Two Plus Two Older Archives  

Go Back   Two Plus Two Older Archives > PL/NL Texas Hold'em > Mid-, High-Stakes Pot- and No-Limit Hold'em
FAQ Community Calendar Today's Posts Search

 
 
Thread Tools Display Modes
Prev Previous Post   Next Post Next
  #11  
Old 04-24-2005, 02:57 AM
iversonian iversonian is offline
Junior Member
 
Join Date: Sep 2003
Posts: 0
Default Re: What\'s My Play?

This problem got my interest. I spent a few minutes working on it and got an answer of the probability of you winning this prop bet at 14%.

Just to clarify the problem, you said that it was 101 games, first to 51. Except with ties, nobody might get to 51 before you hit 101 games. I'm guessing you meant first to 51 regardless. For my program, I assumed every game had a win/lose outcome, with a win% as given by the EV % on twodimes. I'm sure it doesn't change the outcome too much.

The probability of, in 101 games, of winning 0 games, losing 101 (with the nines, p = 0.553 per game) is:

(1-p)^101

The probability of winning one and losing 100 is:
(n choose r) * p^r * (1-p)^(n-r)
where n = games = 101, r = games won = 1
---- (n choose r) = n! / ( r!*(n-r)! )

Probability of winning 51 to 101 games is:
SUM r=51 to 101: (n choose r) * p^r * (1-p)^(n-r) = 0.8579

Probability of AK winning 0 to 50 games should be:
SUM r=0 to 50: (n choose r) * p^r * (1-p)^(n-r) = 0.8579
where p = .447
which checks out.

Public Function PropBet()
Dim n, r As Integer
Dim prob, sum As Double

n = 101
prob = 0.553
sum = 0#

For r = 51 To 101
sum = sum + Factorial((n)) / Factorial((r)) / Factorial((n - r)) * prob ^ r * (1 - prob) ^ (n - r)
Next

MsgBox sum

End Function
Reply With Quote
 


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 12:31 PM.


Powered by vBulletin® Version 3.8.11
Copyright ©2000 - 2024, vBulletin Solutions Inc.