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First time I approach a probability question (:

Throw a dice 4 times. What is the probability 6 be up at-least one time?

Intuitively, I would say: $\frac{1}{6}\times4$.

I would explain as: If you throw one time, probability is $\frac{1}{6}$.

If you do it 4 times, then multiply by 4.

According to the answers I'm wrong.

Can you explain please? thanks in advance.

Billie
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    But suppose you throw it 7 times. Then the probability = 1/6 x 7 > 1? – SDiv Oct 31 '14 at 22:04
  • What are the odds of not getting 6 on a given roll: $5/6$. Now since each roll is an independent event, the probability is $(5/6)^n$, where $n$ is the number of rolls. – Christopher K Oct 31 '14 at 22:22

4 Answers4

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Expanding the comment of Vladimir Vargas about binomial distribution you can, alternatively, use the complementary probability to take zero 6's in the four throws:

$$P(X\ge 1)=1-P(X=0)=1-\binom{4}{0}\left(\frac16\right)^0\left(1-\frac16\right)^4=1-\left(\frac56\right)^4=\frac{671}{1296}\approx52\%$$

Masacroso
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Throw a dice six times - are you certain to get at least one six?

The mean number of sixes in four throws is indeed $\frac 23$. But there are combinations with $2, 3 \text{ or } 4$ sixes, and these reduce the number with just one six (apply the same argument to six throws, where it is more intuitive).

If you get no sixes in four throws, there are $5$ possibilities for each throw, and therefore $5^4=625$ possibilities with no six out of the $6^4=1296$ possibilities in total. So you get the probability by taking the $1296-625 = 671$ possibilities which must include at least one six, out of the $1296$ possibilities altogether.

Mark Bennet
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Use the binomial distribution:

$$\mathbb P (\mbox{"obtaining $x$ times 6 in $n=4$ random experiments"})=\binom{n}{x}p^xq^{n-x},$$ where $p$ is the probability of obtaining '$6$' and $q$ is $(1-p)$.

PS: The probability of obtaining '6' at least one time is the sum of the probabilities of getting '6' one time, getting '6' two times,..., getting '6' four times.

Vladimir Vargas
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Consider the complement problem, there is a 5/6 probability of not rolling a six for any given die, and since the four dice are independent, the probability of not rolling a six is (5/6)^4 = 5^4/6^4 = 625/1296. The probability of rolling at least one six is therefore 1 − 625/1296 = 671/1296 ≈ .517.

adi_226
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