# probability of rolling at least N of C when rolling X dice

I have been looking this up but am struggling to find a straightforward answer. I am still confused about the process.

So basically what is the probability of rolling at least three of a particular number say X when rolling five dice.

I believe the way to go is work out the probability of rolling 5 of the same number then add it too the probability of 4 of the same number and so on.

so 1/6^5 + 1/6^4 + 1/6^3 + 1/6^2.

Is this correct or am I leaving something out?

I originally phrased my question wrong asking the probability of getting x of the same number.

A good example of what I mean is

The probability of rolling at least three 2's when rolling five dice

• Why do you have $\frac16 ^2$ there if you only care about "at least 3 of the same number"? Also, are you saying that $P(\text{all 5 are the same number})=\frac16^5$? And $P(\text{exactly 4 are the same number})=\frac16^4$? This is not true. Dec 2, 2017 at 20:10
• @JohnDoe Yes the power of two was my mistake. and yes too the second part. Dec 2, 2017 at 20:13

We assign a probability distribution to what we want to count. First define the random variable $X_1$ to be the number of $1$'s we throw, $X_2$ to be the number of $2$'s we throw, etc. Then $X_i$ is binomially distributed $B(\frac16,5)$. To compute the probability of throwing exactly $n$ of a particular number, say $N$, we have to find $$P(X_N=n)=P(X_1=n)$$ since $X_1$ has the same distribution as all the other $X_i$'s. Then $$P(X_1=n)={5\choose n}\frac1{6^n}\frac{5^{5-n}}{6^{5-n}}$$as described by the other answer. Then to find the probability of rolling at least $3$ of this number, you work out $$P(X_1=5)+P(X_1=4)+P(X_1=3)$$

• Do you know what I Phrased my question wrong, I will edit it now. Dec 2, 2017 at 20:33
• Please add it to the end instead of changing it so our answers still apply to that portion.
– Remy
Dec 2, 2017 at 20:34

First choose one to be shown $3$ times. There are $6\choose{1}$ ways to choose this die. Then use the standard binomial to get the probability of that selected die showing up exactly $3$ times.

$${6\choose{1}}\cdot{5\choose{3}}\cdot\frac{1}{6}^3\cdot\frac{5}{6}^2 \approx .193$$

And for the same number being shown exactly $4$ times:

$${6\choose{1}}\cdot{5\choose{4}}\cdot\frac{1}{6}^4\cdot\frac{5}{6}^1 \approx .0193$$

And for the same number being shown exactly $5$ times:

$${6\choose{1}}\cdot{5\choose{5}}\cdot\frac{1}{6}^5\cdot\frac{5}{6}^0 \approx 7.713\cdot10^{-4}$$

Then summing these $3$ results we get $p \approx 0.213$

For a fixed value:

We take away the $6\choose{1}$'s since we are no longer choosing $1$ of the $6$ die to be shown at least $3$ times. We have

$${5\choose{3}}\cdot\frac{1}{6}^3\cdot\frac{5}{6}^2 + {5\choose{4}}\cdot\frac{1}{6}^4\cdot\frac{5}{6}^1 + {5\choose{5}}\cdot\frac{1}{6}^5\cdot\frac{5}{6}^0 \approx 0.03549$$

• We cannot just make that calculation. That would be the probability of getting 3 of a particular side followed by 2 of other values. But we are interested in $3$ of one value and $2$ of another in $any$ order. This may be of help.
– Remy
Dec 2, 2017 at 20:22
• Oh I see. Then take away the $6\choose{1}$'s. I will edit my answer.
– Remy
Dec 2, 2017 at 20:38
• You multiply it by $5\choose{3}$ and do the addition separately. So $$\left({5\choose{3}}\cdot\frac{1}{6}^3\cdot\frac{5}{6}^2\right) + \left({5\choose{4}}\cdot\frac{1}{6}^4\cdot\frac{5}{6}^1\right) + \left({5\choose{5}}\cdot\frac{1}{6}^5\cdot\frac{5}{6}^0\right)$$
– Remy
Dec 2, 2017 at 20:51
• Is that what you're asking?
– Remy
Dec 2, 2017 at 20:52
• $5\choose 3$ is notation which means "from a group of 5 items, in how many ways can you pick 3 things". The formula for it is $\frac{5!}{3! 2!}=\frac{120}{6\times 2}=10$. (You know what a factorial is, yes?) Dec 2, 2017 at 21:11