# Probability of getting specific (one) number when 5 dice rolling at the same time?

Firstly, as a beginner I searched a lot of questions here and found out many useful information about probability of dice rolling. There are a lot of questions about 2 or 3 dice and many about 5 dice rolling but I didn't managed to get answers from there.

• QUESTION:

The problem involves rolling 5 dice (all are fair and 6 sided) at the same time.
What is probability of getting at least one number (for example number 3) in each 5 dice rolling?
This means that there are more combinations when this condition is fulfilled:

• one die with number 3, other four dice other numbers
• two dice with number 3, other three dice other numbers
• three dice with number 3, other two dice other numbers
• four dice with number 3, one die other numbers
• all five dice with number 3

I don't know when to use addition theorem of probability or multiplication theorem of probability.

I would like to know the method of calculation and the final result because I have already tried to calculate the probability but with no success (the professor only told me it's incorrect). I won't post it here because it's too long.

NOTE: This is not homework or task for school. In class we only worked simple scenario with 2 dice, I want to learn more about this and I went to consultations with professor and he didn't explained me the scenario with 5 dice)

• may be it's easier to compute the probability of not having a 3.
– wece
Nov 25, 2017 at 13:36
• If you meet terminology "at least" or "at most" then in many cases it is more easy to find the probability of the complement. Nov 25, 2017 at 14:06
• Compare math.stackexchange.com/questions/1632468/… math.stackexchange.com/questions/1701680/… and math.stackexchange.com/questions/115209/… -- there's the question for 3, 4, or 6 dice; you may be able to generalize form those questions and the answer below. Nov 25, 2017 at 14:41
• @DavidK thanks a lot Nov 25, 2017 at 15:16

When doing these kinds of computations in probability, it's often easier to consider the opposite event of what you're actually interested in. Here, the event you care about is "at least one die is a 6" (for example), so the opposite event is "no die is a 6." This means that die 1 is not a 6, and die 2 is not a 6, and dice 3, 4, 5 are all not a 6. The key in this computation is the word "and," because the independence of the dice means we can multiply those individual probabilities. The chance that no die is a 6 is therefore $$\frac 5 6 \cdot \frac 5 6 \cdot \frac 5 6 \cdot \frac 5 6 \cdot \frac 5 6 = \frac{3125}{7776}$$ and the probability of the event you want is thus $1 - \frac{3125}{7776} = \frac{4651}{7776} \approx 0.5981$.

To answer your question a bit more comprehensively, you asked about when to use the addition rule and when to use the multiplication rule. At a basic level, multiplication corresponds to the word "and," and addition corresponds to the word "or." However, there are caveats; multiplication corresponds to "and" when events are independent, and addition corresponds to "or" when events are disjoint. Recognizing when these things happen is one of the main challenges of learning probability theory and comes only with a fair bit of experience in the subject, as far as I can tell.

In this case, the reason I knew to consider the complementary event was that the main event you wanted to consider had to be expressed in terms of several "or" events that were relatively complicated (i.e. "exactly 1 die is a 6," "exactly 2 dice are a 6," etc.) but the opposite event was not complicated in that way.

We have to find the probability of the occurrence of a specific number. Favourable cases are (1,1,1,1,1), (2,2,2,2,2), (3,3,3,3,3), (4,4,4,4,4), (5,5,5,5,5) and (6,6,6,6,6). So number of favourable cases is 6

Now, total number of cases is 6^5 = 216*36 = 7776

Probability of getting specific number when 5 dice are rolling at the same time = 6/7776= 1/1296 = 0.00077

Hope it helps :)