Confusion understanding probability I am a beginner in this forum - please don't judge me too harsh.
I understand that my question is noobie but I read a lot and couldn't understand the concept of summing probabilities.
Here is the problem I cannot understand:
We have a dice with 6 possible outcomes 1...6.
Rolling the dice once, the chance to hit 3 is 1/6
What is the chance to hit 3 if I throw the dice 6 times or 8 times?
Simply summing the probabilities doesn't make sense to me.
I mean 1/6 + 1/6 + 1/6 + 1/6 + 1/6 + 1/6 + 1/6 + 1/6  = 8/6 is greater than 1. How come probability gets greater than 1?
My reasoning must be wrong
 A: You are right that each throw of a die has a $(1/6)$ chance of rolling a $(3)$.  Therefore, your addition is correct that if you roll the die $8$ times, the expected number of times that you will roll a $(3)$ is
$$8 \times (1/6) = (8/6) = (4/3) > 1.$$
Your intuition is also right, that if you roll the die $8$ times, the chance of at least one of the rolls coming up $(3)$ must be less than $(1)$.
This begs the question: if the distribution of $8$ rolls is such that you expect $(4/3)$ of the rolls to be a $3$, how can the chance of not rolling any $(3)$ in 8 rolls, still be positive.
It is because there is a (small) chance, that in $8$ rolls, you may have a $3$ appear two or more times.  It is the possibility of a $3$ appearing two or more times that balances the fact that there is still a positive chance that there will be no $(3)$'s rolled.
However, this is all intuitive hand waving, which doesn't mean much, without math to back it up.
Suppose that you roll a die $8$ times.  There are $(6)$ possibilities for each roll.  Therefore, the total number of possible sequences of $8$ rolls is $6^8$.
For $k \in \{0,1,2,\cdots,8\}$, a natural question is: how many of the $6^8$ sequences will result in exactly $k$ of the rolls coming up $(3)$.
There are $\frac{8!}{k!(8-k)!} = \binom{8}{k}$ ways of selecting $k$ rolls out of $8$, so that those rolls (and only those rolls) come up $3$.
Once the $k$ rolls are selected, you then have $(8-k)$ rolls whose only constraint is that the roll is any number other than a $(3)$.
Therefore, there are exactly $\left[\binom{8}{k} \times 5^{(8-k)}\right]$ sequences of $8$ rolls, in which $3$ comes up exactly $k$ times.
Therefore, the probability of this happening is
$$P(k) = \frac{\binom{8}{k} \times 5^{(8-k)}}{6^8}.$$
You will find that :

*

*$\sum_{k = 0}^8 [k \times P(k)] = (8/6) = (4/3)$, as expected.

*$P(0) = \frac{\binom{8}{0} \times 5^{(8-0)}}{6^8} = \left(\frac{5}{6}\right)^8 > 0.$
A: If you want to have the probability of hitting ONLY once 3, a favoruable event is this
$$\{3,\overline{3},\overline{3},\overline{3},\overline{3},\overline{3}\}$$
where with $\overline{3}$ I mean "not 3"
This probability is $\frac{1}{6}\times\left(\frac{5}{6}\right)^5$ because all this events must happen together.
Now you can understand that the "3" can be hit in any of the six positions, thus this probability must be multiplied by 6
Concluding, the probability to hit exactly one 3 in 6 dice's rolls is
$$6\times\frac{1}{6}\times\left(\frac{5}{6}\right)^5$$
A: If you want the probability that you roll $3$ at least once in $n$ attempts, then you can calculate it as follows.
Let $A$ be the event that you roll $3$ at least once in $n$ attempts.
Then the complement, $A^c$, is the event that you do not roll $3$ on any of the attempts. The probability of $A^c$ is $(5/6)^n$.
Therefore, the probability of $A$ is $P(A) = 1 - P(A^c) = 1 - (5/6)^n$.
For example, if $n = 6$ then $P(A) \approx 0.6651$. If $n=8$ then $P(A) \approx 0.7674$.
