Date: 11/04/97 at 17:29:57 From: dominic Subject: Probability Distributions Dear Dr. Math, Please help me with my problem ... I worry about the following type of random distribution: Supposing a million people, without reference to each other's choices, each choose a 'random' number between 1 and 1,000,000. a percent of numbers will be chosen 0 times b percent of numbers will be chosen 1 time c percent of numbers will be chosen 2 times etc. How can I find a, b, c, d etc., and what is the formula governing these values if the ratio between guessers and numbers is varied? Many thanks if you can help. Best wishes from Dominic Parkinson, UK
Date: 11/04/97 at 19:55:30 From: Doctor Anthony Subject: Re: Probability Distributions If you consider any particular number, say the number 1000, there is no reason for this number to be different from any other number, so what happens to it will be the same as what happens to any other number. To make this question a little more manageable, and avoid unneccessarily large nubers or very small probabilities, suppose we consider 20 people where each has to choose a number at random between 1 and 20 inclusive. Take any number, say 5, and we require the probabilities that this number will be chosen 0, 1, 2, 3, ....., 20 times. This is a binomial probability, with 20 trials and with p = 1/20 at each trial, and q = 19/20 at each trial. P(0) = (19/20)^20 = 0.35848 P(1) = C(20,1) (1/20)(19/20)^19 = 0.37735 P(2) = C(20,2) (1/20)^2 (19/20)^18 = 0.188677 P(3) = C(20,3) (1/20)^3 (19/20)^17 = 0.059582 P(4) = C(20,4) (1/20)^4 (19/20)^16 = 0.0133276 and so on. The probabilities are getting quite small by now. Since any of the 20 numbers will have the same pattern as this one, we can see that the percentage of numbers with no selections = 35.85 percent 1 selection = 37.73 percent 2 selections = 18.87 percent 3 selections = 5.96 percent 4 selections = 1.33 percent and this would continue up to 20 selections, which has a probability of (1/20)^20 = 9.54 x 10^-27 of occurring. If you tried this calculation with 1 million numbers you would be dealing with probabilities that were impossible to work out using a normal calculator, and apart from the percentage getting no selections, all the probabilities would be of the order 1/10000 or less. This problem is a natural for the Poisson distribution. The Poisson, as you know, is a limiting form of the binomial in which n, the number of trials, tends to infinity while p, the probability of success at any trial, tends to 0, but such that the mean of the distribution, np, is a finite number. In this problem with n = 1,000,000 and p = 1/1,000,000 we have np = 1, so mean = 1, and the required probabilities are given by: mean^r e(-mean) e^(-1) P(r) = ---------------- = ------- r! r! P(0) = e^(-1) = 0.36788 P(1) = 1/1 e^(-1) = 0.36788 P(2) = 1/2! e^(-1) = 0.18394 P(3) = 1/3! e^(-1) = 0.06131 P(4) = 1/4! e^(-1) = 0.01533 P(5) = 1/5! e^(-1) = 0.00306 and so on. These are the proportions for any large number of people and the possible numbers from which to choose. You will notice that these probabilities are not all that different from the case n=20 considered before. -Doctor Anthony, The Math Forum Check out our web site! http://mathforum.org/dr.math/
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