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"If a single fair die is to be rolled repeatedly, find the probability that two sixes in a row will come up before a five followed by a six."

I would have thought that the probability is 1/2, as I wouldn't think it mattered what sequence of two numbers you chose like this, all rolls are independent, but my professor says the answer is 5/12.

Could someone please explain to me how this works, why the probability is less than 1/2?

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In order to "produce" two consecutive sixes, you must first wait for the first of these two sixes. And then, as "bad luck" may have it, this may accidentally have been preceeded by a five.

More precisely, while playing the game we are in one of the following states before each roll:

  1. Have just seen a five (i.e. we might see five-six after the next roll)
  2. Have just seen a six (i.e. we might see six-six after the next roll)
  3. Nothing special

The transitions between these are as follows: From state (3), which is also the initial state, we move to state (1) with probability $\frac16$ and to state (2) with probability $\frac16$ and remains in (3) with $\frac46$. From state (2) we move to state (1) with probability $\frac16$, "win" with probability $\frac16$ and move to (3) with $\frac46$. From state (1) we "lose" with probability $\frac16$ and move to (3) with $\frac56$. Note that there is no way to come from (1) to (2) directly. This difference is what breaks the symmetry.

If $p_1,p_2,p_3$ are the probabilities of winning if one starts at state 1,2,3, we obtain the equations $$ \begin{align}p_3&=\frac16p_1+\frac16p_2+\frac23p_3\\ p_2&=\frac16p_1+\frac16\cdot 1+\frac23p_3\\ p_1&=\frac16\cdot 0+\frac23p_3\\ \end{align}$$ You can solve this system and should find $p_3=\frac5{12}$.

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Let $X$ denote the outcome of the first roll and let $E$ denote the event that you describe. Then:

$$(1) \quad P\left(E\right)=\frac{1}{6}P\left(E\mid X=5\right)+\frac{1}{6}P\left(E\mid X=6\right)+\frac{4}{6}P\left(E\right)$$

The third term corresponds with rolling no $5$ and no $6$ at the first roll. Then the first roll is not relevant and somehow you start over again.


$$(2)\quad P\left(E\mid X=5\right)=\frac{4}{6}P\left(E\right)+\frac{1}{6}P\left(E\mid X=5\right)$$

The first term corresponds with throwing no $5$ of $6$ the second roll. The second term with throwing a $5$ at the second roll.


$$(3)\quad P\left(E\mid X_{1}=6\right)=\frac{4}{6}P\left(E\right)+\frac{1}{6}P\left(E\mid X=5\right)+\frac{1}{6}$$

The first term corresponds with throwing no $5$ of $6$ the second roll. The second term with throwing a $5$ then and the third with throwing a $6$ at he second roll.

The $3$ equations lead to $P\left(E\right)=\frac{5}{12}$.

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