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Two fair dice are thrown and their sum is observed. This is done repeatedly. What is the probability that a run of n consecutive 5s occurs before a run of m consecutive 7s?

The probabilty of a run of $5$s occurring and probability of a run of $7$s occurring do not add up to 1. So how exactly to approach the problem?

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2 Answers 2

Hint: throw away everything which is not a possible outcome. For example, to work out the probability that a 5 occurs before a 7, the only events you are interested in are "dice show 5" (probability 1/9) and "dice show 7" (probability 1/6). Which is the ratio between the two events?

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Hint:

Imagine a Markov process with five states:

  1. Previous throw was neither $5$ nor $7$ (this is the starting point)
  2. Previous throw was $5$ but one before was not
  3. Previous $n$ throws were $5$ (an absorbing state)
  4. Previous throw was $7$ but one before was not
  5. Previous $m$ throws were $7$ (another absorbing state)

You can then set up a transition matrix: for example from state 2, the probability of moving to state 3 is $1/9^{n-1}$, of moving to state 4 is $\frac16 \left(1-1/9^{n-1}\right)$ and of moving to state 1 is $\frac56 \left(1-1/9^{n-1}\right)$. Then solve to find the limit of the probabilities of being in states 3 or 5.

For state 1 you have a choice: either you can say the probability of moving to state 2 is $1/9$ and of moving to state 4 is $1/6$ with the probability of staying in state 1 is $1 - 1/9 -1/6 =13/18$, or you can short circuit this by saying for next change of state has probability of moving to state 2 of $\frac{1/9}{1/9+1/6} = 2/5$ and of next moving to state 4 of $3/5$.

As an example to check your result, I think for example if $m=2$ and $n=3$ then the probability of stopping with two consecutive $5$s is $251/340$ and of stopping with three consecutive $7$s is $89/340$.

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