Roll of 3 dice

Of three rolls, there are only two ways when the condition "atleast one even and atleast one odd roll" is NOT satisfied. The condition is not satisfied only when all the rolls are even or all the rolls are odd. With a fair dice, the probability of three even rolls = probability of three odd rolls = 1/2 * 1/2 * 1/2 = 1/8

Hence, the required probability is =1 - 2 * 1/8 = 3/4.

Outcome of 3 consecutive rolls must be one of these(e=even,o=odd)

(e,e,e),(e,e,o),(e,o,e),(e,o,o), (o,e,e),(o,e,o),(o,o,e),(o,o,o)

So the probability to get at least 1 even number and at least 1 odd number=6/8=3/4

There are 216 possible outcomes of rolling the two fair 6-sided dice. (because of 6 x 6 x 6 or 6^3)

By listing all the possible outcomes

You will notice that per number there are 9 possibilities where in the numbers are all even numbers or all are odd numbers

For example:

for the number 1 ( i mean that this is if the outcome if the 1st die is 1)

you may have the outcomes:

(1,1,1) (1,1,3) (1,1,5) (1,3,1) (1,3,3) (1,3,5) (1,5,1) (1,5,3) (1,5,5)

for the different numbers 2 - 6 , there will also be the same number of outcomes but if the number is even, the number used are also even (for exmple is 2 , 246 is a possible outcome.)

So 9 x 6 = 54 ( since there are 6 possible outcomes for the 1st die)

216 - 54 = 162 ( since those 54 outcomes are the unwanted outcomes, so 162 is the number of favorable outcomes.)

Because of the formula: Number of favorable outcomes/Total number of possible outcomes;

the probability is 162/216 = 3/4 or 0.75

The chance that two dices getting $1$ odd and $1$ even is $1/2$ .

The chance that the next dice gets $1$ odd or $1$ even is $1/2$ .

Each combination of even and odd for 2 dice rolls have $1/2 * 1/2 = 1/4$ of chance to be picked at random.

So, following this logic:

1º and 2º dices :

$[e, e],[e,o],[o,e],[o,o] = 1/4$

3º dice :

$[e],[o] = 1/2$

Then :

$[e,e,3º],[e,o,3º],[o,e,3º],[o,o,3º] = 1/4+1/2 = 3/4$

Clarification :

If 3º is even

$[e,e,e]$ is out

$[e,o,e],[o,e,e]$ and $[o,o,e]$ are in

So $3/4$ are in.

Of three rolls, the possibilities are: EOE EEO OEE OOE OEO EOO OOO EEE

We get atleast an even and an odd roll in 6 of the 8 cases above, so the probability will be 6/8, which simplifies to 3/4

Python 2.7:

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from itertools import product
from fractions import Fraction as frac
def even(rolls):
    for roll in rolls:
        if roll % 2 == 0:
            return True
    return False
def odd(rolls):
    for roll in rolls:
        if roll % 2 != 0:
            return True
    return False
def goal(rolls):
    return even(rolls) and odd(rolls)
goals = 0
trials = 0
dice = xrange(1,7)
all_rolls = product(dice,repeat=3)
for rolls in all_rolls:
    if goal(rolls):
        goals += 1
    trials += 1
print "Answer:", frac(goals,trials)