Calculating the probability of two dice getting at least a $1$ or a $5$ So you have $2$ dice and you want to get at least a $1$ or a $5$ (on the dice not added).  How do you go about calculating the answer for this question.
This question comes from the game farkle.
 A: The other way to visualise this would be to draw a probability tree like so:
alt text http://img.skitch.com/20100721-xwruwx7qnntx1pjmkjq8gxpifs.gif
(apologies for my poor standard of drawing :) )
A: To visually see the answer given by balpha above, you could write out the entire set of dice rolls
[1, 1], [1, 2], [1, 3], [1, 4], [1, 5], [1, 6]
[2, 1], [2, 2], [2, 3], [2, 4], [2, 5], [2, 6]
[3, 1], [3, 2], [3, 3], [3, 4], [3, 5], [3, 6]
[4, 1], [4, 2], [4, 3], [4, 4], [4, 5], [4, 6]
[5, 1], [5, 2], [5, 3], [5, 4], [5, 5], [5, 6]
[6, 1], [6, 2], [6, 3], [6, 4], [6, 5], [6, 6]

Total number of possible dice rolls: 36
Dice rolls that contain 1 or a 5: 20
20/36 = 5/9
A: Go backwards: Calculate the probability that neither of them shows a 1 or 5. That means both show a 2, 3, 4, or 6. Thats $(4/6)^2$.
Hence the probability that at least one shows a 1 or 5 is $1-(2/3)^2=5/9$.
A: $$1-\bigg(\frac{2}{3}\bigg)^2=\frac{5}{9}$$
Because $\big(\frac{4}{6}\big)^2$ is the probability that neither of the dice show a $1$ or $5$.
