Two Plus Two Older Archives  

Go Back   Two Plus Two Older Archives > General Gambling > Probability

Reply
 
Thread Tools Display Modes
  #1  
Old 12-21-2005, 12:07 PM
jon_1van jon_1van is offline
Member
 
Join Date: Oct 2002
Location: Silver Spring MD
Posts: 53
Default Rand Var, CI problem

So I'm trying to make a program similar to pokerstove, but for stud instead.


Assume my program generates a string of numbers like : -2, -2, -2, 10, -2, 10..... (ie lose 2 bets with some frequency and gain 10 bets with some frequency)



What I'd like to do is report the expected value of that stream within a certain confidence interval that the user specifies (say within .05 big bets)



Is the following approach valid ::

-Compute sample varience = 1/(n-1) * SUM( Xi - XBAR)
-Sample SD = sqrt(sample varience)

Lets assume the CI is 5 standard deviations big (2.5 SD on either side of the true E[x])

Now all I have to do is generate random number from the stream until the Sample SD = .02 BB.

Is this valid? Is it close enough?
Reply With Quote
  #2  
Old 12-21-2005, 06:09 PM
jon_1van jon_1van is offline
Member
 
Join Date: Oct 2002
Location: Silver Spring MD
Posts: 53
Default bump

anybody?
Reply With Quote
  #3  
Old 12-21-2005, 11:12 PM
alThor alThor is offline
Junior Member
 
Join Date: Mar 2004
Posts: 6
Default Re: bump

Close enough.

1. In principle, you shouldn't decide on an interval size first, and determine n later, on the fly. Instead, determine a (large) sample size first, then construct the interval. But this isn't a biggie. Technically it would induce a bias, but it'd be so small you'd never notice.

2. If you know the output only takes on two possible values, you should use a slightly different sample variance formula. (In effect, divide by n instead of n-1.) Again, this isn't a biggie. When n is large, it is roughly equal to n-1.

alThor
Reply With Quote
  #4  
Old 12-21-2005, 11:27 PM
AaronBrown AaronBrown is offline
Senior Member
 
Join Date: May 2005
Location: New York
Posts: 505
Default Re: Rand Var, CI problem

I don't understand the problem.

If your program generates the stream, then you can compute the expected value unless the program is extremely complex.

If that's impossible, you can use something like your approach, but your formulae are wrong.

If you assume that the individual values are independent and there are enough values that the Normal approximation is value (at least 30, unless you have some extreme values that show up very rarely), then the standard deviation of Xbar is approximately:

{Sum[(Xi - Xbar)^2]}^0.5/n
Reply With Quote
Reply

Thread Tools
Display Modes

Posting Rules
You may not post new threads
You may not post replies
You may not post attachments
You may not edit your posts

BB code is On
Smilies are On
[IMG] code is On
HTML code is Off

Forum Jump


All times are GMT -4. The time now is 01:20 PM.


Powered by vBulletin® Version 3.8.11
Copyright ©2000 - 2021, vBulletin Solutions Inc.