Probability for repeated incidents .

Question

I had problems on understanding the following problem :

A die is continuously thrown until 4 is obtained. What is the probability that the dice is thrown an even number of times?

Here , a solution states that probability is : $$\frac{4}{6}\times \frac{2}{6}+\frac{4}{6}\times \frac{4}{6}\times \frac{4}{6}\times \frac{2}{6} + ...$$ $$=\frac{2}{5}$$

I didn't understand why such process was used and what's the the concept behind it?

Another problem :

(ii) If a biased coin is tossed two times repeatedly , then what is the probability of occurring HH?

Can I do the problem (ii) similarly like (i) as : $$\frac{1}{4}+\frac{3}{4}\times \frac{1}{4}+\frac{3}{4}\times \frac{3}{4}\times \frac{1}{4}+ ...$$

Answer 1

I don't understand the formula used either.

Method I: infinite series.

The probability that the die is thrown exactly $2n$ times is $ \left(\frac 56 \right)^{2n-1}\times \frac 16$ so the answer is $$\frac 16 \times \sum_{n=1}^{\infty} \left(\frac 56 \right)^{2n-1}=\frac 5{11}$$

Method II: recursion

Let $p$ denote the answer. The first roll is either a $4$ or it isn't. Hence $$p=\frac 16\times 0+\frac 56\times (1-p)\implies p=\frac 5{11}$$

Answer 2

Are you sure the first problem doesn't say "a die is thrown until a number higher than 4 is obtained"? That would fit with the probability given, where the first term is the probability of exactly two throws (1-4 then 5-6), the next term four throws (1-4, 1-4, 1-4, 5-6) and so on.

For the problem as you've stated it, the actual answer (by a similar method) is $\frac 5{11}$, since in this case at each throw the probability of stopping is $\frac16$ and the probability of continuing is $\frac56$.

The second problem doesn't make much sense. How biased is the coin? When do you stop tossing it? If you keep going forever you will (with probability $1$) get HH eventually.

Answer 3

1. See, that the probabilty of throwing $4$ in $k$-th throw ($k\geq 1$)is equal to: $$p(k)=\frac{1}{6}\left(\frac{5}{6}\right)^{k-1}$$ So we have to sum up all the probabilities for even $k$: $$P=\sum_{n=1}^{\infty}p(2n)=\sum_{n=1}^{\infty}\frac{5}{6}\left(\frac{5}{36}\right)^{2n-2}=\sum_{n=0}^{\infty}\frac{5}{36}\left(\frac{25}{36}\right)^{n}=\frac{5}{36}\frac{1}{1-\frac{25}{36}}=\frac{5}{11}$$

2. Tossing a coin two times makes the probability of $HH$ equal to $p(H)^2$.