A fair 6 sided dice is rolled 4 times. What is the probability that at least 3 of the numbers will be either 1 or 6?

I'd really love a sanity check here as I walk through what I believe is the solution.

Total possible outcomes = $6^4 = 1296$

Possible combinations of 3 rolls being either 1 or 6 = $({}_4C_3)\cdot2 = (4)\cdot2 = 8$

Also take into account all 1's and all 6's = $1 + 1 = 2$

Answer = $\frac{8+2}{ 1296} = \frac{10}{1296} = \mathbf{\frac{5}{648}}$

Really appreciate the help! :)

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It's good for a question to have an informative title, however, you should always put the question into the body, and the title itself should be fairly concise. –  tomasz Jul 26 '13 at 19:08
Thank you, I appreciate the comment. Was my first posting to the site, and I'll be sure to take your advice next time :) –  Ben Jul 27 '13 at 16:55

I would use a binomial probability: \begin{align*} P(\text{at least 3 are 1 or 6}) &= P(\text{exactly 3 are 1 or 6}) + P(\text{exactly 4 are 1 or 6}) \\ &= {}_4C_3 \left(\dfrac{2}{6}\right)^3\left(\dfrac{4}{6}\right)^1 + {}_4C_4 \left(\dfrac{2}{6}\right)^4\left(\dfrac{4}{6}\right)^0 \\ &= 4 \left(\dfrac{1}{3}\right)^3\left(\dfrac{2}{3}\right) + \left(\dfrac{1}{3}\right)^4 \\ &= \dfrac{8+1}{81} \\ &= \dfrac{1}{9} \\ \end{align*}

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Your sample space of $1296$ can be replaced by a far smaller and more convenient sample space.

But as an exercise we carry out the count of how many of the ordered quadruples satisfy the requirement of at least three $1$ and/or $6$.

First we count the number of cases where every entry is one of $1$ or $6$. There are plenty of choices other than $1111$ and $6666$, like $6111$. The first toss can take on anyone of $2$ values, and for each such value, there are $2$ possibilities for the second toss, and so on for a total of $2^4$.

Next we count the number of cases where we have exactly three $1$ and/or $6$.

Where the oddball throw occurs can be chosen in $\binom{4}{1}$ ways. For each of these ways, the number on the oddball throw can be chosen in $4$ ways. And now the remaining three slots can be filled with $1$ or $6$ in $2^3$ ways, for a total of $128$.

Thus $16+128$ elements of our sample space of $1296$ are "favourable." Now we can write down the probability.

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Hey André, I appreciate the time you took to write that all out! The only move you made that I didn't follow was how you got to 128. 4 choices for the first slot, 2 for the second, third and fourth would lead me to do 4*2*2*2 = 32 where did I loose you? –  Ben Jul 26 '13 at 21:42
The slot where the roll that's neither $1$ or $6$ can be any of the $4$. I called the number of choices $\binom{4}{1}$. The number we put there can be any of $2,3,4,5$, $4$ choices. Then the $2^3$ that you understand. So $\binom{4}{1}(4)(2^3)=128$. –  André Nicolas Jul 26 '13 at 21:50
I see. It's basically 4(4*2*2*2) since the 4*2*2*2 can occur in 4 different combinations. Got it :) –  Ben Jul 27 '13 at 16:51
Yes, that's right, there are two $4$'s in the game, one a position $4$ and the other a number on the die $4$. –  André Nicolas Jul 27 '13 at 16:57

$1$ or $6$ has probability $\cfrac 13$

All four $1$ or $6$ has probability $\left(\cfrac 13\right)^4=\cfrac 1{81}$

Precisely three $1$ or $6$ has probability (choosing a place out of four for the "other" value) $\dbinom 41\cfrac 23\left(\cfrac 13\right)^3=\cfrac 8{81}$

Add the two to get $\cfrac 9{81}=\cfrac 19$

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To the top answer one must also add the possible combination of result with the value of remaining die which can be any value. So your total favourable outcome becomes 10 X 6 = 60. So actual probability of getting either 1 or 6 is 60/1296 = 5/108

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