Originally posted by CalTrader
[BMathematical Expectation is the value of the probability density (or distribution) at x - where x is a continuous or discrete random variable - and is computed by either a discrete sum or integral of xF(x) where F9x) is the probability density (distribution) function.
Coin flips of a balanced coin can be modeled by assuming a binomial distribution for the probability density function. That is your random variable has a binomial distribution for its probability distribution. [/B]
You are right: there are two random variables resp their distributions involved.
The original questions refers to the probability of events in flipping coins, which may be interpreted as a representation of winners and losers. That is the first random variable/distribution of interest: the distribution of winners(losers).
In trading, this distribution may be misleading: you might have a system with 80% winners, but lose money. So the second random variable/distribution is introduced: the distribution of profits(losses). This should be of greater interest than the former one when doing simulations, running tests etc. The value of expectancy of this distribution will tell you more than the value of expectancy of the winner/loser distribution.
@jperl
Sorry for introducing new terms and concepts.
http://www.statsoft.com/textbook/stathome.html offers an electronic textbook with an explanation of concepts.
The coin flip is described by a binomial distribution (you have only two results). The value of expectancy is computed as (n*p). n = number of throws, p probability of outcome. p does not have to be 0.5. So with 100 flips you get a value of expectancy (big surprise) 100*0.5 = 50. The variance you asked for is computed as (n*p*(1-p)). That is (100*0.5*(1-0.5)) = 25. For practical purposes you have to take the square root of the variance, in your case this is 5. Add/subtract this value to the value of expectancy to arrive at an interval your result will fall into with a certain probability.
This means if you do a lot of series of 100 coin flips, in at least 75% of these series you will have between 40 and 60 heads, in 89% of these series between 35 and 65 heads, and in 94% between 30 and 70 heads (this is just chebycheff's inequality applied). So getting 80 heads in 100 flips looks rather unlikely.
The probability of an event for the binomial distribution is described by B(x,n,p) = (n over x)*(p**x)*((1-p)**(n-x)). (n over x) is called the binomial coefficient computed as n!/((n-x)!*x!). n! is defined as n! = 1*2*.....(n-1)*n, 0!=1!=1.
If you want to know the probability of getting at least 80 heads in a row of 100 flips you have got to sum B(100,80,0.5), B(100,81,0.5), .... B(100,100,0.5). Doing this by hand will be rather tedious and error-prone, there are tables and programs to do this. The result is something like 0.000000000557954 (hope I did not mistype it..).
Regards
Bernd Kuerbs