Dice Numbers Creating Triangles

Potsawee has a fair six-sided die. He throws the die 3 times, and the numbers shown on the upper side each time are a , b , c a,b,c respectively. Find the probability that the lengths ( a , b , c ) (a,b,c) are able to create an isosceles triangle . Give your answer to three significant figures.


The answer is 0.319.

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2 solutions

Jessica Wang
Jul 19, 2015

First of all, there are 6 × 6 × 6 = 216 6\times 6\times 6=216 total values of a , b , c a,b,c .

When the lengths a , b , c a,b,c are able to create an isosceles triangle, we can divide it into two different situations:

[ 1 ] . a = b = c [1]. \; a=b=c : We have 6 6 values.

[ 2 ] . [2]. There are exactly two sides which are of the same length. Without loss of generality, assume a = b c a=b\neq c .

When c = 1 c=1 , a = b = 2 , 3 , 4 , 5 , 6 a=b=2,3,4,5,6 ;

When c = 2 c=2 , a = b = 3 , 4 , 5 , 6 a=b=3,4,5,6 ;

When c = 3 c=3 , a = b = 2 , 4 , 5 , 6 a=b=2,4,5,6 ;

When c = 4 c=4 , a = b = 3 , 5 , 6 a=b=3,5,6 ;

When c = 5 c=5 , a = b = 3 , 4 , 6 a=b=3,4,6 ;

When c = 6 c=6 , a = b = 4 , 5 a=b=4,5 .

We have 21 21 values of a , b , c a,b,c .

Therefore, the probability required is 6 + 3 × 21 216 0.319 \frac{6+3\times 21}{216}\approx \boxed{0.319} .

Nice and simple solution; I'm really bothered I forgot the equilateral triangles.

Tiago Abi-Ramia - 5 years, 10 months ago

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The question doesn't actually say to include them, and not all definitions of isosceles include equilateral triangles. Therefore, you shouldn't be bothered.

Axelrod Polaris - 5 years, 10 months ago

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@Axelrod Polaris Thank you for the reminder. I have added a link to the "isosceles triangle" in the question, which gives a definition.

Jessica Wang - 5 years, 10 months ago

Should not the number of ways of obtaining 2 equal faces with 3 dices be 90?

guido barta - 5 years, 4 months ago
Bill Bell
Aug 3, 2015
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from itertools import product

def isIsosceles(a,b,c):
    a,b,c=sorted([a,b,c])
    if a==b: 
        return a+b>c
    elif b==c: 
        return True
    else:
        return False

overallCount=0
specificCount=0
for a,b,c in product(range(1,7),range(1,7),range(1,7)):
    overallCount+=1
    specificCount+=isIsosceles(a,b,c)
print float(specificCount)/overallCount

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