could solve this: Choose a 7 digit number uniformly from the range 0000000 through 9999999. Now choose a second. What is the probability that the rearranged digits of the first are the same as those of the second? Even this gets messy. For example, if the first number I choose is 1234567, then any reordering of those seven digits would make a match. And there are 7*6*5*4*3*2*1 = 5040 such combinations. But if the number is 1111111, there is no other match except for 1111111. This could happen, I suppose, if you and your brother lived in different area codes. So to solve the problem, you've got to find the number of numbers with no duplications, with one pair, with two pairs, with three pairs, with one triple, with two triples, with a triple and a pair, with a triple and two pairs, with a quad, with a quad and a pair, with a quad and a triple, with a 5-match, with a 5-match and a pair, and with a 6-match. Multiply each of these probabilities by the probabilities of getting a match in each case - another ugly calculation. What I'd do is write a little computer program to count them up since working this out by hand would be extremely laborious. To show you what I mean, let me just work it out for 2 digit numbers. There are ten pairs: 00, 11, ..., 99, so there are 90 non-pairs. The probability is 9/10 that you'll get a non-pair and 1/10 that you'll get a pair. If you get a pair, there's only 1 chance in a 100 of a match. If you get a non-pair, there are 2 chances in a 100 of a match (for example, if you get 23, it matches either 23 or 32). So the probability of a match is: (1/10)*(1/100) + (9/10)*(2/100) = 19/1000 (Nineteen times in a thousand.) -Doctor Tom, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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