# dice odds of a specific number rolling at least once

I'm trying to figure out the odds of rolling a 1 (or any other number) at least once with 3 dice (or any other amount).

I was pretty sure that for any specific outcome, you treat each die roll as an independent event and multiply the probabilities, and I saw some other answers on this site that confirm this. So the odds of getting exactly one 1 would be:

(1) (not 1) (not 1)

(1/6) (5/6) (5/6)

11.57%

And then you add the probabilities of the different outcomes (exactly one 1, two 1s, three 1s), which, using the method above, give me:

11.57%

2.31%

0.46%

But 14.34% is a smaller chance than of rolling a 1 with one die (16.67%). How do my odds of rolling a 1 decrease with more dice? What am I doing wrong?

• Your not taking combinations into account. One way to roll a single 1 is to roll 1-2-3, yet another is to roll 2-1-3. Your calculation $\frac{1}{6}\frac{5}{6}\frac{5}{6}$ needs to be multiplied by the number of combinations. Jan 29, 2016 at 20:48

The easiest way to figure out odd for events with the words "at least" is often to figure out the contrary. So here the contrary is : "you have no 1 on any of your 3 dice" and this probality is easy to figure out. It is : $\left(\frac{5}{6}\right)^3=\frac{125}{216}$

So the probality of your event is : $1-\frac{125}{216}=\frac{91}{216}$

The issue you are having is that (1)(not 1)(not 1) is calculating the probability that the first dice is a 1, and the second and third dices are not. This is not the only way to get just one 1, it could either be 1st, 2nd or 3rd and hence we have three ways and we multiply by 3, i.e P(one 1)=3*(1/6)(5/6)(5/6)=75/216.

Similiarly, with two 1's, you've calculated the probability that the first and second are 1 and the third is not (1)(1)(not 1). There are three ways that one of the dice aren't 1 and so we multiply by 3 to get 3*(1/6)(1/6)(5/6)=15/216.

The probability that all are 1 is .46% or 1/216 and when we add them altogether we have 91/216 or 42.129%.

However, the easier way to calculate this question is to ask the following, the chance of me getting at least one 1, is the same as the chance of me no 1's. The chance of getting no 1's is (5/6)(5/6)(5/6)= 125/216 and so the chance of you getting no 1's is 1-(125/216)=91/216=42.129%.

If you are wondering how to get the 3's when the question is more complicated, they are the coefficients in the relevant line of the binomial expansion, each to the right of the the minimal value you require (in your case you wanted at least one, the relevant line is 1 3 3 1 and so you count from the second giving 3,3,1) This is because the three rolls are Bernoulli trials, making this experient a binomial one.

Exactly one 1 is not equivalent to the specific result (1,0,0) as you're calculating it. Because exactly 1 also includes (0,1,0) and (0,0,1). Similarly for two 1's. So you should multiply both of those probabilities with $3$. Note that your distribution here is the binomial distribution and you're missing the binomial part in its definition.

Let $X$ be the number of 1's rolled.

Your initial calculation $P(X = 1) = \frac{1}{6} \cdot \frac{5}{6} \cdot \frac{5}{6} = \frac{25}{216}$ gives you the probability of rolling exactly one 1 on the first die; you need to multiply this by 3 to account for possibly rolling a 1 on the second die or third die. While we can apply a similar procedure to accurately calculate $P(X = 2)$, then calculate $P(X \geq 1) = P(X = 1) + P(X = 2) + P(X = 3)$, it's easier to proceed by realizing that $P(X \geq 1) = 1 - P(X = 0)$; we just need to calculate the odds of rolling no 1's, then subtract that from 1.

\begin{align} P(X = 0) &= (\frac{5}{6})^{3} = \frac{125}{216} \\ \\ P(X \geq 1) &= 1 - \frac{125}{216} = \frac{91}{216} \\ \\ &\approx 42.1\% \end{align}