# Probability of a specific event when rolling a 6 sided die five times

I am stuck on computing probability problem that goes as follows:

I am rolling a 6 sided die 5 times in total. What is the probability that I roll at least one 3, but no 1 or 2 at all (= probability of rolling at least 3 but at least one dice has to be 3).

Examples that satisfy my condition:

• 35456
• 33454
• 45366
• ...

Examples that do not satisfy my condition:

• 45445 (satisfy at least 3, but does not satisfy at least one dice has to be 3)
• 32653 (does not satisfy at least 3, satisfy one dice has to be 3)
• ...

The part of rolling at least one 3, as far as I know, should be opposite event to no 3 at all: $$1 - \left(\frac56\right)^5 = 0.5981.$$

I am stuck on the second part.

As suggested by @Dougal, I post my answer here:

The probability of obtaining at least one 3 and all values bigger than 2 in a six sided dice is equal to the probability of obtaining all values over 2 except the combinations in which all values are bigger than 3. $$P(5\text{ rolls}>2)-P(5\text{ rolls}>3)=\left(\frac{4}{6}\right)^5-\left(\frac{3}{6}\right)^5\approx0.1004$$

The problem can also be solved as the probability of obtaining all values over 2 and getting at least a 3 in those 5 rolls. This can be expressed as: $$P(5\text{ rolls}>2)\cdot P(\text{at least a }3|5\text{ rolls}>2)= \left(\frac{4}{6}\right)^5\cdot\sum_{i=1}^{5}\binom{5}{i}\left(\frac{1}{4}\right)^i\cdot\left(\frac{3}{4}\right)^{5-i}$$

I believe the first solution is easier to understand, but, both provide the same result.

• I included your suggestion in the response. I also updated the ending part because fabiob's answer was erased. May 14, 2018 at 17:40

Let,

$\\ A = Event \space all \space dice \space > \space 2 \\ B = Event \space a \space 3 \space is \space rolled\\$

We are interested in computing $P(A \cap B) = P(A) \space P(B \space | \space A)$

$P(A) = (\frac{2}{3})^{5} = 0.1316872$

$P(B \space | \space A) = 1 - (\frac{3}{4})^{5} = 0.7626953$

Then,

$P(A \cap B) = P(A) \space P(B \space | \space A) = (0.1316872) \space (0.7626953) = 0.1004372$

We can simulate this with R to convince ourselves of the result:

> set.seed(0x12345678)
> hits <- 0
> runs <- 1000000
> for (i in 1:runs) {
+   s <- sample(1:6, 5, replace=TRUE)
+   if ( 3 %in% s & !(2 %in% s) & !(1 %in% s)) {
+     hits <- hits + 1
+   }
+ }
> hits / runs
 0.10021

• This is the cleanest approach. Why not calculate $P(A) \space P(B \space | \space A)$ as fraction, not decimal?
– smci
May 14, 2018 at 2:48
• @smci: I wouldn't at a glance know how big $32/243$, $781/1024$ or $781/7776$ are, but I know how big $0.1004372$ is. May 14, 2018 at 9:42
• @PhilipC: me neither, but my point is some factors of your numerators and denominators from $P(A) P(B | A)$ cancel. I wouldn't have converted the individual terms to decimals before multiplying them.
– smci
May 14, 2018 at 9:43
• $P(A) P(B | A)$ = (32/243) * (781/1024) = 781/(32*243) = 0.1004372
– smci
May 14, 2018 at 9:49