# Dice Probability

Printable View

• November 4th 2008, 12:13 PM
ndcruz
Dice Probability
• November 4th 2008, 12:26 PM
joksy
The propability is not my strength but, I remember that we have 1/36 to have 2 and 1/36 to have 12 as overall outcome.

so for not 2 could be $1 - \frac{1}{36}$

for 4 $\frac{3}{36}$

and for 6 $\frac{5}{36}$

=> think to 2 you have 1/6 for the first dice and 1/6 for the second so 1/36

Hope that it is useful (Sleepy)
• November 4th 2008, 12:32 PM
Soroban
Hello, ndcruz!

It's just matter of counting, isn't it?

There are 36 possible outcomes.

There are: . $1 + 3 + 5 \:=\:9$ ways to get a sum of 2, 4 or 6.

So there are: . $36 - 9 \:=\:27$ ways for some other sum.

Therefore: . $P(\text{sum not 2, 4, or 6}) \;=\;\frac{27}{36} \;=\;\frac{3}{4}$