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Four standard dice are rolled.

What are the chances of:

1.) Rolling three sixes and one five? 2.) Two sixes and two fives? 3.) Exactly one six? 4.) Four different numbers?

Here is my thought process for them, please help me develop an intuition for these sort of questions and/or correct me:

1.) Is it safe to assume that on the first dice I roll and get a five. The probability of that is $\frac{1}{6}$, and then the probability of rolling three sixes is $\frac{1}{6^3} $, with the final probability being $\frac{1}{6^4}$.

2.) Again... I'm not sure I'm doing this correctly. Wouldn't it be the same as 1?

3.)Exactly one six would be $\frac{1}{6}*\frac{5^3}{6^3}$

4.) Four different numbers would be $\frac{6}{6}*\frac{5}{6}*\frac{4}{6}*\frac{3}{6}$

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There are $6^4=1296$ different outcomes.


Rolling $3$ sixes and $1$ five

There are $\frac{(3+1)!}{3!\cdot1!}=4$ ways to do it, hence the probability is $\frac{4}{1296}$


Rolling $2$ sixes and $2$ fives

There are $\frac{(2+2)!}{2!\cdot2!}=6$ ways to do it, hence the probability is $\frac{6}{1296}$


Rolling exactly $1$ six

There are $\binom41\cdot(6-1)^{4-1}=500$ ways to do it, hence the probability is $\frac{500}{1296}$


Rolling $4$ different numbers

There are $\binom64\cdot4!=360$ ways to do it, hence the probability is $\frac{360}{1296}$

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  • $\begingroup$ Thank you for the response! Any chance you could explane the second and third one? $\endgroup$
    – user300011
    Mar 2, 2016 at 21:25
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    $\begingroup$ @RonaldB Second one $\frac{(2+2)!}{2!~2!}$ counts the ways select dice to show the two sixes and two fives. It is $\binom{4}{2}$. Third one: There are $\binom 4 1$ dice that could show the six, and $(6-1)^{4-1}$ ways the other dice could show not-six. Fourth one: There are $\binom{6}{4}$ ways to have four distinct numbers to show, and there are $4!$ ways these could be shown on the dice. $\endgroup$ Mar 3, 2016 at 0:40
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1.) Is it safe to assume that on the first dice I roll and get a five. The probability of that is $\frac{1}{6}$, and then the probability of rolling three sixes is $\frac{1}{6^3} $, with the final probability being $\frac{1}{6^4}$.

Sure, but don't ignore the other ways to roll 1 five and 3 sixes. The five could be rolled on the first, second, third, or fourth die, and the sixes on the remaining three.

$$\frac 4{6^4}$$

2.) Again... I'm not sure I'm doing this correctly. Wouldn't it be the same as 1?

Yes, but this time you have to count ways that two dice can show the fives.   There are $^4\mathrm C_2$ ways that could happen.

$$\frac {6}{6^4}$$

3.)Exactly one six would be $\frac{1}{6}*\frac{5^3}{6^3}$

Again, that is for the six in a particular die.   To account for the fact that it could be any of the four, multiply.

$$\frac{4\cdot 5^3}{6^4}$$

4.) Four different numbers would be $\frac{6}{6}*\frac{5}{6}*\frac{4}{6}*\frac{3}{6}$

Indeed. $\frac{^6\mathrm P_4} {6^4}$

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