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# Please Help with Probabilities Formula for the following problem....

Started by
Mark Giles
, Apr 10 2007 06:08 AM

30 replies to this topic

### #1 of 31

Posted April 10 2007 - 06:08 AM

Simple....I have three 6-sided dice. I want to know the total number of unique results I can obtain by repeatedly throwing all three dice.

111

112

113

114

115

116

121

etc, etc....

I'm guessing it's 6^3, but can someone confirm and prove this?

Thanks for you help!

111

112

113

114

115

116

121

etc, etc....

I'm guessing it's 6^3, but can someone confirm and prove this?

Thanks for you help!

### #2 of 31

Posted April 10 2007 - 06:12 AM

6^3 is correct

### #3 of 31

Posted April 10 2007 - 06:17 AM

Do you have anything I can use to support this? I want to be able to explain it to my friend too.

### #4 of 31

Posted April 10 2007 - 06:21 AM

Ok, nevermind. I figured it out. Every dice increases the power.

One die would give you 6 possibilities or 6^1

Two would give you 6^2

and so on...

One die would give you 6 possibilities or 6^1

Two would give you 6^2

and so on...

### #5 of 31

Posted April 10 2007 - 06:29 AM

This one's easy. Any aging role-playing gamers could explain this one (and draw the bell curve) in their sleep.

Range 3 to 18

Center of bell curve 10.5

etc.

etc.

etc.

(God, I'm a geek.)

Range 3 to 18

Center of bell curve 10.5

etc.

etc.

etc.

(God, I'm a geek.)

Lurking at HTF Since 2001

### #6 of 31

Posted April 10 2007 - 07:06 AM

The formula is simply n1 x n2 x ..... where n is the number of choices on each die (this allows for difference sided dice being used in the same process). The fun sets in when you have different biases in individual dice. And this is a major reason why I was so glad to give up probability and stick to statistics.

### #7 of 31

Posted April 10 2007 - 08:52 AM

Originally Posted by andrew markworthy The formula is simply n1 x n2 x ..... where n is the number of choices on each die (this allows for difference sided dice being used in the same process). The fun sets in when you have different biases in individual dice. And this is a major reason why I was so glad to give up probability and stick to statistics. |

So let me ask this then. Lets say there were six 6-sided dice and you can toss any number of dice you want, as long as you tossed at least one.

Based on your formula, the amount of possible outcomes would this be....

n=6

(n) + (n1 x n2) + (n1 x n2 x n3) + (n1 x n2 x n3 x n4) + (n1 x n2 x n3 x n4 x n5) + (n1 x n2 x n3 x n4 x n5 x n6)

?? Is that correct?

And no, Im not cheating on a test....

### #8 of 31

Posted April 10 2007 - 09:40 AM

Yes but your notations are a bit sloppy there and mostly pointless with your n=6 condition.

--

H

--

H

### #9 of 31

Posted April 10 2007 - 09:51 AM

x[n] = 6^n + x[n-1]

### #10 of 31

Posted April 10 2007 - 11:52 AM

sounds like you need help with your home work ??

**The Sonodome - circa 2001****The Newest Sonotube - circa 2001****Gregg's DVDs updated...sometimes****Lion Audio Video Consultants** usually current

### #11 of 31

Posted April 10 2007 - 12:12 PM

No, actually a co-worker asked me the original question earlier in the day and I wanted to be able to explain it to her. The second one was asked because that's the way she first presented the question to me. My curiosity was taking over. I took Prob. & Stats years and years ago and remember it as well as I remember the one semester of Japanese as a freshman.

Not to mention, I'm bored today....

Not to mention, I'm bored today....

### #12 of 31

Posted April 10 2007 - 02:49 PM

It's actually not 6 to the 3rd power, since the original question said "unique".

### #13 of 31

Posted April 10 2007 - 03:28 PM

Originally Posted by Chris Lockwood It's actually not 6 to the 3rd power, since the original question said "unique". |

### #14 of 31

Posted April 11 2007 - 12:06 AM

Originally Posted by Sami Kallio It's 6^3 since 1,1,2 and 2,1,1 are unique. |

What's your formula for, Kami?

x(n) = 6^n + x(n-1) = (3 dice, 6 sides)

x(n) = 6^3 + 6^2 + 6^1 + 6^0 + 6^(-1) + ...

or maybe you just meant

x(n) = 6^3 + 6^2 + 6^1 + 6^0 = 259

???

As for unique: is 1,1,2 = 1,2,1 = 2,1,1 or are these unique? If I'm rolling identical dice, I can't distinguish these outcomes. They are not unique in typical gameplaying situations.

### #15 of 31

Posted April 11 2007 - 12:57 AM

The "unique" in the orginal question threw me off too. Now that we get the question (which I think is better phrased as non-unique), I think what you want is a Permutation: n!/(n-r)!. Or in this case 6!/(6-3)! = 120.

### #16 of 31

Posted April 11 2007 - 01:08 AM

Although Mark asked for the number of "unique results" in his original post, he also listed both 112 and 121 in his list, so I imagine that 6^3 is the answer he's looking for.

### #17 of 31

Posted April 11 2007 - 02:01 AM

Originally Posted by DaveF What's your formula for, Sami? |

Originally Posted by DaveF As for unique: is 1,1,2 = 1,2,1 = 2,1,1 or are these unique? If I'm rolling identical dice, I can't distinguish these outcomes. They are not unique in typical gameplaying situations. |