# Easy question on probability

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 $$\frac {1}{4}$$?

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