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
 A: 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,
$P(2 Swords) = \frac{1}{2}\times\frac{1}{2}\times(1-\frac{1}{2})\times(1-\frac{1}{2})\times6=\frac{3}{8}$
$P(2 Helmets) = \frac{1}{3}\times\frac{1}{3}\times(1-\frac{1}{3})\times(1-\frac{1}{3})\times6=\frac{8}{27}$
$P(2 Skulls) = \frac{1}{6}\times\frac{1}{6}\times(1-\frac{1}{6})\times(1-\frac{1}{6})\times6=\frac{25}{216}$
A: In poker, a pair is a hand with exactly two cards of the same rank.  Therefore, I would interpret the question to mean:  Find the probability of obtaining two cards of one type and one card of each of the other types.
A pair of swords:  Say the dice are painted blue, green, red, and yellow.  
For the favorable cases, two of the four dice must display a sword, one must display a helmet, and one must display a skull.  Choose which two of the four dice display a sword and which of the two remaining dice displays a helmet.  The remaining die must display a skull.  Since the probability that a given die displays a sword is $3/6 = 1/2$, that a given die displays a helmet is $2/6 = 1/3$, and a given die displays a skull is $1/6$, the probability of a pair of swords is
$$\binom{4}{2}\binom{2}{1}\left(\frac{1}{2}\right)^2\left(\frac{1}{3}\right)\left(\frac{1}{6}\right)$$
The calculations for the probability of obtaining a pair of helmets and for the probability of a pair of skulls are analogous.  
Since there are only three distinguishable outcomes for each die roll, it is not possible to obtain four different results.  Therefore, the probability that at one of those outcomes will appear at least twice when four dice is rolled is $1$.
