It’s been a while, so let’s do a probability problem. I found this one in the youtube channel “MindYourDecisions“. If it rains on a day, the probability of rain tomorrow increases by 10%; if not, it reduces by 10% for the next day. What is the probability that it will rain forever within a few days if the chance to rain today is 60%?
Let Px be the chance of raining forever, starting from a day x% rain. We can write the following equations. Note that P100 means it will rain today and every day from there. On the other hand, P0 suggest no rain today and, as a result, it continues.
P100 | 1 |
P90 | 0.9P100 + 0.1P80 |
P80 | 0.8P90 + 0.2P70 |
P70 | 0.7P80 + 0.3P60 |
P60 | 0.6P70 + 0.4P50 |
P50 | 0.5P60 + 0.5P40 |
P40 | 0.4P50 + 0.6P30 |
P30 | 0.3P40 + 0.7P20 |
P20 | 0.2P30 + 0.8P10 |
P10 | 0.1P20 + 0.9P0 |
P0 | 0 |
Substituting the end value (P0 = 0) for the term P10 and working upwards,
P100 | 1 | |
P90 | 0.9P100 + 0.1P80 | 0.998P100 |
P80 | 0.8P90 + 0.2P70 | 0.98P90 |
P70 | 0.7P80 + 0.3P60 | 0.93P80 |
P60 | 0.6P70 + 0.4P50 | 0.82P70 |
P50 | 0.5P60 + 0.5P40 | 0.67P60 |
P40 | 0.4P50 + 0.6P30 | 0.51P50 |
P30 | 0.3P40 + 0.7P20 | 0.35P40 |
P20 | 0.2P30 + 0.8P10 | 0.22P30 |
P10 | 0.1P20 + 0.9P0 | 0.1P20 |
P0 | 0 |
The value of the last term, 0.998P100 can be evaluated by substituting P100 = 1. Repeating the exercise, now downwards,
P100 | 1 | ||
P90 | 0.9P100 + 0.1P80 | 0.998P100 | 0.998 |
P80 | 0.8P90 + 0.2P70 | 0.98P90 | 0.98 |
P70 | 0.7P80 + 0.3P60 | 0.93P80 | 0.91 |
P60 | 0.6P70 + 0.4P50 | 0.82P70 | 0.75 |
P50 | 0.5P60 + 0.5P40 | 0.67P60 | 0.55 |
P40 | 0.4P50 + 0.6P30 | 0.51P50 | 0.34 |
P30 | 0.3P40 + 0.7P20 | 0.35P40 | 0.18 |
P20 | 0.2P30 + 0.8P10 | 0.22P30 | 0.08 |
P10 | 0.1P20 + 0.9P0 | 0.1P20 | 0.02 |
P0 | 0 |
Therefore, the required probability is 75%.