Hard Dice Probability Question

The probability that the new number is greater is equal to the probability that the new number is lesser. Let it be be $A$ . Let the probability that the new number is same be $B$

$2A + B= 1$

The probability $B$ that the new number is the same is $\frac{1}{6}*\frac{1}{6}\ =\frac {1}{36}$

Solving we have $A= \frac{35}{72}$

Satyen's solution is simplest I guess. To elaborate on that lets look at it this way:

If original number was 11 (1,1) then there are 35 numbers greater than it (12,13,...21,22,....66)

If original number was 12 (1,2) then there are 34 numbers greater than it (13,14,.....66)

& so on till 66 which cannot have a number greater than that.

So our required number of cases are 35+34+33+.....1=35x36/2=35x18

Now total number of cases for original number can be 36, for which any 36 new(re-rolled) numbers are possible, which makes it 36x36

Our required probability=

35x18/36x36=35/72

Python 2.7:

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# method 1: monte carlo
from random import randint
from time import time
def roll():
    return randint(1,6)
def array_to_int(array):
    return int(''.join(str(i) for i in array))
def goal(first, new):
    return array_to_int(new) > array_to_int(first)
def more_than_first():
    first = []
    for i in xrange(4):
        first.append(roll())
    new = first[:]
    i1 = randint(0,3)
    i2 = i1
    # make sure dice are different
    while i1 == i2:
        i2 = randint(0,3)
    new[i1] = roll()
    new[i2] = roll()
    return goal(first,new)
trials = 0.0
count = 0.0
end = time() + 10
# run simulations for 10 seconds
while time() < end:
    if more_than_first():
        count += 1
    trials += 1
prob = count/trials
print "Probability is around", prob

# method 2: exact fraction
from fractions import Fraction as frac
from itertools import product
count = 0
trials = 0
first_rolls = product(xrange(1,7),repeat=4)
for first in first_rolls:
    for i1, i2 in product(xrange(4),repeat=2):
        if i1 < i2:
            second_rolls = product(xrange(1,7),repeat=2)
            for roll_a,roll_b in second_rolls:
                new = list(first[:])
                new[i1] = roll_a
                new[i2] = roll_b
                if goal(first,new):
                    count += 1
                trials += 1
prob = frac(count,trials)
print "Exact probability:", prob
answer = prob.numerator+prob.denominator
print "Answer:", answer