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I am trying to calculate the probability above, and I think that my logic is a bit off. I thought that $$P(1s=a, 2s=b)=\frac{(\frac{1}{6})^{a+b}\times(\frac{4}{6})^{100-a-b}*100!}{a!b!(100-a-b)!}$$

What is incorrect with this logic? I am confused whether $\frac{1}{6}^b$ should be used or $\frac{1}{5}^b$.

shurup
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    There isn't any logic shown here: it's just a formula. If you would explain how you arrived at it, you might be able to answer this question yourself. – whuber Nov 23 '20 at 16:17
  • For material about this distribution please see https://stats.stackexchange.com/questions/431476 and https://stats.stackexchange.com/questions/4649 (which fully answers your question). – whuber Nov 23 '20 at 21:41

2 Answers2

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100 rolls of a fair die can be considered as a multinomial distribution. For simplicity, we will consider 3 categories: a roll of a 1, a roll of a 2, and everything else.

Let $\pi_i$ be the probability for each of the three categories. The density is

$$ \dfrac{\Gamma(101)}{\Gamma(a+1) \Gamma(b+1)\Gamma(101-a-b)} \pi_1^a \pi_2^b \pi_3^{100-a-b} = \dfrac{100!}{a!b!(100-a-b)!} \pi_1^a \pi_2^b \pi_3^{100-a-b} $$

You will notice that is is very similar to what you have posted. The probabilities for each category are 1/6, 1/6, and 4/6 respectively. Note, the probabilities for each category must sum to 1.

As to your point

I am confused whether $\frac{1}{6}^b$ should be used or $\frac{1}{5}^b.$

Why would you think the probability of rolling a 2 is 1/5?

Demetri Pananos
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Here's an alternative solution that relies on the Binomial distribution, which is somewhat more familiar to many readers. In $n$ independent trials with success probability $p$, the probability of exactly $x$ successes is given by $$P(X = x) = \binom{n}{x}p^x(1-p)^{n-x}, \ x=0,\ldots n$$

Let $X$ and $Y$ be the number of ones and twos rolled, respectively. Then we have $$P(X=A \ \cap \ Y=B) = P(X=A|Y=B)P(Y=B)$$ Since $Y$ is a binomial random variable (with parameters $n=100$ and $p=1/6$), the probability $P(Y=B)$ is straightforward. Conditional on $Y=B$, $X$ is still Binomial with parameters $n=100-B$ and $p=1/5$. To be verbose, there are $n-B$ trials for which we don't know the outcome and the remaining numbers $\{1,3,4,5,6\}$ are equally likely. This gives a final solution \begin{align*} P(X=A \ \cap \ Y=B) &= \left\{\binom{n-B}{A}\left(\frac{1}{5}\right)^A\left(\frac{4}{5}\right)^{n-B-A}\right\}\left\{\binom{n}{B}\left(\frac{1}{6}\right)^B\left(\frac{5}{6}\right)^{n-B} \right\} \end{align*} With a bit of manipulation, you can make this answer look the same as the answer given by @DemetriPananos.

knrumsey
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