A couple are on a seven-day holiday in the back of beyond, with no access to radio, TV or the internet. Specifically, they have no means of accessing a weather forecast.
The weather in this particular wilderness is quite changeable, with no two days ever the same, although each day is pretty consistent. That is, if a day starts out nice, it will stay nice; if you wake up to cold temperatures it will be cold all day.
The couple have a tricky problem. They have seven activities planned to match the seven days of the holiday, and none of these activities is fixed to any specific day. The highlight of the holiday is a walk to the top of a mountain that needs to happen on a day with good weather. Ideally they want to go on the best day that they can, as the views are spectacular and it will be much more pleasant if it is warm.
The difficult thing, and one that has troubled them in their preparations for the holiday, is how to choose the day for their mountain hike.
If they wake up and the weather is fairly good do they go on the trek, or do they wait? They could decide to wait and miss the best day of the week; they could decide to go and the next day is glorious!
So what strategy should the couple adopt to maximise their chances of going on as good a day as possible?
The best strategy is to go on the first day that is better than the previous day.
This doesn't guarantee that they go on the best day (and if the days get progressively worse they will go on the worst day), but across all the combinations of seven days this will improve their chances.
There are 5,040 different combinations over seven days, but using four days there are 24 which will demonstrate nicely. If one is the best day and four the worst . . .
- 1 2 3 4 go on 4
- 1 2 4 3 go on 3
- 1 3 4 2 go on 2
- 1 3 2 4 go on 2
- 1 4 2 3 go on 2
- 1 4 3 2 go on 3
- 2 1 3 4 go on 1
- 2 1 4 3 go on 1
- 2 3 1 4 go on 1
- 2 3 4 1 go on 1
- 2 4 1 3 go on 1
- 2 4 3 1 go on 3
- 3 1 2 4 go on 1
- 3 1 4 2 go on 1
- 3 2 1 4 go on 2
- 3 2 4 1 go on 2
- 3 4 1 2 go on 1
- 3 4 2 1 go on 2
- 4 1 2 3 go on 1
- 4 1 3 2 go on 1
- 4 2 3 1 go on 2
- 4 2 1 3 go on 2
- 4 3 1 2 go on 3
- 4 3 2 1 go on 3
Using this approach the couple would go on the best day (1) ten times, the second best day (2) eight times, the third best day (3) five times, and the worst day (4) only once. Nearly half the time (10/24 or 42%) they would go on the best day, and three-quarters of the time (18/24 or 75%) they would go on one of the two best days. A significant improvement over 25% and 50% if they picked a random day. This strategy holds good for seven days as well.