Dice rolling probability. A die (6 face) is thrown 3 times. What is the probability that the second and third rolls are larger then the first roll.
Thank you
 A: Well, the first die has 6 possibilities.
It can roll a 1, 2, 3, 4, 5, or 6
If it rolls a 6, there is no chance of a higher roll
A 5 has a 1/6th chance of higher 
A 4 had a 2/6th chance of higher
A 3 has a 3/6th chance of higher 
A 2 had a 4/6th chance of higher
A 1 has a 5/6th chance of higher 
For both the second and the third one to be higher than the first, we have to roll them (2&3) and they both need to be higher, so for each on, the chance of higher is squared.
If the first roll is X then the chance that both the next ones are higher is Y
X=1 Y=25/36
X=2 Y=16/36
X=3 Y=9/36
X=4 Y=4/36
X=5 Y=1/36
X=6 Y=0
The Y values are the chance of doing this with each first roll. Each first roll has a 1/6 probability of happening, so we have to divide the Y's by 6 before we add them
(25+16+9+4+1+0)/216
55/216
There is a 55/216 chance of getting the second two rolls higher than the first roll
A: D is the number of sides of the dice (maybe 6 on standard dice) so
$$P=\frac{1}{D}\sum_{k=1}^{D-1}\left(\frac{D-k}{D}\right)^2$$
A: Let $E$ denote the event you mention and let $X$ denote the result of the first
roll. Then:
$$P\left(E\right)=\sum_{i=1}^{6}P\left(E\mid X=i\right)P\left(X=i\right)$$
Here $P\left(E\mid X=i\right)$ denotes the probability that the event happens under condition that the first result is $i$. 
The second and third roll must then result in a number that exceeds $i$ so we come to: $$P\left(E\mid X=i\right)=\left(1-\frac{i}{6}\right)^{2}$$
Also it is clear that: $$P(X=i)=\frac{1}{6}$$ for each $i\in\{1,\dots,6\}$ so we end up with:
$$P\left(E\right)=\frac{1}{6}\sum_{i=1}^{6}\left(1-\frac{i}{6}\right)^{2}=\frac{1}{6}\sum_{i=1}^{5}\left(1-\frac{i}{6}\right)^{2}$$
Counting in opposite direction is handsome here and leads to: $$P\left(E\right)=\frac{1}{6}\sum_{i=1}^{5}\left(\frac{i}{6}\right)^{2}=\frac{1}{216}\sum_{i=1}^{5}i^{2}=\frac{55}{216}$$
A: Just to clarify between Laertes' and Asimov's solutions, here is a Sage simulation of this:
N=1000000
count=0
for i in range(N):
    dice1 = ZZ.random_element(1,7)
    dice2 = ZZ.random_element(1,7)
    dice3 = ZZ.random_element(1,7)
    if dice2>dice1 and dice3>dice1:
        count = count+1
print count/N.n()   

Output:
0.254369000000000

This agrees with Asimov's solution: $\frac{55}{216} \approx 0.255$, but disagrees with Laertes' solution: $\frac{75}{432} \approx 0.174$.
