3
$\begingroup$

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?

$\endgroup$
7
  • $\begingroup$ 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. $\endgroup$
    – José Carlos Santos
    Commented Aug 9 at 13:44
  • $\begingroup$ Added the question to the description, hope that makes it better? $\endgroup$
    – Vaibhav Dixit
    Commented Aug 9 at 13:46
  • $\begingroup$ Yes, it does.${}$ $\endgroup$
    – José Carlos Santos
    Commented Aug 9 at 13:49
  • $\begingroup$ 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. $\endgroup$
    – lulu
    Commented Aug 9 at 13:57
  • $\begingroup$ 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. $\endgroup$
    – lulu
    Commented Aug 9 at 14:05

1 Answer 1

Reset to default
3
$\begingroup$

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 $s = \frac{17}{3}$

$\endgroup$
2
  • 1
    $\begingroup$ Your equations are correct but the answer is incorrect, it should be 17/3 with a = 5 b =4 $\endgroup$
    – Vaibhav Dixit
    Commented Aug 9 at 15:42
  • $\begingroup$ @VaibhavDixit: Thanks, there was a typo , corrected $\endgroup$
    – true blue anil
    Commented Aug 9 at 15:46

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .