Probability of Increasing Order in Multiple Dice Rolls

[Mandelbroth]

Homework Statement

A friend proposed this question to me. It looks like a homework problem, though, so I'll put it here to avoid any conflict with forum rules.

Suppose you roll a fair 9 sided die (with sides 1 through 9) 4 times. Let A be the value of the first roll, B be the value of the second roll, C be the value of the third roll, and D be the value of the fourth roll.

Find the probability that A\leq B\leq C\leq D

The Attempt at a Solution

I figure that the probability that A\leq B is \frac{9+36}{81}=\frac{45}{81} because there are 9 ways that they can be equal and half of the remaining 72 outcomes have A less than B. However, I'm having difficulty extending this to greater numbers of rolls.

Can someone please help guide me toward my next step? Thank you, in advance.

 

[Ray Vickson]

Homework Statement

A friend proposed this question to me. It looks like a homework problem, though, so I'll put it here to avoid any conflict with forum rules.

Suppose you roll a fair 9 sided die (with sides 1 through 9) 4 times. Let A be the value of the first roll, B be the value of the second roll, C be the value of the third roll, and D be the value of the fourth roll.

Find the probability that A\leq B\leq C\leq D

The Attempt at a Solution

I figure that the probability that A\leq B is \frac{9+36}{81}=\frac{45}{81} because there are 9 ways that they can be equal and half of the remaining 72 outcomes have A less than B. However, I'm having difficulty extending this to greater numbers of rolls.

Can someone please help guide me toward my next step? Thank you, in advance.

Letting $E = \{A \leq B \leq C \leq D\}$ we have
P(E) = \sum_{a=1}^9 P(A=a) P(E | A=a),
and for a = 1,2,3, ..., 9 we have
P(E|A=a) = P(a \leq B \leq C \leq D ).

Let $F_a = \{ a \leq B \leq C \leq D\}$. We have
P(F_a) = \sum_{b=a}^9 P(B=b) P(F_a | B = b) ,
and for b = a, ..., 9 we have
P(F_a | B = b) = P(b \leq C \leq D).
Keep going like that, or else try to find a nice formula for $Q(b) \equiv P(b \leq C \leq D).$ Then $P(E) = \sum_{a=1}^9 \sum_{b=a}^9 P(A=a)P(B=b) Q(b).$

 

[haruspex]

Another approach is to break it down according to numbers of rolls the same. There are 9*8*7*6 (ordered) rolls in which they're all different, and 1/24th of these will be in the desired order. ⁴C₂*9*8*7 with one pair the same, of which 2/24 are in the right order. Then there's the two pairs case, the 3 of a kind, and four of a kind.