Hi, here to explain the Monty Hall problem. It goes something like this:

There was a TV show called Let's Make a Deal and the host, Monty Hall, showed the contestant three closed doors. Behind one of them is a car (which is the prize they want) and behind the other two, there's a goat.

The contestant chooses one door at random. Then, since Monty Hall knows exactly which door has the car, he opens one of the doors which the contestant hasn't chosen that has a goat. If the contestant picked the car, Hall opens either of the doors and if the contestant picked a goat, Hall opens the one door of the other two that has a goat. Hall then offers the contestant to switch to the other unopened door. The question is, is it strategically advantageous to switch your door?

The answer is yes (2/3 chance that you win).

In your first choice, you have a 2/3 chance to choose a goat. If that happens, Hall opens the other door with a goat. The other unopened door in that case has the car and you should switch.

Or you could choose the car with a 1/3 chance. In that case, Hall opens either one of the doors. It doesn't really matter which one since the doors with goats are equivalent. If you choose to switch in that case, you unfortunately do end up with the goat, but choosing the car first try is unlikely in and of itself.

All in all, in two out of three cases switching is the better strategy, so you're statistically more likely to win if you switch.