What is the probability of constructing a random triangle after the third rolling of the die?

It seems this is a duplicated question. So I repeat here the answer I gave earlier in a similar thread.

There is actually no need to compute one by one each valid instance. The probability you are looking for is

P = 1/2 + 1/(2n²) where n is the number of faces of the die.

For a standard die (n = 6), we have P = 1/2 + 1/72 = 37/72.

Some details might be worth mentioning. let N be the total number of valid instances. We can rewrite N = E + 3I + 6S where

E = Total number of equilateral triangles

3I = Total number of Isosceles

6S = Total number of scalene

The factors 3 and 6 take into account permutations.

Of course, E = n but S and I will depend on the parity of n

If n is even then

I = (3n - 4)(n / 4)

S = n(2n - 5)(n - 2)/24

if n is odd then

I = (3n - 1)(n - 1)/4

S = (2n - 1)(n - 1)(n - 3)/24

The above expressions for S are valid only if n > 3. Otherwise take S = 0.

It turns out that 3I + 6S = n(n² - 1)/2 whether n is even or odd. Hence the final result

E + 3I + 6S = N = (n/2)(n² + 1)

and hence the probability P = N/(n^3) = 1/2 + 1/(2n²)

In summary, for a standard die (n = 6) we have

P = 37/72

P(Equilateral) = 1/36

P(isosceles) = 7/24

P(scalene) = 7/36

Kosto V. Mitov

The probability is 111/216.

Aliakbar Haghighi

Thanks Kosto. I have a different answer. Could you show me how you obtained yours?

I have just count the cases which satisfy the triangle inequalities. I counted them by EXCEL.

Thanks again. I did the same rather than formulated. However, I also used the condition that no side's measure should exceed the sum of the other two. Based on this assumption, I had 120 choices. Perhaps, I have miscounted, do you agree?

By the way what formula did you use for Excel?

Hiqmet Kamberaj

I also found 111/216, using the rule that each side must be less than the sum of other two.

Thanks very much Professor Kamberaj. Yes it is the same answer Professor Mitov submitted a few days ago. I aapreciate your time.

Thanks very much Lehtihet for an elaborated answer.

Dear Aliakbar, you are most welcome.

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