Matching sets on Dice

ArmoredThirteen

New Member
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...).

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

New Member
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

TS Contributor
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

New Member
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).