We have 5 dice, 3 outside of a cup, 2 inside of a cup. The three outside are number 6.
We only throw the ones inside the cup. What’s the probability of throwing a poker (4 dice with 6s) or a full house (3 dice with 6, and a par of anything else).
Still debating my friends that its more probable to throw a poker.
Cheers
Let X and Y be independent random variables corresponding to the die rolls (uniformly distributed over outcomes 1,...,6).
Let's consider two mutually exclusive events: Let A be the event where at least one die is 6: $$A = \{X = 6\} \vee \{Y = 6\}.$$
By the Inclusion-exclusion principle, $$P(A) = P(X=6) + P(Y=6) - P(X=Y=6) = 1/6 + 1/6-(1/6)^2 = 11/36.$$
Now, let B be the event where both dice are any other number: $$B = \{X = Y \wedge X\neq 6\}.$$ Hence, $$P(B) = \sum_{i=1}^{5}P(X=i)P(Y=i) = \sum_{i=1}^{5}(1/6)(1/6) = 5/36.$$
Finally, as A and B are mutually exclusive, we have that $$P(A \vee B) = P(A) + P(B) = 11/36 + 5/36 = 4/9.$$