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A fair die is tossed repeatedly. The experiment ends as soon as the last six outcomes form the pattern 131131

What is the expected length (i.e. the number of rolls of the die) of this experiment?

Matthew Conroy
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bogus
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  • I remember that the geometric distribution is used to calculate the first occurence, but I'm not sure how to apply it here. Obviously the probability for both a 1 and a 3 is 1/6 – bogus Dec 07 '13 at 20:30
  • Maybe you should start with simpler problems. What is the expected number of rolls until a 1 appears? What is the expected number of rolls until the sequence 13 appears? Can you do those? – Matthew Conroy Dec 07 '13 at 20:42
  • E(x) = 1/p and p = 1/6, so the expected number of rolls until a 1 appears is 6. For 13, p is 1/6^2, so the expected number of rolls is 36 – bogus Dec 07 '13 at 20:46
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    So, what is stopping you from doing the 131131 problem? Consider, though, the expected number until you get 66: http://math.stackexchange.com/questions/192177/how-many-times-to-roll-a-die-before-getting-two-consecutive-sixes/192211#192211 – Matthew Conroy Dec 07 '13 at 21:09
  • In case my assumption was correct, the expected length of the experiment is 6^6=46656. Right? – bogus Dec 07 '13 at 21:14
  • let us continue this discussion in chat – Matthew Conroy Dec 07 '13 at 21:26

2 Answers2

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Conway's algorithm would compare identical terms on the left and right of 131131:

  • length $1$: yes 1 both sides
  • length $2$: no 13 on left, 31 on right
  • length $3$: yes 131 both sides
  • length $4$: no 1311 on left, 1131 on right
  • length $5$: no 13113 on left, 31131 on right
  • length $6$: yes 131131 both sides

So you have yes for $1$, $3$ and $6$, and $6$ faces on a die so the expected time is $6^1+6^3+6^6=46878$

Henry
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I thought I would add my solution to this here as well, in case it may help others. There are other ways to do this. One is a way to do this using Markov Chains. This will work even in cases when the dice is not a fair dice.

Markov Chain Method

You can construct the chain as follows (I just used different colors to disambiguate):

enter image description here

Let state 0 be nothing i.e. similar to starting afresh.

Note that state 131131 is absorbent i.e. $P($state 131131 to state 131131$)\ = 1$.

Let $\psi(i)$ be the expected number of throws to reach state 131131 from state $i$

We now have equations as follows:

$\psi(13113) = 1 + \frac{5}{6}\psi(0)$

$\psi(1311) = 1 + \frac{1}{6}\psi(1) + \frac{4}{6}\psi(0) + \frac{1}{6}\psi(13113)$

$\psi(131) = 1 + \frac{1}{6}\psi(13) + \frac{4}{6}\psi(0) + \frac{1}{6}\psi(1311)$

$\psi(13) = 1 + \frac{5}{6}\psi(0) + \frac{1}{6}\psi(131)$

$\psi(1) = 1 + \frac{1}{6}\psi(1) + \frac{4}{6}\psi(0) + \frac{1}{6}\psi(13)$

$\psi(0) = 1 + \frac{5}{6}\psi(0) + \frac{1}{6}\psi(1)$

Solve the system of equations for $\psi(0)$ i.e. the expected number of throws to reach 131131 from state 0. You will get:

$\psi(0)=46878$

Extra

Additionally, if you want a bunch of intermediate results, you get:

$\psi(1)=46872$

$\psi(13)=46842$

$\psi(131)=46656$

$\psi(1311)=45576$

$\psi(13113)=39066$

Rahul P
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