[Math] Chance of having two consecutive rainy days

combinatoricsprobability

I have the question

If the chance of rain vs. sun is 50:50 what is the chance of having 2 consecutive rainy days in a 5-day period?

I would solve it by saying that two consecutive rainy days can only happen with

RRSSS
SRRSS
SSRRS
SSSRR

and there is a total of $2^5 = 32$ combinations.

So the probability is

$$
P = \frac{4}{2^5} = \frac{4}{32} = \frac{1}{8}
$$

but what if I have to determine the chance of having 321 rainy days in a 31289-day period?

How do I more systematically determine the number of outcomes?

Best Answer

You can solve this using recurrence relation.

Let $a_n$ denote the number of combinations of $n$ days such that:

  • No $2$ consecutive days are rainy
  • The last day was rainy

Let $b_n$ denote the number of combinations of $n$ days such that:

  • No $2$ consecutive days are rainy
  • The last day was not rainy

Then the number of combinations of $5$ days such that no $2$ consecutive days are rainy is $a_5+b_5$.

Now:

  • $a_1=1$
  • $b_1=1$
  • $a_n=b_{n-1}$
  • $b_n=b_{n-1}+a_{n-1}$

So we can compute $a_5$ and $b_5$ gradually:

  • $a_2=1$
  • $b_2=1+1=2$
  • $a_3=2$
  • $b_3=2+1=3$
  • $a_4=3$
  • $b_4=3+2=5$
  • $a_5=5$
  • $b_5=5+3=8$

So the number of combinations of $5$ days such that no $2$ consecutive days are rainy is $8+5=13$.

And the probability of a combination of $5$ days such that no $2$ consecutive days are rainy is $13/32$.

Note that the probability in the general case is $\frac{F_{n+1}+F_n}{2^n}$, where $F_k$ is the $k$th Fibonacci number.

You can use the closed form of $F_n=\left[\frac{\phi^n}{\sqrt5}\right]$ in order to calculate this probability easily.

Related Question