I have a basic background in probability, which is apparently not enough.
I roll N dice.
A negative result occurs if two are the same. An extremely negative result occurs if three are the same. The more dice I roll, the better, unless a negative result occurs. How many should I roll to have a better than 50/50 chance of avoiding the negative result? How many to avoid the extremely negative result?
And how do I calculate it 
One die, the chance of a negative result is zero.
Two dice, the chance of a negative result is 1/6.
Roll die A, and it comes up with the number ‘a’
Die B has one ‘a’ face and 5 non-‘a’ faces.
For three dice:
Roll A, and you get ‘a’
Then roll B, and you either get ‘a’ (1/6) or ‘b’ (a non-‘a’ number) (5/6)
If B came up ‘a’, then roll C and you get either ‘a’ (1/36) or ‘c’ (5/36)
If B came up ‘b’, then roll C and you get either ‘a’ (5/36) or ‘b’ (5/36) or ‘c’ (20/36)
So your choices are:
aaa (doubly negative result) - 1/36
aac (negative result) - 5/36
aba (negative result) - 5/36
abb (negative result) - 5/36
abc (good result) - 20/36
So for three dice, your chances of a negative result are 1/36, of a doubly negative result are 15/36 = 5/12, and of a good result are 20/36 = 5/9.
I’m not sure of a more general formula, since doing it by hand like that is going to get real tedious real soon. Does this help at all, by at least giving you a place to start?
Quick stab at it:
1 and only 1 pair=[61(5factorial/(N-1)factorial) ] / 6^^N (^^ being used here as power)
1 and only 1 set of triples=[611*(5factorial/(N-2)factorial)]/6^^N
Explanation for triple:
6=First die can be anything
1=2nd die=first
1=3rd die=first
5factorial=don’t want additional doubles or triples in remaining dice
/(N-1)factorial=if 5 dice thrown, for example, can’t do 5 factorial, need to end up with 61154 so the second factorial cancels the extra positions
What does this mean? Is there some other aspect of what you’re trying to optimize? If you’re goal is to avoid matches, then the more dice you roll, the worse off you are. If there is some upside to rolling more that you haven’t mentioned, we’d need to know what it is.
This is just a variant of the “birthday problem”, which we’re discussing over here if you’re interested. Basically, the probability of getting a “non-bad” result for three dice is 5/9 (6/6 * 5/6 * 4/6), and for four dice is 5/18 (6/6 * 5/6 * 4/6 * 3/6.) Generally, for n dice the probability of no matches is 6!/((6 - n)! 6[sup]n[/sup]). As for the probability of getting exactly one pair of matching dice, unless I’ve done my math wrong, it’s n(n-1)*6!/(2 (7 - n)! 6[sup]n[/sup]).
Here’s my results, all tallied:
N no matches one pair three or more matches
1 1 0 0
2 5/6 1/6 0
3 5/9 5/12 1/36
4 5/18 5/9 1/6
5 5/54 25/54 4/9
6 5/324 25/108 61/81
Huh? 6 dice and the chance of getting no matches is 5/324? i think you skipped the sanity check…
No, that 5/324 is right. There are 6[sup]6[/sup] = 46656 ways to roll the 6 dice, and only 6! = 720 of them have no matches. 720/46656 = 5/324.
I think he’s trying to say “What’s the most I can roll and still have less than a 50% chance of getting a match?” And also, “What’s the most I can roll and still have less than a 50% chance of getting two matches?”
Am I understanding the question(s) correctly?
The problem is "“What’s the smallest number of dice I can roll and have less than a 50% chance of getting any kind of a match?”
I came up with the same answer as MikeS, and the answer is n = 4.
<role playing geek>
Can we assume 6 sided dice? It isn’t explicitly stated how many side the dice are or even that they are all the same.
</rpg>
Tho I reluctantly admit that all 6 sided dice is the most common configuration.
Brian
I didn’t say 6 sided dice because I figured it was unnecessary to know the number of faces to derive a general formula.
MikeS has answered the pair question with a nice table (thanks :)) but I will clarify it. Each die rolled gives me a “power” (or points if you will), so the more dice rolled the better. But each pair hurts me and a triple kills me, invalidating the roll.
Once the birthday problem was suggested I found a really nice explanation of the triple problem here. Apparently the (1 - no matches - one pair) approach does not work because it includes cases with multiple pairs. Given the formula on that page I calculated 6 dice to still be under a 50% chance of rolling a triple, but 7 dice to be pushing my luck.
Thanks all!