# what is the probability of rolling a 5 sided die 5 times and getting a 1 any number of times

I am so confused; wouldn't it just be 1/5(5)? The probability of rolling a six-sided dice five times and getting a 1 at least one time is 5/6, but how would this work for a five-sided dice?

• Compute the probability of not getting a $1$ in the $5$ rolls, and then compute $1-$ that probability.\ Commented Feb 27, 2022 at 17:49
• What does 1/5(5) mean? Actually hard to think of a more confusing way to write something so short...
– lulu
Commented Feb 27, 2022 at 17:50
• It appears the probability that you are computing is the probability of rolling any one five different outcomes. i.e. $(1/5)*5$.Of course this should come out to $1$ as this is a guarantee on a five-sided die with five different outcomes Commented Feb 27, 2022 at 17:52
• Since no one else has pointed it out, the probability of rolling a six-sided die five times and getting a 1 at least one time is not 5/6. You're starting with a false premise, so that's why you're finding it difficult to generalize to a different case. Commented Feb 28, 2022 at 2:06

Suppose the die has $$S$$ sides and is thrown $$n$$ times. At any given throw, the chance of getting a 1 is $$\frac{1}{S}$$ and the chance of getting something else than 1 is $$\frac{S-1}{S} = 1 - 1/S$$. Let $$E$$ denote the event where at least a 1 is thrown after $$n$$ throws. The complement of $$E$$ is the event where every single throw is not a 1. Hence

$$P(E) = 1- P(E^c) = 1 - (1-1/S)^n$$

Setting $$S = 5$$ and $$n=5$$, we get $$P(E) = 1 - (1-1/5)^5 = 1 - (4/5)^5$$

Your dice has $$5$$ possible outcomes, each with a probability of $$\frac{1}{5}$$. So on each throw $$P(1) = \frac{1}{5}$$. The probability of rolling the dice $$5$$ times and getting $$1$$ at least one time can be easily calculated if you notice that the probabilities fit a binomial distribution.

$$P(x = 0, n=5, p=0.2) = \binom{5}{0}0.2^0(1-0.2)^5 = 0.8^5 = 0.32768$$

This means the probability of not getting any $$1$$ is $$0.327$$. Therefore the probability of getting $$1$$ at least one time is $$1-0.327 = 0.67232 \approx 67.2\%$$

• A very approachable explanation of the binomial distribution and its uses in probability can be found here: istat924158936.wordpress.com/2022/02/16/… If you are looking for something less beginner friendly and more technical, just check the Wikipedia page. Commented Feb 27, 2022 at 18:12

The probability of rolling a six-sided dice five times and getting a 1 at least one time is 5/6

Okay, so the probability of rolling a six-sided die seven times and getting a 1 at least one time is $$7/6$$?

But probabilities can't be larger than $$1$$.

You cannot add probabilities unless events are disjoint (i.e. can't both happen at once). Here, it can happen that you get a 1 on more than one roll. Your calculation is double counting sequences with more than one $$1$$ in them.

In the simple case where we roll the six-sided die twice, write out each possible pair of rolls explicitly. There are $$36$$ options. $$10$$ include exactly one 1. There is $$1$$ option where both rolls land on 1. So the probability of getting a $$1$$ at least once is $$11/36\neq 2/6$$.

This generalises to the inclusion-exclusion principle.