links that might provide some clue

http://www.math.uah.edu/stat/buffon/Triangles.html

www.jstor.org/stable/3617359‎

P = 37/72

There are 216 possibilities and 111 of which will yield a feasible solution.

.

ugly, painful way :

d1, d2, d3 drawn independently from the uniform distribution on {1, 2, 3, 4, 5, 6} (considered as a subset of the integers)

d(1)= d(3)) from those distribution functions !

Direct counting and checking as proposed above is certainly easier and faster !

Except for the case of dices with very very many faces ...

(funny problem : what is the limit behaviour of this "triangulation probability" as the number of faces grows to infinity ?

intuition says that it should be the same as obtained from sampling three times independently from the uniform distribution on [0, 1] but intuition and combinatorics do not always mix ...)

.

The following are some of the feasible triads:

(1, 1, 1),

(1, 2, 2),

(1, 3, 3),

(1, 4, 4),

(1, 5, 5),

(1, 6, 6),

(2, 1, 2),

(2, 2, 1), (2, 2, 2), (2, 2, 3),

(2, 3, 2), (2, 3, 3), (2, 3, 4),

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

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

(2, 6, 5), (2, 6, 6),

(3, 1, 3),

(3, 2, 2), (3, 2, 3), (3, 2, 4),

...... ........ .......

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

There are 111 such triads.

Dear Fabrice

Nice post !!. Thanks for the reference.

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Dear Hemanta,

It is much easier to divide the cases into two groups and to take advantage of the symmetry of the problem. Indeed, feasible solutions do not depend on the ordering.

1) First group

This group deals with triangles having (at least) two equal sides. The following table gives the corresponding feasible solutions.

111

221 222 223

331 332 333 334 335

441 442 443 444 445 446

551 552 553 554 555 556

661 662 663 664 665 666

The six diagonal elements are unique and represent equilateral triangles. On the other hand, to each of the 21 off-diagonal elements will correspond two other valid solutions obtained by simple permutation. For example, to the case 331, we deduce 133 and 313.

Therefore, the total number of valid solution for this first family is

6x1 + 21x3 = 69

2) Second group

This group deals with triangles with all sides having different length. It is much easier to handle this group by limiting the search to the cases where the sides a, b and c are ordered : a < b < c. Then, we subdivide this group into sub-groups or classes.

For example, consider the class (123) which deals with the case where the sides are three successive integers. In this class, we have three valid solutions 234, 345, and 456. The following listing gives all the valid solutions within each class (shown in parenthesis).

(123) = 234 345 456

(124) = 346

(125) = /

(126) = /

(134) = 245 356

(135) = /

(136) = /

(145) = 256

(146) = /

(156) = /

For each of these 7 valid solutions will correspond 5 others obtained by permutation. For example, from the solution 234 we deduce 243,324,342,423,432.

Therefore, the total number of valid solution for this second family is

7x6 = 42

Hence, the overall total number is 69 + 42 = 111

Thank you all very much. The Ugly method is interesting, but I did not see the entire solution. Is it too ugly?

Of course counting is the simplest. I was wondering, however, if one could find a formulation avoiding tedious counting process!

Dear Lehtihet,

Three days ago, I happened to have found that the question was lying unanswered for nearly a month! Just when I enumerated the configurations, not in your way, but in my crude method, I have seen that your short answer was posted. So without going into grouping as you have done, I just wrote my answer!

Thanks Hemanta,

In fact, I did not notice this thread until you have up-voted this question.

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Dear Aliakhbar

Actually, you don't need to count one by one the number of valid instances.

The probability you are looking for is given by P = N/(n^3)

Here, N is the number of valid triangles whereas n is the number of faces in your dice (n = 6 for a standard dice). The question is how to compute N for a given n.

Answer N = (1/2)(n^3+n) so that P = 1/2 + 1/(2n²)

For example, if your dice has n = 20 faces, then the probability would be

P = 1/2 + 1/800

.

regarding the ugly way, i just outlined the solution and i am way too lazy to work out the complete calculation ! it is just "easy but tedious" as most publications say ...

Halim's solution above is far more elegant !

.

.

Dear Fabrice

Thanks for your very kind words -:).

You are right, the calculation is not difficult but can be extremely tedious and messy depending on how we handle it.

If we follow the example I gave previously for the case of a die with n = 6 faces then we can take advantage of the symmetry and the ordering. By doing so, we simplify somewhat the problem and we get as well some more detailed information.

Let N be the total number of valid triangles. Based on the previous example, it is interesting to write N as follows:

N = A + 3B + 6C where

A = the number of equilateral triangles (A = n).

3B = number of valid Isosceles triangles.

6C = number of valid triangles with all sides distinct in length.

The numbers 3 and 6 account for the permutations as shown in my above post.

Of course A = n. On the other hand, expressions for B and C will depend on the parity of n.

If n is even then

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

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

if n is odd then

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

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

Of course, the expressions for C are valid only if n > 3. Otherwise take C = 0.

We may then calculate easily the probability of obtaining, say, a isosceles triangle

P(isosceles) = 3B/(n^3)

It is interesting to note that 3B + 6C is always equal to n(n² - 1)/2 whether n is even or odd. Hence the final result

A + 3B + 6C = N = (n/2)(n² + 1)

Dear Lehtihet,

I have always been of the view that mathematics is the best kind of poetry if you can find the rhythm in it!

This solution of yours is poetry! Equilateral, isosceles and scalene, all considered separately, and then the number of possibilities added together leads to this solution. Beautiful!