# How many times do we have to roll a fair 6-sided die till we roll two numbers in a row that differ by 2?

How many times do we have to roll a fair 6-sided die till we roll two numbers in a row that differ by 2?

I tried two approaches but couldn't get the answer.

First, naively I thought only an even number of rolls were possible and created a geometric distribution with $$\frac{8}{36}$$ probability of the rolls that differ by two. This is obviously wrong because this can happen even at 3, 5... rolls and so on.

For this, I tried to come up with a recursive solution but that didn't work out.

Any help?

• Welcome to MSE. A question should be written in such a way that it can be understood even by someone who did not read its title. Commented Aug 9 at 13:44
• Added the question to the description, hope that makes it better? Commented Aug 9 at 13:46
• Yes, it does.${}$ Commented Aug 9 at 13:49
• I'd suggest using states, to exploit the symmetry. There are three active states, which we can denote as $\{1,2,3\}$ exploiting the symmetry that $1$ and $6$ are essentially the same, as are $2,5$ and $3,4$. Now look at the transitions between those states and use them to write a system of linear equations.
– lulu
Commented Aug 9 at 13:57
• Note: might be able to reduce this to two states by remarking that $1,2,5,6$ share the property that only one successor ends the game while $3,4$ each have two winning successors. That might be simpler.
– lulu
Commented Aug 9 at 14:05

Following Lulu's second suggestion rather than having more states, let the starting state be s, and the two active states be $$a =1,2,5,6,\;b = 3,4$$, then
$$s = \frac46(a+1) + \frac26(b+1) \tag1$$ $$a =\frac46(a+1) +\frac16(b+1) +\frac16\times1 \tag2$$ $$b = \frac26(a+1) +\frac26(b+1) +\frac26\times 1\tag 3$$
Solving the set of linear equations, we get $$s = \frac{17}{3}$$