# 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.

• Would you reword this question? How are you going to get a 1 with two dice? Do you need the die to sum to quantities between 2 and 5, inclusive? Jul 21 '10 at 13:20
• It's nice to see three different approaches. Jul 21 '10 at 14:15
• @Tom it's not a sum of 1 or 5, it's a 1 or a 5 showing on either die. The game 'farkle' cares about the values on individual dice, not the sum of values showing in most cases. Jul 21 '10 at 16:37
• Hopefully Pieces or someone else can edit the question to make that clear. Jul 21 '10 at 16:51

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 $1-(2/3)^2=5/9$.

• it is usually a good idea working backwards in basic probability problem, I agree :-)
– mau
Jul 21 '10 at 16:19
• What does the dollar sign stand for? Also why is the probability of not getting a 1 or a 5 (4/6)^2? Jul 21 '10 at 18:45
• @Pieces: the dollar sign just delimits mathematical expressions. It's a syntax that comes from Tex/LaTeX. Jul 21 '10 at 18:55
• @Pieces: if you want the TeX formatting to light up, I have created a GreaseMonkey script: userscripts.org/scripts/show/81977 Jul 21 '10 at 19:02
• @Pieces Daniel is right, that's to delimit LaTeX expressions, for the case that LaTeX will be supported later on (which is not improbable if the site makes it). Re your other question: The probability of not getting a 1 or 5 on one die is the probability to get a 2, 3, 4 or 6, that's 4 out of 6. If you want that on two dice, the probability is (4/6)*(4/6)=(4/6)^2 Jul 21 '10 at 19:02

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 :) )

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

$$1-\bigg(\frac{2}{3}\bigg)^2=\frac{5}{9}$$

Because $$\big(\frac{4}{6}\big)^2$$ is the probability that neither of the dice show a $$1$$ or $$5$$.