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A perfect dice is drawn $4$ times.What's the probability to the same number comes out at least $2$ times?

At first, I applied to binomial law.I made all calculations.

I set up a random variable.The possible values for the variable were $0,1,2,3$ and $4$. My thought was that any number have a probability of $\frac{1}{6}$ to come out.

Then I started to think.In the $4$ launchs two different numbers can come out $2$ times, each one.So my previous thoughts were wrong.

Can you give me an idea on how to solve this problem?

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If you are interested in the probability that there exists a number that comes out at least twice, compute the probability that the four results are different. –  Did Mar 16 '12 at 10:11
    
For the first launch there are $6$ possible results, for the second launch there are $5$, and so on until the fourth launch where there are $3$ possible results.So there are $6 \cdot 5 \cdot 4 \cdot 3=360$.Then the total possible cases are $6^4$, and so the probability is $\frac{360}{6^4}$.Is this? –  João Mar 16 '12 at 10:48

2 Answers 2

up vote 1 down vote accepted

If you find the probability that a $1$ appears at least two times, and similarly a $2$, and a $3$, etc., then you've got the complication that those six events are not mutually exclusive: a $1$ could appear twice and a $2$ twice.

So the simpler way is to find the probability that the event you're looking for does not occur, i.e. no number appears more than once.

The probability that the second number differs from the first is $\dfrac 5 6$.

The probability that the third number differs from the first two is $\dfrac 4 6$.

The probability that the fourth number differs from the first three is $\dfrac 3 6$.

Multiply those: $\dfrac{5\cdot4\cdot3}{6^3} = \dfrac{5}{18}$.

So what you're looking for is $\dfrac{13}{18}$.

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I chosen this answer, because the calculations are well explain.Thanks –  João Mar 16 '12 at 19:26

"Alea iacta est !!"

$$P=\frac{6^4-\frac{6!}{2!}}{6^4} \approx 0.722$$

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Now I understood. I've to find the probability of the opposite event.In this case the opposite event is "come out $1$ time on the best way".But in this case don't I have to find the favorable cases to a given number never come out? –  João Mar 16 '12 at 11:01
    
@João,probability=(number of permutations with repetition - number of permutations without repetition)/(number of permutations with repetition) –  pedja Mar 16 '12 at 11:13
    
Actually $6^3$ is a big enough denominator for the occasion. (See my answer.) –  Michael Hardy Mar 16 '12 at 18:25

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