Quote: Originally posted by Hyperdimension on December 26, 2004
Hi,
The right person to answer your question is Mr. Ion Saliu,
I find an interesting article about binomial distribution, with the next example:
Poisson Distribution
In extreme cases, very small p so that the standard deviation is not much less than the mean, the Gaussian Distribution is not appropriate, but a different approximation is: the Poisson Distribution. Going back to the Binomial Distribution (which is still exact), we only need to worry about values of n much smaller than N.
In the Indiana Lottery, people choose 6 numbers from 1-49. There are
45!/(39! 6!) = 13983816
combinations; more than one person can buy a ticket with the same number. Suppose 28 million tickets are sold in a given week, what is the probability for zero winners? one winner? etc...
One number is selected, we have to assume that all numbers are purchased with equal likelihood, so that on any given purchase the probability that it will be the winner is about 1 in 14 million (that is p=1/14000000). We have:
a= N p = 28000000 (1/14000000) = 2
Pa(n) ≅ e-a an/n!
P2(0) ≅ e-2 20/0! = e-2 = 0.1353
P2(1) ≅ e-2 21/1! = e-2 = 0.2707
P2(2) ≅ e-2 22/2! = e-2 = 0.2707
P2(3) ≅ e-2 23/3! = e-2 = 0.1804
P2(4) ≅ 0.0902
P2(5) ≅ 0.0361
P2(6) ≅ 0.0120
P2(7) ≅ 0.0034
P2(8) ≅ 0.0009
P2(9) ≅ 0.0002
P2(10) ≅ 0.00003
We see that it is most likely that there are 1 or 2 winners, but 0, 3, and 4 would not be surprising. The probability for more than 10 winners is about 0.000008 (be very suspicious if this occurs!).
What is the probability to sell 50 million tickets without a winner? The expected number of winners is
a = N p = (50000000)(1/14000000) = 3.57
The chance to get zero is:
P3.57(0) ≅ e-3.57 3.570/0! = e-3.57 = 0.028
That is, about 1/35 (not completely unlikely).
What is the probability to sell 100 million tickets without a winner? The expected number of winners is
a = N p = (108)(1/14000000) = 7.1
The chance to get zero is:
P7.1(0) ≅ e-7.1 7.10/0! = e-7.1 = 0.00079
We don't expect to see this very often (1/1265).
Sounds interesting