We didn’t receive a lot of reader-feedback to last month's Geek Challenge. Some questioned the assumptions involved, and some guesses were taken. Here is my analysis of the problem:
The Deer Dilemma asks about the probability of collision between a car traveling an unspecified speed and a number of deer traveling at unspecified speed and route. The primary challenge is figuring out how to estimate the chance of collision of a single deer.
To solve this, consider the front bumper of the car. This is where a collision between the car and a deer would occur. It moves along a path as a function of time. At any point along the path there is a collision if there is a deer present at the same time as the bumper.
Since the deer move around randomly, what can be said about the chance of the deer being in any one place at any one time? Their random movement means the deer have the same chance of being in any one spot as any other spot at one given time, and they have the same chance of being in one given spot at one time as any other time. So, as the car traverses the same length of road it does not matter when it does so, the chance of hitting any one deer within the distance is the same. Likewise, the probability remains the same regardless of the speed of travel, since the probability of impact is based on distance.
The simplest case to calculate is where the car travels the route at an infinite speed, and this probability is equal to that of other speeds. If the car were to travel the entire route in a flash, the probability of hitting a single deer is equal to the probability of the deer standing on the road within the car’s reach at that instant. Since the probability of a single deer being at any place in any instant is the same as being in any other place, this probability is the ratio of the area of the car’s path to the area of the whole forest.
So, the probability of hitting any one deer P1 can be expressed by the following formula:
P1 = 6 miles / (6 ft. + 3 ft.) = 31680 ft. / 9 ft. = 0.000284
Where 6 miles is the width of the forest, 6 feet is the width of the car, and 3 feet is the width of the deer.
The forest has 1000 deer, not 1 deer, so the chance of collision is much higher. One way to calculate this is to multiply the individual probability by 1000, resulting in P=0.284, or 28%. However, this is the expected number of strikes, not the probability of hitting at least one.
A car hitting a deer is the inverse result of the car coming through unscathed. So the chance of missing any one deer is 1-0.000284 or 0.9997159. The chance of missing all the deer is 0.9997159^1000 = 0.753.
Therefore, the chance of hitting a deer is 1 - 0.753 = 0.247, and the best answer is B, 25%.