Matching sets on Dice

[ArmoredThirteen]

For those of you who know dice, I have this question:
You roll multiple dice (with the same number of sides), what are the chances of getting matching sets?

For instance: You roll 10 dice that have 6 sides each. What is the chance that 2 of them are the same (does not matter the number, just that they are the same). What is the chance that 4 of them are the same?

For this I need something universal, a formula that can have varying numbers.


I got this formula from a friend, but it doesn't appear to be working correctly:

n = Total number of dice being rolled.
k = Number of dice that come up with desired result (the number of matching dice).
p = Chance of desired outcome occurring on one die, as a number between 0 and 1 (for six sided dice, this would be .166666).

n!*(p^k)*((1-p)^(n-k))
-----------------------
k!*(n-k)!

 

[Dason]

Your friend is wrong. I don't know the exact correct formula in general and it would probably not be the easiest thing to get. But what your friend gave you is the pmf for a binomial distribution with success probability p. This clearly isn't the case that you're in (for example you can't have 1 die that matches...).

 

[ArmoredThirteen]

No ideas?

If it helps at all, to get an example:
This is based on an RPG that I play called Godlike. For a description and links, here it is on Wikipedia.

 

[Dason]

Like I said it's probably pretty difficult to get a 'universal' form for what you're asking for.

If you have a specific form in mind (i.e. you tell us what n and p are) then it would be pretty easy to run some simulations to give you approximations to the true probabilities.

 

[SmoothJohn]

For instance: You roll 10 dice that have 6 sides each. What is the chance that 2 of them are the same (does not matter the number, just that they are the same).

This one's easy. The probability is 1, by the pigeonhole principle.

John

 

[BGM]

This one's easy. The probability is 1, by the pigeonhole principle.

John

I guess thats exactly the point needs to be clarified.

If the question is asking for at least have 2 dice having the same
number, then the probability, as suggested is one.

But, if the same number occurred 3 times or more are not count, then the
required probability is completely different.

Also, if the question is asking for exactly 1 pair only, then the probability
is zero by the pigeonhole principle again.
(with 6-faces dice, only 2-7 dices are possible)

 

[ArmoredThirteen]

It would be the chance that one set of exactly two shows up. There could be a set of three in there, too, but that doesn't matter. And the pair could be of any number (1-6, in the example that needs clarifying).

So, example, rolls that would qualify:
1, 1, 2, 2, 2, 3, 4, 5, 5, 5.
This has a pair of ones. It has a triplet of 2s and 3s, but that doesn't matter.

1, 3, 3, 4, 5, 5, 5, 5, 5, 6.
Now, a pair of 3s, and a bunch of 5s (the 5s, again, don't matter).