Game: I roll a die 4 times. What is the probability that I get a strictly increasing sequence of numbers.
My initial thought is as follows: we condition on R1 (the first roll) being a 1, 2, or 3 (which happens with probability 1/2).
Now, we look at R2 - there is a 1/6 probability that R1 = R2 and 5/6 probability that R2 is different from R1. In the case that R2 is not equal to R1, by symmetry we get that R2>R1 with a probability of 1/2. Therefore, the probability that R2>R1 is 5/12.
Continuing in the same manner...
There is a probability of 2/6 that R3 will be equal to R1 or R2. Therefore, there is a 4/6 probability that it is different. Using the same argument as above, there is a 1/6 probability that R3 > R2 > R1. Continuing similarly,
P(R4 > R3 > R2 > R1) = (1/2) * (5/12) * (4/36) * (3/144)
However, I have a feeling that I am doing something terribly wrong. Any help would be appreciated.