Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

If you roll 4 dice, what is the probability of at least 1 three appearing?

I'm not sure if I did the calculation correctly but here it is:

$\text{P(At least 1 three)} = P(1\mbox{ three}) + P(2\mbox{ threes}) +P(3\mbox{ threes}) +P(4\mbox{ threes})$

$(^4C_1)+(^4C_2)+(^4C_3)+(^4C_4)/6^{4} = 5/432$

I am not sure if I have calculated the probability correctly (probably not) but the probability of $5/432$ seems very low.

Logically speaking, if you have more dice, you have more chances of a three appearing on at least one of them right?

So why is the probability lower than the probability of getting a 3 on one dice (1/6)?

share|cite|improve this question
A side comment on English - dice is actually the plural form of the singular noun die (so dices$\to$dice). – Sanath K. Devalapurkar Feb 3 '14 at 6:08
Probability of no $3$ is $(5/6)^4=625/1296$. So at least one $3$ has probability $1$ minus this, roughly $1/2$. – André Nicolas Feb 3 '14 at 6:11
up vote 7 down vote accepted

You want probability of atleast one theree in 4 rolls that means total probability{1} - none of the die has 3.Hence the probability comes out to be $ 1-(\frac{5}{6})(\frac{5}{6})(\frac{5}{6})(\frac{5}{6})=\frac{671}{1296}$

share|cite|improve this answer

In a comment, I gave a quickie way of finding the answer, which, as you noted, should not be low. Let us do it in the way that you used. In structure the calculation is the same as yours, but there are differences of detail.

The probability of exactly $1$ three is $\binom{4}{1}\cdot \frac{1}{6}\cdot\left(\frac{5}{6}\right)^3$.

The probability of exactly $2$ threes is $\binom{4}{2}\cdot \left(\frac{1}{6}\right)^2\cdot\left(\frac{5}{6}\right)^2$.

We can find similar expressions for the probability of exactly $3$ threes, exactly $4$ threes, and add.

share|cite|improve this answer

Probability of exactly $k$ threes is $$ \binom{4}{k}\left(\frac16\right)^k\left(\frac56\right)^{4-k} $$ Your probabilities are off by a factor of $5^{4-k}$ which is why your answer came in low.

By the binomial theorem, we have $$ \sum_{k=0}^4\binom{4}{k}\left(\frac16\right)^k\left(\frac56\right)^{4-k}=\left(\frac16+\frac56\right)^4=1 $$ Thus, adding up the probabilities for $1$, $2$, $3$, and $4$ threes gives the complement the probability of getting $0$ threes.

This relates the answers of Devgeet Patel and André Nicolas.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.