# Probability that all three dice show different numbers exactly two times when rolled three times simultaneously

Three different dice are rolled simultaneously, three times simultaneously. The find the probability that all of them show different numbers exactly two times.

Will total cases will be $(6^3)^3$? For favorable cases I think we can select two cases when we get different numbers on all three dice, then we will have $6 \times5 \times 4$ favorable cases for these two cases, then in last cases at least two numbers must be same. Is my approach correct? Could someone help me with this?

• So far, it's right. – saulspatz May 7 '18 at 17:23

For a single roll of three dice, there are $6^3$ possible outcomes. There are $6 \cdot 5 \cdot 4$ ways to obtain three different outcomes (for each of the six outcomes of the blue die, there are five outcomes of the green die that are different from the outcome of the blue die, and four outcomes of the red die that are different from both the outcomes on the blue and green dice). Hence, the probability of obtaining three different outcomes when three dice are rolled simultaneously is $$p = \frac{6 \cdot 5 \cdot 4}{6^3}$$ What we want is the probability that in exactly three trials of rolling three dice simultaneously, exactly two of those trials produce three different outcomes on the dice. We can use the binomial distribution, which states that the probability of exactly $k$ successes in $n$ trials, each of which has probability $p$ of occurring is $$\Pr(X = k) = \binom{n}{k}p^k(1 - p)^{n - k}$$ where $\binom{n}{k}$ is the number of ways exactly $k$ successes could occur in $n$ trials, $p^k$ is the probability of $k$ successes, and $(1 - p)^k$ is the probability of $n - k$ failures. In your problem, $n = 3$, $k = 2$, and $p$ was worked out above.