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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.

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    Hopefully Pieces or someone else can edit the question to make that clear.2010-07-21

3 Answers 3

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

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    @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`2010-07-21
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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 :) )

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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