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When 100 fair dices are thrown the probability of getting $n$ "6" can be modelled by

$$P=\binom{100}{n}\left(\frac{1}{6}\right)^n\left(\frac{5}{6}\right)^{100-n}$$

What is the value of $n$ such that the probability is maximum?

I graphed the function and know the answer is $16$, but I am wondering why.

Aka_aka_aka_ak
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user362325
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2 Answers2

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Let $\displaystyle P_n=\binom{100}{n}\left(\frac{1}{6}\right)^n$

Note that

\begin{align*} &\quad\binom{100}{n}\left(\frac{1}{6}\right)^n\left(\frac{5}{6}\right)^{100-n}>\binom{100}{n+1}\left(\frac{1}{6}\right)^{n+1}\left(\frac{5}{6}\right)^{100-(n+1)}\\ \Longleftrightarrow &\quad \frac{100!}{n!(100-n)!}\times 5> \frac{100!}{(n+1)!(99-n)!}\\ \Longleftrightarrow &\quad \frac{5}{100-n}> \frac{1}{n+1}\\ \Longleftrightarrow &\quad n> \frac{95}{6}=15+\frac{5}{6}\\ \end{align*}

So $P_0\le P_1\le P_2\le\cdots\le P_{16}>P_{17}>P_{18}>\cdots>P_{100}$.

$P_n$ is the largest when $n=16$.

CY Aries
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  • Why $P_0\le P_1\le P_2\le\cdots\le P_{16}$ ? – user362325 May 13 '17 at 02:38
  • $P_{n}>P_{n+1} \Longleftrightarrow n> \frac{95}{6}$ is equivalent $P_{n}\le P_{n+1} \Longleftrightarrow n\le \frac{95}{6}$. Actually, as $\frac{95}{6}$ is not an integer, we can prove that $P_{n}< P_{n+1} \Longleftrightarrow n< \frac{95}{6}$. – CY Aries May 13 '17 at 02:46
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What you are asking for is the mode of the binomial distribution. The answer and proof is well-documented in the corresponding Wikipedia article:

Wikipedia: Binomial Distribution

heropup
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