Dice rolling problem

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January 12th 2010, 07:52 PM
Zhankfor

Dice rolling problem

Hi

A friend and I have been trying to figure out this seemingly simple problem for a while and couldn't find any help with it anywhere online. The problem is as follows:

What are the odds of rolling a seven at least once (that is, once, twice OR three times) in three rolls of two dice?

The closet we could come up with is (1/6)+(1/12)+(1/18) which gives the logical looking number of 30.5%, but (we think) that assumes you stop rolling if you roll a seven. At this point I can't even tell if that matters.

(If you're wondering, the impetus for this question is the boardgame Settlers of Catan - I wanted to know, with three people playing, what the probability is of a 7 being rolled between the end of your turn and your next turn, i.e. 3 rolls)

Thanks a lot.
January 12th 2010, 09:49 PM
Soroban

Hello, Zhankfor!

Quote:

What is the probabiity of rolling a "7" at least once in three rolls of two dice?

For each roll, there are $6^2\,=\,36$ outcomes.
Six of them result in a "7": . $P(7) \:=\:\frac{6}{36} \:=\:\frac{1}{7}$
Hence, the probability of not getting a 7 is: . $P(\text{Other}) \:=\:\frac{5}{6}$

The opposite of "at least one 7" is "no 7's" ... that is, "3 Others"

. . $P(\text{3 Others}) \:=\:\frac{5}{6}\cdot\frac{5}{6}\cdot\frac{5}{6} \:=\:\frac{125}{216}$

Therefore: . $P(\text{at least one 7}) \;=\;1 - \frac{125}{216} \;=\;\frac{91}{216} \;\approx\;42\%$
January 13th 2010, 05:55 AM
Zhankfor

Thanks a lot! I had at one point gotten to that basic idea (subtract the Others from all possibilities) but I couldn't figure out how to translate that into numbers.