Zane, a contestant on a game show, has a big decision to make. There are three doors in front of him
and behind each door is a prize. The prize behind one of the doors is a shiny red sports car while the
prize behind each of the other two doors is a goat (yes, a goat). The game show host asks Zane to
choose a door. Zane squirms nervously before finally announcing his choice.
"Zane, I am feeling generous" says the game show host, "I will open one of the two doors you didn't choose
which I know to have a goat behind it." So he does, and now there are only two closed doors, the one Zane
picked and one other. The host continues "I will give you a chance to change your mind and choose the
other closed door instead. Or, if you prefer, you may keep your original door choice."
What should Zane do?
a) Keep his original door choice
b) Choose the other door
c) It makes no difference either way
: Imagine the same scenario, but with 100 doors. Behind one of the doors is a car and behind
each of the other 99 doors is a goat. The game show host then opens 98 of the 100 doors leaving only two
closed doors (the one Zane picked and one other). The host makes the same offer (allowing Zane to change
his pick if he wishes).
: Zane should choose the other door.
: By choosing the other door, Zane will have a two in three chance of winning the car!
This is extremely counterintuitive and may baffle even the brightest minds. After the game show host
opens a door revealing a goat, logically there is a 50% chance that the car is behind the door Zane
picked (there are two doors and the car is behind one of them). But the opening of one of the doors by
the game show host doesn't change the odds that the door Zane picked originally has a car behind it
(those odds REMAIN one in three even though there are two doors left). This may be very difficult to see.
Imagine if Zane played this game three times and always picks door #1. Each time the host opens one of the
two doors that Zane did not pick (door #2 or door #3) and always a door with a goat behind it. On
average, one time the car will be behind door #1, one time the car will be behind door #2, and one time
the car will be behind door #3. If Zane never changes his pick, he will win the car one time out of three
(when it's behind door #1). If he always switches his pick, he will win the car two times out of three
(when the car is behind door #2 OR door #3).
Don't believe it? OK, imagine a similar scenario, only this time there are 100 doors instead of 3. Behind
one of the doors is a car and behind each of the other 99 doors is a goat. Let's say Zane picks door #77.
The game show host then opens 98 of the 100 doors (always making sure to only open doors with goats behind
them). After this is done, there will be two closed doors remaining, let's say #77 and #99. Does it still
seem like that there is 50% chance that the car is behind the door #77 that Zane originally picked? If
he played 100 times and always picked door #77 and never changed his pick, on average he'd win the car
only one time out of 100. That means that if you switched you'd win 99 times out of 100.
Stil not convinced? Do a simulation with the orginal three door scenario or write out the possibilities on
paper (car behind door #1, car behind door #2, and car behind door #3) and see how often Zane would win
the car by never switching versus always switching.
Do you have a suggestion
for this puzzle (e.g. something that should
be mentioned/clarified in the question or solution, bug, typo, etc.)?