# 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

• Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or closed. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers. Commented Jun 16, 2021 at 8:22

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.
• $$P(0) = \frac{\binom{8}{0} \times 5^{(8-0)}}{6^8} = \left(\frac{5}{6}\right)^8 > 0.$$

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

$$6\times\frac{1}{6}\times\left(\frac{5}{6}\right)^5$$

• Use \left( and \right) instead of ( and ) wherever the size of brackets is to be adjusted. For example, $\left(\dfrac56\right)$ vs $(\dfrac56)$. Commented Jun 16, 2021 at 8:32
• @ultralegend5385: yes, of course....I was in a hurry. Thanks for your comment. Commented Jun 16, 2021 at 8:39

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 $$P(A) \approx 0.7674$$.