# Dice throwing and calculating different probabilities

Question:

In a dice game that is played during the Chinese Moon 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)
1. Compute the probability that each of the combinations described above appears.
2. Suppose there are ten people playing this game and in each round, everyone gets a turn. By sheer luck, you have the first turn at each round. Now, there is only one first prize so the first person who throws the right combination wins the prize. What is the probability that you'll win it in the fifth round?

Solution:

1. Probabilities:

• $$\binom{6}{6} \times (\frac{1}{6})^6 \times (1-\frac{1}{6})^{6-6} + \binom{6}{6} \times (\frac{1}{6})^6 \times (1-\frac{1}{6})^{6-6} = \frac{2}{6^6}$$ (1st prize)
• $$6 \times \Big(\binom{6}{5} \times (\frac{1}{6})^5 \times (1-\frac{1}{6})^{6-5}\Big)$$ (2nd prize)
• $$6! \times (\frac{1}{6})^6$$ (3rd prize)
• $$\binom{6}{3} \times (\frac{1}{6})^3 \times (1-\frac{1}{6})^{6-3}$$ (4th prize)
2. To win at the beginning of the $$5$$th round of $$10$$ people no one should win the 1st prize in $$40$$ consecutive dice throwing and I should win in the $$41$$th dice throwing.

• $$\Big(1- \frac{2}{6^6}\Big)^{40} \times \Big(\frac{2}{6^6}\Big)$$

Am I on the right track?

• Yes, your work is correct. Jan 26 at 10:59
• The reasoning seems correct and simulations also agree with your results! Jan 26 at 11:08