# Thread: simple probability

1. ## simple probability

Hi everyone,

Im trying to brush up on some maths, and looking at probabilities.

For the life of me I cant work this relatively simple question I set my self up with.

Player 1 has a six sided dice, with each side numbered 1 to 6
Player 2 has a four sided dice, with each side numbered 1 to 4

Each player throws their dice once.

I am trying to work out the probability that player 2 will throw the highest value.

My attempt is 0.33, but looking for some validation.

Thanks

2. There are a total number of 24 possible rolls for the two die, and p2 only throws a higher number 6 times

P1 P2
1 1
1 2$\displaystyle \leftarrow$
1 3$\displaystyle \leftarrow$
1 4$\displaystyle \leftarrow$
2 1
2 2
2 3$\displaystyle \leftarrow$
2 4$\displaystyle \leftarrow$
3 1
3 2
3 3
3 4$\displaystyle \leftarrow$
4
4
4
4
5
5
5
5
6
6
6
6

3. Originally Posted by unigee
Player 1 has a six sided dice, with each side numbered 1 to 6
Player 2 has a four sided dice, with each side numbered 1 to 4
Each player throws their dice once.
I am trying to work out the probability that player 2 will throw the highest value.
The word ‘dice’ is the plural of the word ‘die’.
So each player throws his die.

The probability that player 2 will throw the highest value acts like this.
$\displaystyle P(x=k)$ is the probability that player 2 has k as the highest value.
$\displaystyle P(x=1)=0$
$\displaystyle P(x=2)=\left(\frac{1}{4}\right)\left(\frac{1}{6}\r ight)$
$\displaystyle P(x=3)=\left(\frac{1}{4}\right)\left(\frac{2}{6}\r ight)$
$\displaystyle P(x=4)=\left(\frac{1}{4}\right)\left(\frac{3}{6}\r ight)$
Add them up.