Question

Rolling four six-sided dice twice, what is the probability of the second total being larger than the first?

Answer 1

the total ranges from 4 to 24 i.e 21 outcomes
First Calculating, probability for each case
Let T2 be second total
When T2 is 4, the probability is 0, since no total is lower than that.

When T2 is 5, probability(P5) is (1/21)*(1/21) since only when first sum is 4, our condition is met

T2 = 6,P6=2/21*(1/21)(for T1 ={4,5})
T2 = 7,P7=3/21*(1/21)(for T1 = {4,5,6})
T2 = 8,P8 = 4/21*(1/21) (for T1 = {4,5,6,7})
similarly
.
.
T2 = 23,P23 = 19/21*(1/21)
T2 = 24,P24 = 20/21*(1/21)

The multiplication term (1/21) accounts for the probability that we get the said sum for T2

Final probability is the sum : (1/21)*(1+2+3+4..+20)/21 = 210/(21*21)=.4762