Probability, expectation, significance and related statistics

[Matt_T_hat]

Let us assume a tombola is being run. All the tickets ending 5 or 0 are winners all the others are users.

Assume that no tamperig has taken place and that a negligable number of wins have taken place. Now as we do not know the number of tickets in the barrel we will simplify and assume that they are effectivly infinite as removal will have little impact.

Now we know that there are only ten possible endings for the numbers and that the numbers on tickets are distributed evenly. Therefore in a well shaken tombloa there is a 2 in 10 or 1 in 5 change of winning.

Win = 0.2
Loose = 0.8

Now suppose that a person has 18 goes on the tombla what is the probability that they will loose all 18 draws?

I make the figure 0.018014398509481984 or 0.018 to mke it readable. (1.8%).

My first question is am I correct? I applied (4/5)^18 to get this number.

Now I remember back when I was getting a low grade (a basic pass) at statistics A-Level tat we could apply a test (Students T test if I recall the name correctly).

Could some one run me through working out the following:

H1: Drawing 100% non winners in 18 draws is significantly unlikely
H0: This is just a thing tha happened call it luck if you wish but there is nothing unusual about this.

Useing the 12%, 10%, 5%, 2% and 1% confidence levels for these.

Further more I recall a function called expectation. Given a known probability and a known number of events following the known probabilities one could create an estimation of the expected outcome which equates to where the mean should fall with an infinite number of itterations.

I can not remember how to calculate this value.

Can anyone else help me satisfy my mathmatical inquisitivness on an event I witnessed today?

[Code Doc]

What you have in your tombola example is is a binomial distribution where there are 0 successes in 18 trials and a probability of a success is 0.2.

Excel reveals 0.018014399. Note that your formula works only because there are no successes. However, you really should examine the binomial distribution, which is much more powerful and allows you to determine the probability of one success, two successes, ... 18 successes.

[TBeck]

I believe the expectation you are refering to is another term for the Expected value. http://en.wikipedia.org/wiki/Expected_value

edit: in adition to the sumation way of calculating, each disribution has a simplified formula for calculating it, for example a binomial is:
E(x) = np
where n is the number of trials and p is the probability of success