Cows and Cars

In television contests, participants are often asked to choose one or more doors from a set, with various prizes hidden behind them. In this problem, we consider a simplified version of such a contest. The presenter offers you the following game:

There are three doors in front of you. Behind two of them are cows, and behind the third is a prize — a car. After you choose a door, but before you open it, the presenter gives you a hint by opening one of the other doors and revealing a cow. The presenter never opens the door you have chosen, even if it hides a cow. You can then decide whether to keep your original choice or switch to the other unopened door. You win whatever is behind the door you finally choose.

Surprisingly, in this example, the probability of winning the car is $2/3$ if you always switch your choice after the presenter reveals a cow. The reason is simple: if your first choice is one of the two cows, switching always leads you to the car, since the presenter must open the other cow. If your first choice is the car, switching leads you to a cow, and you lose the prize. Thus, in two out of three cases, switching wins the car. If you never switch, your probability of winning is only $1/3$.

In this problem, you must calculate the probability of winning the car in a generalized version of this situation.

• The number of cows is given.

• The number of cars is given. The total number of doors is equal to the sum of the number of cows and the number of cars.

• The number of doors hiding cows that the presenter opens is given. The presenter may open several doors containing cows before you have to decide whether to switch your choice.

Assume that you always switch your choice to one of the other unopened doors after the presenter opens the specified number of doors containing cows.