# Number of outcomes of two dice, three throws

• May 25th 2009, 02:56 PM
seavari
Number of outcomes of two dice, three throws
What is the maximum number of outcomes of a pair of dice thrown three times? For example:

1st Set: 2, 10, 4
2nd Set: 2, 2, 2
3rd Set: 10, 4, 2

Is there a universal formula I can use to calculate this problem with different variables? (For example, pair of dice thrown five times, three dice thrown six times, etc.)

Any help would be appreciated!
• May 25th 2009, 03:18 PM
Plato
Quote:

Originally Posted by seavari
What is the maximum number of outcomes of a pair of dice thrown three times? For example:
1st Set: 2, 10, 4
2nd Set: 2, 2, 2
3rd Set: 10, 4, 2
Is there a universal formula I can use to calculate this problem with different variables? (For example, pair of dice thrown five times, three dice thrown six times, etc.)

It is not at all clear as to what your question means.
If we toss a pair of dice, then add the showing numbers we have eleven outcomes: 2 to 12.
If we repeat this three times, we get a set of ordered triples.
There are then $11^3$ possible triples.

Is that the meaning of this question?
If not please try to explain further.
• May 25th 2009, 03:35 PM
seavari
Quote:

Originally Posted by Plato
It is not at all clear as to what your question means.
If we toss a pair of dice, then add the showing numbers we have eleven outcomes: 2 to 12.
If we repeat this three times, we get a set of ordered triples.
There are then $11^3$ possible triples.

Is that the meaning of this question?
If not please try to explain further.

(I'm sorry, it's been a really long time since I last tried to grapple a math problem. ^^; )

I think you answered my question though...so there could be 1,331 possible outcomes? (For example, "2, 3, 4" being one possible outcome.)

If I repeated four times instead of three, it would be 11^4?

If I used just one die and cast three times, would it be 6^3?
• May 25th 2009, 03:39 PM
Plato
Quote:

Originally Posted by seavari
I think you answered my question though...so there could be 1,331 possible outcomes? (For example, "2, 3, 4" being one possible outcome.)
If I repeated four times instead of three, it would be 11^4?
If I used just one die and cast three times, would it be 6^3?

Yes that is correct.
If repeated N times instead of three, it would be $11^N$?
If just one die and cast N times, would it be $6^N$?
• May 25th 2009, 03:42 PM
seavari
You have made my day! Thank you so much! I didn't know it would be so simple...you are a hero!