Probability of rolling 3 and 4 in a row with 4 6-sided dice So I was playing a game and was wondering what the probability was to roll numbers in a row. 

Four fair six-sided dice are rolled. What is the probability that three of the numbers will be in a row. Also, that all 4 of them will be in a row.

I've tried solving it but I couldn't get it. How can I solve this? I know that with 4 dice there are 1296 combinations but I have not come up with a way to determine the outcome.
 A: For four in a row, there are three possible lowest numbers.  There are $4!$ ways to roll the required set of four numbers for each one, so the chance is $\frac {3 \cdot 4!}{6^4}=\frac {72}{1296}=\frac 1{18}$  
Three in a row is harder.  There are four possible lowest numbers of the run.  Presumably we are prohibited from having a run of four.  The fourth die can then match one of the three we have. There are three ways to choose which one, then $\frac {4!}2$ ways to order the throw for $4 \cdot 3 \cdot 12=144$ rolls.  We can also have four distinct numbers with three in a row.  Three are six rolls that satisfy this, $1235, 1236,2346,1345, 1456,2456$.  These six possibilities can be arranged in $4!=24$ ways for a total of $144$.  The total probability is then $\frac {144+144}{1296}=\frac 29$
A: The only wanted sequences are 1234, 2345, 3456.
Wlog let's look at the sequence 3456.
In 4 out of 6 cases we get one of those numbers with the first roll.
Again wlog let's say we rolled a 4.
In 3 out of 6 cases we hit one of the remaining 3 numbers with the second roll.
In 2 out of 6 cases we hit yet another one with the third roll. In one out of 6 cases we hit the remaining number in the last roll.
So we hit the sequence 3456 in 4*3*2*1=24 cases. Same goes for the other 2 sequences. So in total we have 3*24 wanted outcomes out of 1296, which is a probability of 1/18 or 0.0555...
