Probability of three dice throws in strictly increasing order

Question

I have a math problem which states the following:

"Roll a 6-sided die. Then roll a 10-sided die. Then roll a 20-sided die.

Will the results be in strictly increasing order?

Example: [1,2,3] is true where [1,3,3] is false."

My assignment is to assign a probability to the statement being true.

I have figured this problem has $n_{\textrm{permutations}} = 6 \cdot 10 \cdot 20 = 1200$ permutations, but I don't know an elegant way to calculate the number of these permutations that are in strictly increasing order. I guess it is some kind of combinatorics problem with a k-word of length 3 etc. but have not managed to get across the finish line.

Appreciate all help.

Answer 1

Apply total probability theorem (1s level branches for the 6-sided die, then 2nd level branches for the 10-sided die, and 3rd level branches for the 20-sided die, all with same respective probabilities 1/6, 1/10 and 1/20):

$\sum_{i=1}^6 \frac{1}{6} \left( \sum_{j=i+1}^{10} \frac{1}{10} \left(\sum_{k=j+1}^{20} \frac{1}{20} \right) \right) =$

$\frac{1}{6 \cdot 10 \cdot 20} \sum_{i=1}^6 \left( \sum_{j=i+1}^{10} \left( \sum_{k=j+1}^{20} 1 \right)\right) =$

It takes time to do the math, but if you use the computer, then you get 506 over 1200.

The maths imply a sum of arithmetic sequence within the $k$ summands (from $j+1$ to 20), then a sum of squares of numbers in the $j$ summands, and a sum of cubes of numbers in the $i$ summands.

A little bit of it:

• $\sum_{k=j+1}^{20} 1 = 20-j$.
• $\sum_{j=i+1}^{10} \left( \sum_{k=j+1}^{20} 1 \right) = \sum_{j=i+1}^{10} (20 - j) = 20(10-i) - \sum_{j=i+1}^{10} j = 20(10-i) - \frac{(11+i)(10-i)}{2}$

Answer 2

It is quite easy to count the number of strictly increasing outcomes. Just split them into four cases:

Case I: All three are in the range $1-6$
Case II: The first two are in the range $1-6$ and the third is in the range $7-20$
Case III: The second and third are in the range $7-10$
Case IV: The second is in the range $7-10$ and the third is in the range $11-20$

For instance, Case II contains $\frac12(6\times 5) \times 14$ outcomes.

Answer 3

If $\ N_1\ $, $\ N_2\ $ and $\ N_3\ $ are the numbers appearing on the 6-sided, 10-sided and 20-sided dice respectively, then the probability you're asking for is \begin{align} P\left(N_1<N_2<N_3\right)&=\sum_{n_1=1}^6\sum_{n_2=n_1+1}^{10}\sum_{n_3=n_2+1}^{20} P\left(N_1=n_1,N_2=n_2,N_3=n_3\right)\\ &=\sum_{n_1=1}^6\sum_{n_2=n_1+1}^{10}\sum_{n_3=n_2+1}^{20}\frac{1}{1200}\\ &= \frac{1}{1200} \sum_{n_1=1}^6\sum_{n_2=n_1+1}^{10}\left(20-n_2\right)\\ &= \frac{1}{2400} \sum_{n_1=1}^6\left(10-n_1\right)\left(29-n_1\right)\\ &= \frac{1}{2400} \sum_{n_1=1}^6\left(290-39n_1+n_1^2\right)\\ &= \frac{1}{2400}(1740-39\cdot3\cdot7+7.13)\\ &= \frac{253}{600} \end{align}