September 29th, 2014, 05:49 PM  #1 
Member Joined: Aug 2014 From: Minnesota Posts: 38 Thanks: 2  Dice, is this right?
Please check my math or at least my logic.... Dice game using only one kind of six sided die. 1 side = Full Moon 2 sides = Half Moon remaining sides (3) are blank I'd like to solve for the probability of getting at least one full moon, with the understanding that two half moons (across two dice) are just as good as one full moon. Probability A (at least one full moon) Probability B (any two (or more) half moons) I have determined the probability of A for rolling anything from 1 to 6 dice. 1 Die = 0.17 2 Dice = 0.31 3 Dice = 0.42 4 Dice = 0.52 5 Dice = 0.60 6 Dice = 0.67 I have done the same for B, again for 1 to 6 dice. 1 Die = 0.00 2 Dice = 0.11 3 Dice = 0.26 4 Dice = 0.41 5 Dice = 0.54 6 Dice = 0.65 Now, here's where I'm a little shaky.... For one die, it's impossible to get two half moons P(B) = impossible simultaneous P(A) and P(B) = impossible Therefore: P(A) = chance of at least one moon For TWO dice, it's possible to get EITHER 1 full moon or 2 half moons, therefore the results are mutually exclusive. simultaneous P(A) and P(B) = impossible Therefore: P(A)+P(B) = chance of at least one moon For THREE (or more) dice, results are not mutually exclusive simultaneous P(A) and P(B) = possible Therefore: ( P(A)+P(B) )  ( P(A) X P(B) ) = chance of at least one moon. Do I have this right? For the record, here are the probabilities I came up with of at least one moon occurring. 1 Die = 0.17 2 Dice = 0.42 3 Dice = 0.57 4 Dice = 0.71 5 Dice = 0.81 6 Dice = 0.88 Thanks for the help. 
September 30th, 2014, 05:22 AM  #2 
Math Team Joined: Apr 2010 Posts: 2,778 Thanks: 361 
I'd split in two cases, for n dice: 1. no full moon and no half moon 2. no full moon and one half moon All other cases give at least one full moon. This gives me the following chances for n dice accordingly. 1 0.166666666666667 2 0.416666666666667 3 0.625 4 0.770833333333333 5 0.864583333333333 6 0.921875 7 0.955729166666667 8 0.975260416666667 9 0.986328125 10 0.992513020833333 
September 30th, 2014, 06:26 AM  #3 
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms 
Right. This gives the formula: $$ P_n=1\frac{2n+3}{3\cdot2^n} $$ for the probability of getting at least one moon in ether of those ways. 
September 30th, 2014, 11:19 AM  #4 
Member Joined: Aug 2014 From: Minnesota Posts: 38 Thanks: 2 
Okay, I see that you guys have the correct answers. I even took the example of 3 dice and painstakingly worked out 216 possible outcomes and then came up with the same .625 outcome. Mathematically, as for how either of you got to that answer, I'm barely on the radar. Likewise, with regard to my own approach, I still do not understand where I went wrong. But that's okay! Luckily, I'm not a proud man. Being my high school and even college days are well behind me, I'm happy to accept the spoon feeding GreatHouse gave me in the formula he provided. Any chance you could give me two more? 1) Same exactly as original problem with 1 face = moon, 2 faces = half moon, but on an 8 sided die instead of a six sided. 2) Again, same exactly as original problem with 1 face = moon, 2 faces = half moon, but on a 10 sided die instead of a six sided. Much appreciated. 
September 30th, 2014, 12:25 PM  #5  
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms  Sidenote: that part of my username is all one word (no capital "H"), my last name actually. Quote:
$$ \left(\frac{s3}{s}\right)^n + n\cdot\frac2s\cdot\left(\frac{s3}{s}\right)^{n1} $$ which you can simplify to $$ \frac{2n+s3}{s}\left(\frac{s3}{s}\right)^{n1} $$ if you like.  
September 30th, 2014, 04:51 PM  #6 
Member Joined: Aug 2014 From: Minnesota Posts: 38 Thanks: 2 
CRGreathouse, Thanks again for your reply. First, sorry about the last name. That's one correction I find easy to make. Before I dig any further into the new equations you gave me, if n = number of times dice are being rolled and s = number of sides of each individual die, then I'm having trouble getting the same results as the formula you offered earlier. For example, using your new formula for a 6 sided die rolled 3 times I get .375 but I believe I should be getting .625 as I did with your first equation tailor made for six sided dice. I get the feeling that instead of being pleased to excuse my dear Aunt Sally, I'm actually violating her.... Even so, I've worked it 34 times and I am still getting .375. 
October 1st, 2014, 06:05 AM  #7  
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms  Quote:
Quote:
Let's see what turns up: Quote:
$$ \frac{2n+63}{6}\left(\frac{63}{6}\right)^{n1}=\frac{2n+3}{6}\left(\frac{1}{2}\right)^{n1}=\frac{2n+3}{3}\cdot\frac{1}{2^n} = \frac{2n+3}{3\cdot2^n}. $$ Looks good to me.  
October 1st, 2014, 06:30 AM  #8 
Math Team Joined: Apr 2010 Posts: 2,778 Thanks: 361 
The given formula computes the chance of getting no full moon. 1  the formula gives the desired result, $\displaystyle 1\frac{2n+s−3}{s}\left(\frac{s−3}s\right)^{n− 1}$.

October 1st, 2014, 06:30 AM  #9 
Member Joined: Aug 2014 From: Minnesota Posts: 38 Thanks: 2 
CR Greathouse, Thanks 4 reply. Luckily I'm anonymous, so there's little shame in displaying my algebraic ignorance. Looking at your last equation let's say there are 4 steps, each separated by an "=" sign. I don't understand how you got from step 2 to step 3 because I'm confused as to how to treat the exponent, particularly with regards to how you pealed the 1 from it. Can you dumb it down a little more? edit: Hoempa set me straight on how to use the formula, so the above question is more academic than practical for me now. Last edited by johnmushroom; October 1st, 2014 at 06:39 AM. 
October 1st, 2014, 06:32 AM  #10  
Member Joined: Aug 2014 From: Minnesota Posts: 38 Thanks: 2  Quote:
LIGHTBULB! Thanks Hoempa, and the VBA code you gave me from the other thread is still proving very useful!  

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