dice probability 2 dice are the same in 5

Question

I have 5 dice that I roll. I want to find that at least 2 values are the same and at least 3 values are the same.

So for k=2: $P(2) = 1-P(\text{all values distinct}) = 1-\frac{6*5*4*3*2}{6^5}$ but for k=3 I thought it's $P(3)= P(2)-P(\text{exactly 2 values are the same})$ where $P(\text{exactly 2 values are the same}) = 6*\binom{5}{2}* (1/6)^2 *(5/6)^3$ but I get $P(3)$ negative... can you please help?

Answer 1

Your equation for the probability of at least three being the same is incorrect. P[at least three are the same]=p[at least two are the same]+p[exactly three are the same]. What you have is p[at least two are the same]+p[exactly 2 are the same] which won't work.

Or to put it another way, if p[x] is the probability that exactly x dice are the same, the equation you have is p[1]+p[2]+p[2] instead of p[1]+p[2]+p[3]

Answer 2

You are correct in that $P(X\ge3)=P(X\ge2)-P(X=2)$, but you have not correctly calculated $P(X=2)$

Having two values the same can be either one or two pairs.

Number of ways to make one pair is $6\times\binom{5}{2}\times5\times4\times3=3600$

Number of ways to make two pairs is $\binom{6}{2}\times\binom{5}{2}\times\binom{3}{2}\times4=1800$

Taken together, $P(X=2)=\frac{5400}{7776}$

Now, with $P(X\ge2)=\frac{7056}{7776}$ we have $P(X\ge3)=\frac{7056-5400}{7776}=\frac{1656}{7776}$

Answer 3

When subtracting the scenarios where you have a pair, remember these can also have two pairs and still not have three of a kind.

P(2 dice are the same but not 3 dice) = P(1 pair) + P(2 pairs)

$=\frac{6*5C2*5!/2!}{6^5}+\frac{6*5C2*5*3C2*4/2}{6^5}=\frac{3600}{7776}+\frac{1800}{7776}=\frac{5400}{7776}$