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Question:

If a fair die is rolled, the outcome is one of the numbers $1, 2, \dots, 6$ with probability $\frac{1}{6}$. Consider the experiment where $n$ fair dice are rolled. The outcome can be described as $(x_1, x_2, \dots , x_n)$ such that $X_i$ is the number that shows in the ith die for $i = 1, \dots , n$. Express, as a function of $n$, the probability that $min\{x_1, \dots , x_n\} = 3$.

Solution:

  • Probability of having minimum 6 is equal to $(\frac{1}{6})^n$ (all $6$s)
  • Probability of having minimum 5 is equal to $(\frac{2}{6})^n - (\frac{1}{6})^n$ (excluding all 6s)
  • Probability of having minimum 4 is equal to $(\frac{3}{6})^n - (\frac{2}{6})^n - (\frac{1}{6})^n - (\frac{1}{6})^n$ (excluding mix of $\{5,6\}$, excluding all $5$s and excluding all $6$s)
  • Probability of having minimum 3 is equal to $(\frac{4}{6})^n - (\frac{3}{6})^n - (\frac{2}{6})^n - (\frac{1}{6})^n- (\frac{1}{6})^n- (\frac{1}{6})^n$ (excluding mix of $\{4,5,6\}$, excluding mix of $\{5,6\}$, excluding all $4$s, excluding all $5$s and excluding all $6$s)

Am I on the right track? or completely off?

Node.JS
  • 1,119

2 Answers2

2

I would not say you were completely off, but the last two probability values are not correct.

As a sanity check, you can find that $\left(\frac{3}{6}\right)^n - \left(\frac{2}{6}\right)^n - \left(\frac{1}{6}\right)^n - \left(\frac{1}{6}\right)^n$ is negative when $n = 1$, so there's something wrong there.

Steps for a correct solution:

  1. Let $F_m$ denote the probability that $\min\{x_1, x_2, \dots, x_n\} \geq m$. Can you find an expression for $F_m$ in terms of $m$ and $n$? (You have done this in your attempt above.)
  2. Use that to find $F_3$ and $F_4$ in terms of $n$.
  3. The answer to the original question is $F_3 - F_4$. Why?
VTand
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Another formulation would be to use the Multinomial Theorem
We want none from $1$ and $2$, $k$ from $3$, $(n-k)$ from $4,5,6$ with respective probabilities $\frac26, \frac16, \frac36$

Thus in full form, $$P(n) = \sum_{k=1}^n {\frac{n!}{0!k!(n-k)!}(2/6)^0(1/6)^k(3/6)^{n-k}}$$

which, of course, can be simplified as $$P(n) = \sum_{k=1}^n {\frac{n!}{k!(n-k)!}(1/6)^k(1/2)^{n-k}}$$

true blue anil
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