[Information - Statistics] Analyzing C9++ and family - a missing equation

**Brilliant Dynamite Neon** · 04-17-2014, 03:17 AM

The most elegant way I can think to phrase this problem is:

Within a set number X of random drawings, what is the chance of producing results of Y% probability Z times within the set?

This is critical to analyze the possibilities of "generative" setups in Mafia, like C9++. You don't need to be too knowledgeable about the game to do this (although The Werewolf Game on this forum is an offshoot of it.) I'm just culling an example from it... so that it's clear what I'm talking about.

http://wiki.mafiascum.net/index.php?titl...

7 random drawings are made. I want to find out the probability that four out of those seven are Cs (a result which has a 15% chance of being drawn on any given drawing)

Therefore, in this particular instance:
X is 7
Y is 15
Z is 4

What is the general equation needed to solve this problem (and what is the answer in this particular instance, if it's not too much trouble)?

**Guest15937** · 04-17-2014, 07:11 AM

I believe the formula you are looking for is: Y^Z * (1-Y)^(X-Z) * C(X,Z)

So in this case, the probability is (0.15^4) * (0.85^3) * 35 = 1.088%.

**Reincarnate** · 04-17-2014, 07:40 AM

^this

**Brilliant Dynamite Neon** · 04-17-2014, 06:16 PM

Is C a form of nCr?

EDIT: yes it is, just found out. Thanks a lot!

04-17-2014, 03:17 AM	#1
Brilliant Dynamite Neon nobody fiffers anymore. Join Date: Dec 2006 Location: 6-arrow land Age: 31 Posts: 2,440	[Information - Statistics] Analyzing C9++ and family - a missing equation The most elegant way I can think to phrase this problem is: Within a set number X of random drawings, what is the chance of producing results of Y% probability Z times within the set? This is critical to analyze the possibilities of "generative" setups in Mafia, like C9++. You don't need to be too knowledgeable about the game to do this (although The Werewolf Game on this forum is an offshoot of it.) I'm just culling an example from it... so that it's clear what I'm talking about. http://wiki.mafiascum.net/index.php?titl... 7 random drawings are made. I want to find out the probability that four out of those seven are Cs (a result which has a 15% chance of being drawn on any given drawing) Therefore, in this particular instance: X is 7 Y is 15 Z is 4 What is the general equation needed to solve this problem (and what is the answer in this particular instance, if it's not too much trouble)? __________________ ~ OFFICIALLY RETIRED FROM FFR THE GAME AND FFR TWG ~ Thanks for the memories, folks. u da bes

04-17-2014, 07:11 AM	#2
Guest15937 One-handed elite Join Date: May 2008 Posts: 1,464	Re: [Information - Statistics] Analyzing C9++ and family - a missing equation I believe the formula you are looking for is: Y^Z * (1-Y)^(X-Z) * C(X,Z) So in this case, the probability is (0.15^4) * (0.85^3) * 35 = 1.088%. __________________ The renegade has betrayed me.

04-17-2014, 07:40 AM	#3
Reincarnate x'); DROP TABLE FFR;-- Join Date: Nov 2010 Posts: 6,332	Re: [Information - Statistics] Analyzing C9++ and family - a missing equation ^this

04-17-2014, 06:16 PM	#4
Brilliant Dynamite Neon nobody fiffers anymore. Join Date: Dec 2006 Location: 6-arrow land Age: 31 Posts: 2,440	Re: [Information - Statistics] Analyzing C9++ and family - a missing equation Is C a form of nCr? EDIT: yes it is, just found out. Thanks a lot! __________________ ~ OFFICIALLY RETIRED FROM FFR THE GAME AND FFR TWG ~ Thanks for the memories, folks. u da bes Last edited by Brilliant Dynamite Neon; 04-17-2014 at 06:18 PM..

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