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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)?

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    $\begingroup$ A side comment on English - dice is actually the plural form of the singular noun die (so dices$\to$dice). $\endgroup$ – user122283 Feb 3 '14 at 6:08
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    $\begingroup$ Probability of no $3$ is $(5/6)^4=625/1296$. So at least one $3$ has probability $1$ minus this, roughly $1/2$. $\endgroup$ – André Nicolas Feb 3 '14 at 6:11
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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}$

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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.

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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.

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