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I know this is a trivial question but I want to make sure I'm not missing anything: We have a biased 6-sided die, which brings any of the 6 numbers with equal probability in the first roll, but in the second and all subsequent rolls, brings the previous result with probability $\frac {1}{2}$ and all others with probability $\frac {1}{10}$.

The question is: Suppose we get a 4 in the first roll; what is the probability we also get a 4 in the 3rd roll? In the 4th? And so on.

If we get a 4 in the 1st roll, then for the 2nd roll we have $\frac {1}{2}$ probability to get a 4 and $\frac {1}{10}$ for all other numbers in the 2nd roll.

So in the 3rd roll, we already have the results of the previous roll of getting a 4 with probability $\frac {1}{2}$, so now the probability is $\frac {1}{4}$?

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There are actually two cases:

(1) Get a 4 in the 2nd roll and the 3rd roll

(2) Get a 4 in the 3rd roll but not in the 2nd roll

The probability of (1) is $\frac{1}{4}$ as you have correctly calculated.

In the case of (2), it is $\frac{1}{2} \times \frac{1}{10} = \frac{1}{20}$ because the probability of not getting a 4 in the 2nd roll is 1/2 and getting a 4 in the 3rd roll is 1/10.

Therefore, the probability of getting a 4 in the 3rd roll is $\frac{1}{4} + \frac{1}{20} = \frac{3}{10}$.

We can generalize this by finding the "recurrence formula" for the probability of getting a 4 in the $n$th roll, which we will define $p_n$ to be. That is:

$p_n = \frac{1}{2} p_{n-1} + \frac{1}{10} (1 - p_{n-1}) = \frac{2}{5} p_{n-1} + \frac{1}{10}$

Also, $p_1 = 1$ since we supposed that we got a 4 in the first roll.

Solving this yields $p_n = \frac{25}{12}(\frac{2}{5})^n + \frac{1}{6}$.

So, for example, $p_4 = \frac{25}{12}(\frac{2}{5})^4 + \frac{1}{6} = \frac{11}{50}$.

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