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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.

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You don't have symmetry as used in your third paragraph. You know the first die was $1$, $2$, or $3$. The probability that the second roll exceeds the first in this case is strictly greater than $1/2$ (since, with probability $1/2$, the second roll is one of $4$, $5$, or $6$). – David Mitra Apr 22 '13 at 18:29
up vote 6 down vote accepted

You need them all to be different (which happens with probability $\frac{5\cdot 4 \cdot 3}{6^3}$), and given this, you need them to come out in the exact right order, which has probability $\frac{1}{4!}$. In total, then: $$ P(R_1<R_2<R_3<R_4) = \frac{5\cdot 4 \cdot 3}{6^3\cdot 4!} = \frac{5}{2\cdot 6^3}\approx 0.01157 $$

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I got to the same number...but yours is a much better (faster) way of getting there :) – david Apr 22 '13 at 18:25

Hint: getting a strictly increasing sequence of$\, \,$4 numbers is just the same as calculating the number of subsets with $\,4\,$ elements that a set with $\,6\,$ elements has: whenever you have such a $\,4$-subset you have an increasing sequence, and different subsets fix to different increasing sequences.

Can you take it now from here?

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