Please Help with Probabilities Formula for the following problem.... (1 Viewer)

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!

6^3 is correct

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

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...

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.)

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....

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

--
H

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

sounds like you need help with your home work ??

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....

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

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.

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.

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.

As I understood, they are unique. Just like a lock on your suitcase has three numbers you roll and the unique combinations are 10^3. Dice #1 and dice #2 for example are not interchangeable. If they are then 6^3 is not the answer.

Yes, if the three dice are distinguishable from each other, the answer is 6^3 = 216.

If the dice are NOT distinguishable, i.e. if the outcomes of 112 and 121 and 211 are all considered to be the same, then you need to account for those repeat possibilities in the right way, which I think can be done as follows:

EDIT: Screwed this up the first time. Corrections below.

In the total of 216, multiple-counting of the same possibility can happen in two different ways.

One is if two different numbers show up on the three dice, e.g. the 112/121/211 possibility mentioned before. We also have to take into account the 122/212/221 option here, since we are classifying this as "what if the numbers 1 and 2 are the only ones that appear." As we can see from this example, this will result in one possibility counting six times (in the 216 total), when we only want it to count twice.

The other type of multiple counting is if three different numbers show up on the three dice, e.g. 123/132/213/231/312/321. This will result in a sextuple counting, when we only want it to count once.

So the multiple counted possibilities that we want to throw away are:

"6 choose 2" (binomial coefficient) ways of getting two different numbers, each of which contributes 4 excess possibilities (e.g. we'll keep 112 and 122, but we want to discard 121/211 and 212/221).

"6 choose 3" ways of getting three different numbers, each of which contributes 5 excess possibilities (e.g. we'll keep 123, but discard the other 5 permutations)

The total we want then turns out to be (see the link above for how to calculate the binomial coefficient):

6^3 - (6 choose 2)*4 - (6 choose 3)*5 = 216 - 15*4 - 20*5 = 56.

Ok, I'm sorry about the confusion in the beginning. 112 and 121 shoudnt be counted twice. So 216 isnt correct. I'll take some time and let your formula sink in Haggai. Thanks a lot for the breakdown!

In that case Al's solution is correct, which is basically the same as 6*5*4 = 120

For the six dice you get 6! = 720