Probability of getting a pair of symbols when throwing four dice marked with swords, helmets, and skulls I have 4 dices with those faces display:
1-2-3 = sword,
4-5 = helmet,
6 = skull
When rolling 4 dice, what are the probabilities of getting a PAIR of swords ? helmets ? skulls ?
I tried to solve this by using a regular dice at the beginning, by calculating the probability of doing at least one pair.
1 - (6/6)(5/6)(4/6)(3/6) = 26/36 = 13/18

But in our case I want a pair of:


*

*skulls with 1/6, so 13/18 * 1/6 = 13/108

*helmet with 1/3, so 13/18 * 1/3 = 13/54

*sword with 1/2, so 13/18 * 1/2 = 13/36


But I am pretty sure I am mistaken

I'm assuming you mean getting the probability of two swords, of two helmets, and of two skulls separately. Here, the individual probabilities are:
$P(Sword) = \frac{3}{6} = \frac{1}{2}$
$P(Helmet) = \frac{2}{6} = \frac{1}{3}$
$P(Skull) = \frac{1}{6}$
Here, we use combination instead of permutation for we do not care about the order of desired outcome appearing. Since we want the outcome to occur twice out of four rolls, our combination value is $4C2 = 6$. Hence,
$P(2 Swords) = \frac{1}{2}\times\frac{1}{2}\times(1-\frac{1}{2})\times(1-\frac{1}{2})\times6=\frac{3}{8}$
$P(2 Helmets) = \frac{1}{3}\times\frac{1}{3}\times(1-\frac{1}{3})\times(1-\frac{1}{3})\times6=\frac{8}{27}$
$P(2 Skulls) = \frac{1}{6}\times\frac{1}{6}\times(1-\frac{1}{6})\times(1-\frac{1}{6})\times6=\frac{25}{216}$
