# Probability of Runs ,when the probability of success and failure dont add up to 1.

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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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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