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  #1  
Old March 29th, 2007, 03:38 PM
ob1wan ob1wan is offline
 
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Dice Stats

I have a question was trying to figure out the odds for different dice rolls. If I roll 4 dice and I want to know the odds of getting 1 or more skulls is this the correct formula, (0.5) + (0.5*0.5) +(0.5*0.5*0.5)+(0.5*0.5*0.5*0.5)93.75%=?
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  #2  
Old March 29th, 2007, 04:08 PM
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The easiest way to figure this out would probably to figure out the probability of not rolling any dice and subtracting it from 1. The way you did it seems to work but it would be quicker to just take 0.5 to the 4th power and subtract it from 1 like this:

1 - (0.5)^4 = Probability of rolling no skulls
1 - 0.0625 = 0.9375 = 93.75%

This saves you the trouble of figuring out probability of finding out each individual roll outcome. I'm not entirely sure if the way you did it is correctly done since there are dice multiple combinations that can lead to a the different dice outcomes. I'll look it up in my statistics book and if anyone knows off the top of your head they can help you unless I find it first.
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Old March 29th, 2007, 04:12 PM
TheRealQ TheRealQ is offline
 
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One of the math wizards will probably be able to answer this better but I would do it by figuring out the opposite and subtracting it from 100%.

Chance of getting all blanks: .5x.5x.5x.5=.0625 thus the chance of getting at least one skull would be 93.75% or 1-.0625

Edit: Kalislord beat me to it, posting while I was typing.

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  #4  
Old March 29th, 2007, 04:30 PM
ob1wan ob1wan is offline
 
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the one problem is what if i wanted to find the odds of getting 2 or more skulls I dont think that formula would work. but I'm not sure if (0.5*0.5) +(0.5*0.5*0.5)+(0.5*0.5*0.5*0.5)=43.75% is right either
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  #5  
Old March 29th, 2007, 05:22 PM
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KalisLord KalisLord is offline
 
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Alright after reviewing some statistics here is the deal. You coincidentally came up with the right answer but if you were figuring out the probability of say 2 or better you would have done it incorrect using your methodology I believe. Someone correct me if I am wrong on this, but the way you did it doesn't account for the normal distribution of the curve.

The issue is that you didn't account for the probability of each happening. For instance the probability of rolling 0 is 1/16, 1 is 4/16, 2 is 6/16, 3 is 4/16, and 4 is 1/16 on a given roll of 4 dice. Since there is an equal probability of rolling a skull or whatever. If we were figuring it out shields for instance we would have to use an equation like:

[(1/3)^x * (2/3)^(n-x)] * Number of Combinations for that Value = Probability of a Particular Die Roll

n is the total number of dice being rolled and x is the number of shields we want to determine. The Number of Combinations is easiest to figure out, in my opinion, by n! / [x! * (n -x)!]. If you don't have a scientific calculator or don't know how to do factorials you can use the following pyramid:

Dice = 1 ----------------------- 1 1 --------------Total Combinations = 2
Dice = 2 ---------------------- 1 2 1 ------------Total Combinations = 4
Dice = 3 -------------------- 1 3 3 1 -----------Total Combinations = 8
Dice = 4 -------------------1 4 6 4 1 ----------Total Combinations = 16
Dice = 5 ---------------- 1 5 10 10 5 1 ------Total Combinations = 32
Dice = 6 -------------- 1 6 15 20 15 6 1 ----Total Combinations = 64
Dice = 7 ------------1 7 21 35 35 21 7 1 ---Total Combinations = 128
ETC....

In order to use this you just need to go from left to right setting the the first value to 0, the second value to 1, etc. For example if we are rolling 5 dice then the number of combinations for 0 is 1, 1 is 5, 2 is 10, 3 is 10, 4 is 5, and 5 is 1.

There are other ways to figure it out I believe, but I haven't had statistics for a couple years and I almost never need to use it. After you figure out the probability of a certain die roll you can just add up the ones you need to determine what you need. For instance if you want to know how many odd numbers of skulls you would roll then just add up the probabilities of odd numbers of skulls. I hope you get the idea and I need to run to dinner so I will leave it at that.

Alright I hope I explained that in a fashion that wasn't too confusing. Hopefully someone can help you if you have any questions or I just completely screwed up.

Quote:
Edit: Kalislord beat me to it, posting while I was typing.
I seem to be on top of my game.
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  #6  
Old March 29th, 2007, 06:53 PM
ob1wan ob1wan is offline
 
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[(1/3)^x * (2/3)^(n-x)] * Number of Combinations for that Value = Probability of a Particular Die Roll

okay after messing around for a long time on the number of combinations for that value you if your doing 2 or more you have to add all the possiblitly for two and after
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  #7  
Old March 29th, 2007, 08:24 PM
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Yes you are correct you just add up the probabilities you want. I just want to make sure you realize that equation is useful for Defence Dice only. I realized I didn't actually give you the Attack Dice formula in the same manner. It's basically the same as the defence dice version, but simpler. Since there is equal probability of both outcomes we can add the exponents of both parts which leaves us the following equation:

(1/2)^n * Number of Combinations for that Value = Probability of a Particular Die Roll

So in summary here are the formula's you need:

Attack Dice: (1/2)^n * Number of Combinations for that Value = Probability of a Particular Die Roll

Defence Dice: [(1/3)^x * (2/3)^(n-x)] * Number of Combinations for that Value = Probability of a Particular Die Roll

Generic Formula: [(Probability of Outcome A)^x * (1 - Probability of Outcome A)^(n-x) * Number of Combinations = Probability of Value

Number of Combinations = n! / [x! * (n -x)!]

n = total number of dice being rolled
x = the number of Outcome A we want

Let me know if this isn't clear or if you have any other questions.
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  #8  
Old March 30th, 2007, 01:00 AM
ob1wan ob1wan is offline
 
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having a problem with rolling 6 dice and getting 2 or more shields getting [(1/3)^2 * (2/3)^(6-2)]*57 =125%. Not sure where I messed up.
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  #9  
Old March 30th, 2007, 07:03 AM
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Your number of combinations is way too high. It should only be 15.

6! / (2! * (6-2)!)
6! / (2! * 4!)
720 / 2 * 24 = 15 = Number of Combinations for 2 shields on 6 dice
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  #10  
Old March 30th, 2007, 09:41 AM
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  #11  
Old March 30th, 2007, 12:11 PM
ob1wan ob1wan is offline
 
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Dice = 6 -------------- 1 6 15 20 15 6 1 ----Total Combinations = 64
Shields---------------- 0 1 2_ 3_ 4, 5, 6 ---- If i wanted to see when theres two or more I would add 15 + 20 + 15 + 6 + 1 =57. If I only used 15 then it would be the odds of rolling only 2
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  #12  
Old March 30th, 2007, 03:59 PM
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Unfortunately you cannot add up the combinations after the one you pick. If we were doing attack that would be fine, but because the probability for each combination is different when we are doing the attack we cannot. For instance the probability of rolling 2 shields out of 6 dice is about 0.022 before we multiply it by the number of combinations for that number. On the other hand 3 shields out of 6 dice is 0.011. Thus since the they are not equal we can't just add up the number of combinations.
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