Feeling Sleepy

Let $X$ and $Y$ be two random variables defined as follows:

$X$ : Number of rolls until $6$ appears immediately after a $5$

$Y$ : Number of rolls until either $6$ appears in the first roll or $6$ appears immediately after a $5$

Lets relate the expected values of random variable X and Y before and after the most recent throw.

$E[X] = \frac{5}{6}E[X+1] + \frac{1}{6}E[Y+1]$

$E[Y] = \frac{1}{6} + \frac{4}{6}E[X+1] + \frac{1}{6}E[Y+1]$

Solving these two equations yield $E[X] = 36$

Explanation for above two equations

Let $A$ be the event that the most recent roll was a $5$ .

We know that:

$E[X] = E[X|A'] \times P[A'] + E[X|A] \times P[A]$

$E[X] = \frac{5}{6}E[X|A'] + \frac{1}{6} \times E[X|A]$

Now, here's the crucial part,

$E[X|A'] = E[X+1]$

The above is true because given the most recent throw was determined to be not a $5$ , the expected value is same as that of $X+1$ because of the memorylessness property.

Similarly $E[X|A] = E[Y+1]$ because given the most recent roll is determined to be $5$ , the expected value is same as that of $Y+1$ .

Similarly the second equation can be thought of.

You want to get a specific permutation of 2 dice. There are 36 possible permutations. Since you have a 1/36 chance of getting the correct permutation each roll, it is expected to take 36 tries. I don't really understand why it's not 37, though- the first roll doesn't count as a try, so I'm pretty sure you need to add 1.

Since we have six possibilities for any single roll, it takes 6 rolls to get a 5. On getting a 5, we have six different possibilities of the outcome after it as well. One of those is getting a 6. Hence, out of 36 different possibilities for a pair (like 3 then 4), there is one possibility of 5 then 6.

So the maximum time needed to get a (5,6) is 36 seconds.

Yet, the average time would definitely be less than this. It'd be 18.