# Probability Sum of Red and Blue Dice

Two ordinary fair dice are rolled (one blue, one green). Consider the random variables $X=(S,p_x)$ and $Y= (T,p_Y)$ where $S$={a,b} and $T$={c,d}.

The outcome $a$ denotes that the blue dice showed 5 or 6, and outcome $b$ denotes that the blue dice showed 1, 2, 3, or 4. Outcome $c$ denotes that the sum of the faces is 10 or greater, and outcome $d$ denotes that the sum of faces is 9 or less.

I'm just looking to calculate the probabilities for the individual outcomes right now.

Outcome A I calculate that its a 1/3 chance
Outcome B I calculate that it is a 2/3 chance

I am a bit unsure on C, D

Outcome C Heres my thinking for sums that are 10 or greater theirs 3 different possibilities

Blue Die + Green Die
5 + 5
5 + 6
6 + 6

And 2*3 is 6 so there is 6 possibilities for sums greater than 10.

Outcome D
I feel like theirs a better way to approach this part besides writing out all the sums individually..

Blue Die + Green Die
6 + 3
6 + 2
6 + 1
5 + 3
5 + 2
5 + 1
.
.
.
1 + 3
1 + 2
1 + 1

I get something like 24 different ways that the sum is less than 9 but thats not right.. I could really appreciate some help.

• There are indeed six possibilities for the sum to be 10 or greater, but not for the reasoning you give. Looking via brute force is fine for this particular problem since the numbers involved are so small. The possible outcomes leading to a sum of six or more are $(4,6),(5,5),(5,6),(6,4),(6,5),(6,6)$ where the left number represents the blue die and the right the green. It looked as though you were multiplying your earlier answer of 3 by two to account for the order, but notice that rolling a blue 6 and a green 6 is the same as rolling a green 6 and a blue 6. – JMoravitz Oct 13 '17 at 19:25
• For the probability, notice that every $(x,y)$ pair of results is equally likely to occur. How many ways can you roll two fair dice with the dice different colors? Think rule of product. To find the probability then, you take the ratio of the number of good possibilities divided by the total number of possibilities (disclaimer: this only works if each possibility is equally likely to occur). As for the last part of the problem, notice that either the sum is 10 or more or it is 9 or less, so their probabilities should add up to $1$. – JMoravitz Oct 13 '17 at 19:28
• I'm confused how i'd find how many possibilities there are for the sum of less than 9, I can't brute force it and I'd rather not brute force it. I understand that there are 36 possibilities in total since 6 * 6 = 36 – Temirzhan Oct 13 '17 at 19:31
• If there are 6 that don't add up to less than 9, and there are 36 all together, then.... since they should add up to 36..... – JMoravitz Oct 13 '17 at 19:32
• So 30? Alright lets say hypothetically my dice is 24 sided. and I wanted to find the probability of getting a sum greater then equal to 36 and less than equal to 35. How could I compute that without having to bruteforce the sums that are greater than 36. Computationallly bruteforcing becomes a hassle is their a smarter way to do that? – Temirzhan Oct 13 '17 at 19:35

Outcome $C$ can be obatined in $6$ ways $(4,6),(5,5),(6,4),(5,6),(6,5),(6,6)$ out of the $36$ possibilities.
There are $30$ possibilites for event $D$ and you will need to systematically count them ... the diagram above might help.