Dice game in the Mid-Autumn Festival

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I have the following question. I solved them but I need to double check that my answer is correct.

In a dice game that is played during the Mid-Autumn Festival, participants take turn throwing six dice into a large bowl. If certain combinations show up, the person gets a prize. Below, we describe the important combinations and the prize assigned to each combination:

  1. six 1’s or six 4’s show up (1st prize)
  2. exactly five of any number show up (2nd prize)
  3. the numbers 1, 2, 3, 4, 5, 6 show up (3rd prize)
  4. exactly three 4’s show up (4th prize)

What is the probability of winning each of these prizes?

Here are my answers:

  1. $\frac1{6^6}+\frac1{6^6}$
  2. $\frac{\binom61\binom65}{6^6}$
  3. I am confused here. The first solution I thought of was $\frac1{6!}$, but my friend said the solution was $\frac16(1-\frac16)(1-\frac26)(1-\frac36)(1-\frac46)(1-\frac56)$.
  4. $\frac1{6^3}(1-\frac16)^3$
3

There are 3 best solutions below

1
On

There are $6^6$ equally likely outcomes (counting order of the dice). Of these, $6!$ win third prize. So the probability of third prize is $6!/6^6$.

0
On

The second answer is incorrect. It should be $$\frac{6(6)(5)}{6^6} = \frac{180}{46656} = \frac{5}{1296},$$ because there are $6$ ways to choose which number is the $5$-of-a-kind, $5$ ways to choose the singleton (which cannot be the same number as the $5$-of-a-kind), and $6$ positions/permutations of the singleton among the $5$-of-a-kind.

1
On

First Prize: Aye, Okay.

Second Prize: the favoured event is a selection of two numbers, the penttuple and single, and dice for the pentuple.

$$\dfrac{\binom 6 1\binom 5 1\binom 65}{6^6}$$

Third Prize: the favoured event is some permutation of all six distinct numbers. How many ways can that happen out of the $6^6$ total?

$$\dfrac{\bbox[0.25ex, border: 0.1ex dashed green]{\qquad ?}}{6^6}$$

Fourth Prize: the favoured event is a selection of three dice to show 4 and the three other dice must each show one of the 5 not-4 numbers.

$$\dfrac{\binom{6}{3}\bbox[0.25ex, border: 0.1ex dashed green]{\qquad ?}}{6^6}$$