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:
- six 1’s or six 4’s show up (1st prize)
- exactly five of any number show up (2nd prize)
- the numbers 1, 2, 3, 4, 5, 6 show up (3rd prize)
- exactly three 4’s show up (4th prize)
What is the probability of winning each of these prizes?
Here are my answers:
- $\frac1{6^6}+\frac1{6^6}$
- $\frac{\binom61\binom65}{6^6}$
- I am confused here. The first solution I thought of was $\frac1{6!}$, but my friend said the solution was $\frac16(1-\frac16)(1-\frac26)(1-\frac36)(1-\frac46)(1-\frac56)$.
- $\frac1{6^3}(1-\frac16)^3$
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 $6!/6^6$.