# Math Help - Dice probability

1. ## Dice probability

A dice is tossed 6 times. What is the probability that on each toss, the number shown is less than or equal to the previous tosses.

2. I think my previous answer was wrong, let me do more research and then I'll post an answer unless somebody gets it first.

3. Hello, math sucks!

I have an approach to this problem.
But I have no proof of my theory . . .

A die is tossed 6 times. What is the probability that on each toss,
the number shown is less than or equal to the previous tosses?

Consider rolling two times.
The outcomes are:

. . $\begin{array}{cccccc}
{\color{blue}(1,1)} & (1,2) & (1,3) & (1,4) & 1,5) & (1,6) \\
{\color{blue}(2,1)} & {\color{blue}(2,2)} & (2,3) & (2,4) & (2,5) & (2,6) \\
{\color{blue}(3,1)} & {\color{blue}(3,2)} & {\color{blue}(3,3)} & (3,4) & (3,5) & (3,6)\end{array}$

. . $\begin{array}{cccccc}
{\color{blue}(4,1)} & {\color{blue}(4,2)} & {\color{blue}(4,3)} & {\color{blue}(4,4)} & (4,5) & (4,6) \\
{\color{blue}(5,1)} & {\color{blue}(5,2)} & {\color{blue}(5,3)} & {\color{blue}(5,4)} & {\color{blue}(5,5)} & (5,6) \\
{\color{blue}(6,1)} & {\color{blue}(6,2)} & {\color{blue}(6,3)} & {\color{blue}(6,4)} & {\color{blue}(6,5)} & {\color{blue}(6,6)}\end{array}$

Out of the $6^2$ outcomes,
. . a triangular number, $\frac{6\cdot7}{2}$ are desirable.

Roll three times. .There are: . $6^3$ outcomes.
The outcomes will be displayed in a 6×6×6 cube.
Picture a plane "bisecting" the cube through two opposite vertices.

The desirable outcomes are in the the "larger half".
This is a tetradedon with sides of 6.
It contains: $\frac{6\cdot7\cdot8}{3!} = 56$ outcomes.

Roll four times. .There are: . $6^4$ outcomes.
The ouctomes are displayed in a 6×6×6×6 hypercube.
The "larger half" contains: . $\frac{6\cdot7\cdot8\cdot9}{4!} \:=\:126$ outcomes.

Let's skip to six rolls. .There are: . $6^6$ outcomes.
The outcomes are in a 6×6×6×6×6×6 hypercube.
The "larger half" contains: . $\frac{6\cdot7\cdot8\cdot9\cdot10\cdot11}{6!} \;=\;462$ outcomes.

Therefore: . $\text{Prob} \;=\;\frac{462}{46,656} \;=\;\frac{77}{7,776}$

4. Very interesting Soroban! Now we can generalize, if we toss the dice $n$ times, the resulting probability would be
$\frac{\displaystyle\dbinom{n+5}{5}}{\displaystyle 6^n}$