Dice - conditonal probability

Question

I just need a quick solution check.

100 dice are rolled. What is probability of getting die with number 5 at least twice if there is exactly 20 dice with number 6?

My solution: Let $A=${at least $2$ dice with $5$} and $B=${exactly $20$ dice with number $6$}. $$P(A|B)=\frac{P(A\cap B)}{P(B)}$$ and $P(A\cap B)=P(B)-P(B$ and $0$ dice with $5)$ -$P(B$and $1$ die with $5)={100\choose 20}(\frac{1}{6})^{20}(\frac{5}{6})^{80}-{100\choose 20}(\frac{1}{6})^{20}(\frac{4}{6})^{80}-{100\choose 20}{80 \choose 1}(\frac{1}{6})^{21}(\frac{4}{6})^{79}$ and $P(B)={100\choose 20}(\frac{1}{6})^{20}(\frac{5}{6})^{80}$.

Is this correct?

Answer 1

Yes, your calculation is correct.

Let $X$ be a random variable that counts the number of fives rolled. Look at a small case. Suppose we have $n = 3$ dice. What is the probability that, given that exactly one of the dice shows a six, that there is exactly one other die that shows a five? There are $216$ total distinct outcomes, of which there are $\binom{3}{1}\cdot 5^2 = 75$ for which exactly one six occurs. Of these outcomes, how many also have exactly one five? This is also easy to count: $\binom{3}{1}\binom{2}{1} \cdot 4 = 24$. So the resulting probability should be $24/75 = 8/25$. If we were to extend this reasoning to your case, we see that the total number of outcomes for which there are exactly $20$ sixes is $$\binom{100}{20} \cdot 5^{80}.$$ The total desired outcomes for which there is exactly one five is $$\binom{100}{20} \binom{80}{1} \cdot 4^{79}.$$ So the resulting probability is $16 (0.8)^{79}$. You should be able to calculate the other case as well.

Notice that this solution doesn't explicitly calculate the joint probabilities, but instead calculates the conditional probabilities directly through a counting argument--we enumerated the total number of desired outcomes, and divided by the total number of possible outcomes satisfying the given conditions. An alternative method of solution involves recognizing that once the $20$ sixes are given/observed, the only dice of interest, whose outcomes are subject to randomness, are the $80$ remaining ones. But once we know that these dice did not show a six, their individual outcomes are taken from the set $\{1,2,3,4,5\}$, not $\{1,2,3,4,5,6\}$. So we can also express the desired probability as $$\Pr[Y \ge 2] = 1 - \Pr[Y = 0] - \Pr[Y = 1],$$ where $Y$ is a binomial random variable with number of trials (dice) equal to $n' = 80$, and the individual probability of "success" (i.e., rolling a five) is $p' = 1/5$, not $1/6$ as in the unconditional case. So $\Pr[Y = 0] = \binom{80}{0}(1/5)^0 (4/5)^{80}$, and $\Pr[Y = 1] = \binom{80}{1}(1/5)^1(4/5)^{79}$.