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An unbiased die is thrown $n$ times; let $M$ and $m$ denote the maximum and minimum points obtained respectively. Find $P\left( m=2, M=5 \right)$. (Hint: begin with $P\left (m\geq2, M\leq 5 \right)$.)

The question is from elementary probability theory (Kai Lai Chung)

Here is my thought process:

The minimum $m$ and maximum $M$ points obtained in $n$ throws are independent events. So:

$$ P(m\geq2, M\leq5) = P(m \geq 2)P(M\leq5)$$ $$ P(M\leq5) = \left(\frac{5}{6}\right)^n $$ $$ P(m\geq2) = 1-\left(\frac{4}{6}\right)^n $$ $$P(m\geq2, M\leq5) = \left(\frac{5}{6}\right)^n - \left(\frac{5}{9}\right)^n$$

under above conditions, $P(m=2, M=5)$ is just $\frac{1}{6}$ in all combinations. So:

$$ P(m=2, M=5) = \left(\frac{5}{6}\right)^n - \frac{\left(\frac{5}{9}\right)^n}{6} $$

But the answer is $\left(\frac{4}{6}\right)^n - 2\left(\frac{3}{6}\right)^n + \left(\frac{2}{6}\right)^n$.

Can anyone help me figure out where I went wrong?

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    $\begingroup$ The minimum and the maximum are never independent events. $\endgroup$
    – JohnK
    Commented Sep 23, 2015 at 13:46
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    $\begingroup$ @MatthewDrury Uncool to post a possible answer because of the self-study tag? $\endgroup$
    – Antoni Parellada
    Commented Sep 23, 2015 at 14:04
  • $\begingroup$ I didn't think of that as a full answer, just a "try to put your mind here". I will delete. My apologies. $\endgroup$
    – Matthew Drury
    Commented Sep 23, 2015 at 14:06
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    $\begingroup$ @MatthewDrury, based on the correct answer, I reversely think about this question. My though was following: the probability $2=< x =< 5$ was $(\frac{4}{6})^n$. there was another situation only appeared 3 and 4 whose probability was $ (\frac{2}{6})^n$. The minus part in the answer suggest that this part was double counting. But I can not explain it. $\endgroup$
    – wikichung
    Commented Sep 23, 2015 at 14:17
  • $\begingroup$ @wikichung $\frac{4}{6}^n$ is the probability of rolling between a $2$ and $5$, $n$ times in a row. If you do that, does that make $m=2$ and $M=5$? If not, what quantity is $\frac{4}{6}^n$, in terms of $m, M$? $\endgroup$
    – jlimahaverford
    Commented Sep 23, 2015 at 14:50

2 Answers 2

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I'll post an "answer," mainly composed of wikichung's ideas. We independently roll the unbiased die $n$ times obtaining $X_1, \dots, X_n$, with minimum value $m$ and maximum value $M$. What is $P(m=2, M=5)$? You noted that

$$ P(2 \leq X_i \leq 5; \forall i) = \prod_{i=1}^n \frac{4}{6} = \left(\frac{4}{6}\right)^n. $$ Now how does this relate to $P(m=2, M=5)$? It is two big! For example we are counting events for which $m \neq 2$, such as $(5, 5, \dots, 5)$. However, if for each $i$, $2 \leq X_i \leq 5$, then $m \geq 2$ and $M \leq 5$. So we have determined that,

$$ P(m \geq 2, M \leq 5) = \frac{4}{6}. $$

By the same logic for any $a \leq b$:

$$ P(m \geq a, M \leq b) = \frac{b-a+1}{6}. $$

So I will leave the last question to you. How can we rewrite $P(m=2, M=5)$ by adding and subtracting terms of the form $P(m\geq a , M\leq b)?$ This is where you will need inclusion/exclusion.

edit - Additional Information

Define $S(m,M) = |\{$sequences with min $=m$ and max $=M\}|$, and $T(m,M) = |\{$sequences with min $\geq m$ and max $\leq M\}|$. Then we have:

$$ S(2,5) = T(2,5) - T(3,5) - t(2,4) + ?. $$

There are some sequences counted by $T(2,5)$, which are not counted by $S(2,5)$, but which were doubly undone with the two negative terms. What are they?

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  • $\begingroup$ @jlimahaverford $(\frac{2}{6})^n$ stands for {2,2,2,…,5,5,5…}, right? $\endgroup$
    – wikichung
    Commented Sep 23, 2015 at 16:11
  • $\begingroup$ @wikichung It could, but that is not my interpretation of the answer. JohnK I don't quite understand what you mean by endpoints. I'll edit my answer with my interpretation. $\endgroup$
    – jlimahaverford
    Commented Sep 23, 2015 at 18:47
  • $\begingroup$ I meant the double-counting but you got it. $\endgroup$
    – JohnK
    Commented Sep 23, 2015 at 19:10
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