Probability of getting at most 4 and at least one 4 in 5 consecutive die rolls Let's suppose we have one die. We roll it 5 consecutive times. How can we calculate the probability of getting at least one 4 and at most 4 in all the 5 rolls of the die. Let me explain the case with some example successful and failed cases
1 3 4 4 4 -> success
4 1 1 2 1 -> success
4 4 4 4 4 -> success
1 2 3 4 1 -> success
1 2 3 4 5 -> fail
5 6 5 6 4 -> fail
1 2 3 2 3 -> fail
I am not sure if I have been able to ask and explain the question properly. If further clarification is needed please do not hesitate to ask.
Thanks,
 A: Hints:  


*

*How many equally probably ways are there of rolling five (or ten) dice?

*How many equally probably ways are there of rolling five (or ten) dice so that each is no more than $4$?

*How many equally probably ways are there of rolling five (or ten) dice so that each is no more than $3$?

*How many equally probably ways are there of rolling five (or ten) dice so that largest value shown is exactly $4$?
A: Paraphrasing, you want to roll $10$ six-sided dice, and get at least one die showing $4$, and no die show either $5$ or $6$
There are $6^{10}$ distinct sequences of roll without restriction.
How many sequences are there that don't contain $5$ and $6$?
How many sequences are there that don't contain $4$, $5$ and $6$ 
Subtract, divide...
A: Conditional probability:
$$P(A|B) = \frac{P(A \land B)} {P(B)}$$


*

*A is probability of rolling at least one 4

*B is probability of not rolling a 5 or 6


P(B) is simple, $(4/6)^{10}$
P(A|B) means "the probability of A, assuming B is true".  The probability of rolling at least one four, given that you are only rolling values one through four, is the same as if you were rolling a four sided dice:  $1 - (1 - 1/4)^{10}$
$$\begin{align}P(A \land B)  &= P(A|B)P(B)\\
&= (1 - (1 - 1/4)^{10})(4/6)^{10}\\
&= \frac{989527}{60466176}\end{align}$$
