Probability Distributions

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/