From: Josh Cheek Date: 2009-09-26T11:08:50+09:00 Subject: Re: Roulette & rand --000e0cd119fe8bb6a20474718bb5 Content-Type: text/plain; charset=ISO-8859-1 On Fri, Sep 25, 2009 at 5:16 PM, Semih Ozkoseoglu wrote: > > And Josh, > > > Hi, it is giving me the output that I would expect, here is a image > > showing > > how I calculated it, and I ran the program 3 times with results very > > similar > > to what I calculated they should be. > > http://img132.imageshack.us/img132/8939/roulette.png > > I get the exact same results with you but what I dont understand is why > you think that the results are normal. I dont understand why it looses > twice as much when you bet randomly (if its not the problem Paul > mentioned above). It would be great if you can tell me how you > calculated what the results should be. > > Thanks. > Semih > > -- > Posted via http://www.ruby-forum.com/. > > Here is the formatted explanation: http://img38.imageshack.us/img38/8939/roulette.png And for completeness sake, here is a the unformatted explanation: Okay, so you are adding 1 for every win (w), and subtracting one for every loss (l), which gives n*w-n*l , which can be simplified to n(w-l) So for a pair of win and loss, at some given number of attempts, n, we get the formula. [math]f(w,l) = n(w-l)[/math] You want to know why it's value is approximately twice as large, so let's compare the two functions by dividing them. We will calculate f for bet_rand and bet_red, and divide them by eachother. First we need to find the likelihood of winning and losing, our w and l, for bet_rand and bet_red. bet_rand compares color_01 and color_02, each of which are randomly selected from values 0 through 36 (thirty seven possible values). So there is a 1 in 37 chance of color_01 being zero, and for that case, there is a 1 in 37 chance of color_02 being zero. And there is an 18 in 37 chance of color_01 being red, and a 18 in 37 chance of color_02 being red. And there is an 18 in 37 chance of color_01 being black, and a 18 in 37 chance of color_02 being black. So this gives us [math]\frac{18}{37} * \frac{18}{37} + \frac{18}{37} * \frac{18}{37} + \frac{1}{37} * \frac{1}{37}[/math] Which comes out to [math]\frac{649}{1369}[/math] And the likelihood of losing is l = 1 - w = [math]\frac{720}{1369}[/math] Now, for bet_red, color_01 will always be red. So there is a 0 in 37 chance of color_01 being zero, and for that case, there is a 1 in 37 chance of color_02 being zero. And there is an 37 in 37 chance of color_01 being red (because it is manually set to red), and a 18 in 37 chance of color_02 being red. And there is an 0 in 37 chance of color_01 being black, and a 18 in 37 chance of color_02 being black. So this gives us [math]\frac{37}{37} * \frac{18}{37} + \frac{0}{37} * \frac{18}{37} + \frac{0}{37} * \frac{1}{37}[/math] Which comes out to [math]\frac{18}{37}[/math] And the likelihood of losing is l = 1 - w = [math]\frac{19}{37}[/math] So now we have our probabilities to feed the function. Now, we need to find a meaningful way to compare them. What we will do is compare their values for some given n, we noticed that bet_rand seemed to grow about twice as fast as bet_red. So we will divide bet_rand's limit as n approaches infinity by bet_red's, and see if it comes out to about 2. This gives us Our formula [math]d(w,l) = \frac{f(w_{bet\_rand},l_{bet\_rand})}{f(w_{bet\_rand},l_{bet\_rand})}[/math] Fill in the values. [math]= \frac{f(\frac{649}{1369},\frac{720}{1369})}{f(\frac{18}{37},\frac{19}{37})}[/math] Substitute the value of f(w,l) [math]= \frac{n(\frac{649}{1369}-\frac{720}{1369})}{n(\frac{18}{37}-\frac{19}{37})}[/math] At this point, we can see that the value of n is irrelevant, as it cancels itself out. [math]= \frac{(\frac{649}{1369}-\frac{720}{1369})}{(\frac{18}{37}-\frac{19}{37})}[/math] And simplifying, we get. [math]= \frac{71}{37}[/math] [math]\approx 1.91891891891892[/math] So we have shown that we can expect bet_rand to grow (in a negative direction) about 1.92 times quicker than bet_red. --000e0cd119fe8bb6a20474718bb5--