Product of 2 dice

Jesse rolls two fair 6-sided dice. What is the probability that the product of the faces is at least 6?

13 18 \frac{ 13}{18} 1 6 \frac{ 1}{6} 2 3 \frac{ 2}{3} 11 18 \frac{ 11}{18}

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2 solutions

Angela Fajardo
Apr 7, 2015

There are 36 possible outcomes of rolling the two fair 6-sided dice. (because of 6 x 6 or 6^2)

By listing all the possible outcomes:

(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)

Since the problem asks for the probability that the product of the faces is at least 6, you may notice that there are 26 possibilities that show the favored outcome which are:

(1,6)

(2,3) (2,4) (2,5) (2,6)

(3,2) (3,3) (3,4) (3,5) (3,6)

(4,2) (4,3) (4,4) (4,5) (4,6)

(5,2) (5,3) (5,4) (5,5) (5,6)

(6,1) (6,2) (6,3) (6,4) (6,5) (6,6)

By counting them, you are able to get 26 favored outcomes

Because of the formula: Number of favorable outcomes/Total number of possible outcomes;

the probability is 26/36 = 13/18

Yes, that works. Sadly in this problem, there isn't a way to simplify it other than checking every case.

Chung Kevin - 6 years, 2 months ago

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You can at least eliminate most cases by noting than 2 is the only number that can be multiplied by anything other than 1 to get a number less than six. So all viable cases are 1,2,3,4,5 times 1 and 2 times 2; as well as noting that there are 36 total cases. The previously mentioned cases amount to 10 total (2 for each in which the numbers differ and one for each in which they are the same), so 10/36 is the complement of the answer.

Tristan Goodman - 2 months, 2 weeks ago
Brock Brown
Apr 14, 2015

Python 2.7:

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from fractions import Fraction as frac
from itertools import product
count = 0
trials = 0
all_rolls = product(xrange(1,7),repeat=2)
for roll_1, roll_2 in all_rolls:
    if roll_1 * roll_2 >= 6:
        count += 1
    trials += 1
print "Answer:", frac(count,trials)

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