# Rolling 20 times a fair dice

#### mathmari

##### Well-known member
MHB Site Helper
Hey!! Tim rolls $20$ times a fair dice.

• Calculate the probability to get a "$3$" exactly $8$ times.
• The probability to get at least $k$ times a "$5$" is equal to $10,18\%$ Calculate $k$.
• Explain the relation to the term $\displaystyle{\sum_{i=0}^2B\left (10;\frac{5}{6};i\right )}$.

I have done the following:

• The probability to get a "$3$" is equal to $\frac{1}{6}$. If we throw $20$ times the dice the probability to get exactly eight "$3$" is equal to $\binom{20}{8}\cdot \left (\frac{1}{6}\right )^8\cdot \left (1-\frac{1}{6}\right )^{20-8}=\binom{20}{8}\cdot \left (\frac{1}{6}\right )^8\cdot \left (\frac{5}{6}\right )^{12}\approx 0.84\%$.

Is this correct? • We have that \begin{align*}P(X\geq k)=10,18\% &\Rightarrow 1-P(X\leq k)=10,18\% \\ & \Rightarrow \sum_{i=0}^k\binom{20}{i}\cdot \left (\frac{1}{6}\right )^i\cdot \left (1-\frac{1}{6}\right )^{20-i}=89.82\% \\ & \Rightarrow \sum_{i=0}^k\binom{20}{i}\cdot \left (\frac{1}{6}\right )^i\cdot \left (\frac{5}{6}\right )^{20-i}=89.82\% \end{align*} How can we continue to get the value of $k$ ? • Is this the formula we used in the previous question, just with other numbers? #### Klaas van Aarsen

##### MHB Seeker
Staff member
Hey!! Tim rolls $20$ times a fair dice.

• Calculate the probability to get a "$3$" exactly $8$ times.
• The probability to get at least $k$ times a "$5$" is equal to $10,18\%$ Calculate $k$.
• Explain the relation to the term $\displaystyle{\sum_{i=0}^2B\left (10;\frac{5}{6};i\right )}$.

I have done the following:

• The probability to get a "$3$" is equal to $\frac{1}{6}$. If we throw $20$ times the dice the probability to get exactly eight "$3$" is equal to $\binom{20}{8}\cdot \left (\frac{1}{6}\right )^8\cdot \left (1-\frac{1}{6}\right )^{20-8}=\binom{20}{8}\cdot \left (\frac{1}{6}\right )^8\cdot \left (\frac{5}{6}\right )^{12}\approx 0.84\%$.

Is this correct?
Hey mathmari !!

Yep. • We have that \begin{align*}P(X\geq k)=10,18\% &\Rightarrow 1-P(X\leq k)=10,18\% \\ & \Rightarrow \sum_{i=0}^k\binom{20}{i}\cdot \left (\frac{1}{6}\right )^i\cdot \left (1-\frac{1}{6}\right )^{20-i}=89.82\% \\ & \Rightarrow \sum_{i=0}^k\binom{20}{i}\cdot \left (\frac{1}{6}\right )^i\cdot \left (\frac{5}{6}\right )^{20-i}=89.82\% \end{align*} How can we continue to get the value of $k$ ?
Shouldn't it be $P(X\geq k)=10.18\% \Rightarrow 1-P(X{\color{red}<} k)=10.18\%$ ? Can't we calculate those terms until we get to the percentage? Alternatively we can use some tool or table to calculate the inverse cumulative binomial function. • Is this the formula we used in the previous question, just with other numbers? Close, but not exactly.
You effectively wrote that that "at least $k$ times a $5$" is the same as $1$ minus "less than $k$ times a $5$".
Or:
$$P(\text{at least k times 5}) = 1- P(\text{less than k times a 5})$$

The value $B\left (10;\frac{5}{6};i\right )$ is the probability that $i$ times out of $10$ we do not roll a $5$.
In our case "at least $k$ times a $5$" is the same as "less than $10-k$ times not a $5$".
In formula form:
$$P(\text{at least k times 5}) = P(\text{less than 10-k times not a 5})$$ #### mathmari

##### Well-known member
MHB Site Helper
Shouldn't it be $P(X\geq k)=10.18\% \Rightarrow 1-P(X{\color{red}<} k)=10.18\%$ ? Can't we calculate those terms until we get to the percentage? Alternatively we can use some tool or table to calculate the inverse cumulative binomial function. So we have the following:
\begin{align*}P(X\geq k)=10,18\% &\Rightarrow 1-P(X< k)=10,18\% \\ & \Rightarrow P(X<k)=89,82\% \\ & \Rightarrow P(X\leq k-1)=89,82\% \\ & \Rightarrow \sum_{i=0}^{k-1}\binom{20}{i}\cdot \left (\frac{1}{6}\right )^i\cdot \left (1-\frac{1}{6}\right )^{20-i}=89.82\% \\ & \Rightarrow \sum_{i=0}^{k-1}\binom{20}{i}\cdot \left (\frac{1}{6}\right )^i\cdot \left (\frac{5}{6}\right )^{20-i}=89.82\% \end{align*} or not? Could you explain to me further how we could calculate $k$ from here? I got stuck right now. #### mathmari

##### Well-known member
MHB Site Helper
Close, but not exactly.
You effectively wrote that that "at least $k$ times a $5$" is the same as $1$ minus "less than $k$ times a $5$".
Or:
$$P(\text{at least k times 5}) = 1- P(\text{less than k times a 5})$$

The value $B\left (10;\frac{5}{6};i\right )$ is the probability that $i$ times out of $10$ we do not roll a $5$.
In our case "at least $k$ times a $5$" is the same as "less than $10-k$ times not a $5$".
In formula form:
$$P(\text{at least k times 5}) = P(\text{less than 10-k times not a 5})$$ So, do we have that $\displaystyle{\sum_{i=0}^2B\left (10;\frac{5}{6};i\right )}$ is the probability that neither $0$ nor $1$ time out of $10$ we do not roll a $5$, i.e. at least twice out of $10$ we don't roll a $5$, i.e. at most $8$ times out of $10$ we roll a $5$ ? #### Klaas van Aarsen

##### MHB Seeker
Staff member
So we have the following:
\begin{align*}P(X\geq k)=10,18\% &\Rightarrow 1-P(X< k)=10,18\% \\ & \Rightarrow P(X<k)=89,82\% \\ & \Rightarrow P(X\leq k-1)=89,82\% \\ & \Rightarrow \sum_{i=0}^{k-1}\binom{20}{i}\cdot \left (\frac{1}{6}\right )^i\cdot \left (1-\frac{1}{6}\right )^{20-i}=89.82\% \\ & \Rightarrow \sum_{i=0}^{k-1}\binom{20}{i}\cdot \left (\frac{1}{6}\right )^i\cdot \left (\frac{5}{6}\right )^{20-i}=89.82\% \end{align*} or not? Could you explain to me further how we could calculate $k$ from here? I got stuck right now. Yes.
We can write:
$$\sum_{i=0}^{k-1}\binom{20}{i}\cdot \left (\frac{1}{6}\right )^i\cdot \left (\frac{5}{6}\right )^{20-i} =\overbrace{\left (\frac{5}{6}\right )^{20} + \binom{20}{1}\cdot \left (\frac{1}{6}\right )^1\cdot \left (\frac{5}{6}\right )^{19} +\ldots}^{k\text{ terms}} \approx\overbrace{0.26\% + \ldots}^{k\text{ terms}} =89.82\%$$
We can approximate each term... it is quite a few though... Perhaps we should take the other approach. So, do we have that $\displaystyle{\sum_{i=0}^2B\left (10;\frac{5}{6};i\right )}$ is the probability that neither $0$ nor $1$ time out of $10$ we do not roll a $5$, i.e. at least twice out of $10$ we don't roll a $5$, i.e. at most $8$ times out of $10$ we roll a $5$ ?
You are off by $1$ in each case. The sum is up to and including $i=2$, isn't it?

#### mathmari

##### Well-known member
MHB Site Helper
Yes.
We can write:
$$\sum_{i=0}^{k-1}\binom{20}{i}\cdot \left (\frac{1}{6}\right )^i\cdot \left (\frac{5}{6}\right )^{20-i} =\overbrace{\left (\frac{5}{6}\right )^{20} + \binom{20}{1}\cdot \left (\frac{1}{6}\right )^1\cdot \left (\frac{5}{6}\right )^{19} +\ldots}^{k\text{ terms}} \approx\overbrace{0.26\% + \ldots}^{k\text{ terms}} =89.82\%$$
We can approximate each term... it is quite a few though... Perhaps we should take the other approach. Do you mean such a table ? You are off by $1$ in each case. The sum is up to and including $i=2$, isn't it?
Ah yes, so is it as follows?

$\displaystyle{\sum_{i=0}^2B\left (10;\frac{5}{6};i\right )}$ is the probability that neither $0$ nor $1$ nor $2$ time out of $10$ we do not roll a $5$, i.e. at least three times out of $10$ we don't roll a $5$, i.e. at most $7$ times out of $10$ we roll a $5$ #### Klaas van Aarsen

##### MHB Seeker
Staff member
Do you mean such a table ? Such a table yes, except that that one is for the t-distribution instead of the binomial distribution.

$\displaystyle{\sum_{i=0}^2B\left (10;\frac{5}{6};i\right )}$ is the probability that neither $0$ nor $1$ nor $2$ time out of $10$ we do not roll a $5$
$\displaystyle{\sum_{i=0}^2B\left (10;\frac{5}{6};i\right )}$ is the probability that neither $0$ nor $1$ nor $2$ times out of $10$ we do not roll a $5$, ... 