Earlier today I set you a new twist on the Monty Hall problem, which is (according to AI) the most discussed recreational maths puzzle in history.
The standard Monty Hall problem is as follows:
You are a contestant on the US game show Let’s Make a Deal. Monty Hall, the host, asks you to choose one of three doors. Behind one door is a car. Behind the other two are goats. To win the car, which is what you want, you must choose the correct door.
You choose a door. Before your chosen door is opened, however, Monty Hall opens one of the other two doors that he knows has a goat behind it.
He opens the door and it reveals a goat. He now offers you chance to switch your choice to the other door. Do you stick, or do you switch?
The answer is that it makes sense to switch, because you double your odds of winning from 1/3 to 2/3. The puzzle is the subject of debate because the intuitive answer is that switching should not make any difference.
Now to today’s version:
Monty Hall and the golden goat
You are a contestant playing the Monty Hall problem. However, you secretly know one of the goats is the former pet of an eccentric billionaire who lost it and is willing to pay an enormous amount for its return, way more than the car is worth. You really want that goat. The host is unaware of this. After you pick your door, as is traditional, the host opens one door, which he knows doesn’t have the car. He reveals a goat, which you can tell is the ordinary goat and not the secretly valuable one. The host offers to let you switch doors. Should you?
Solution No, you should not!
What is interesting here is that when you want to win the car it makes sense to switch, but when you don’t want to win the car it makes sense to stick!
The easiest way to explain the answer is to look at frequencies. Suppose the game is played a large number of times, say 3,000 times.
The expected number of games in which your initial choice is the (door with the) car is 1,000. Then the expected number of times the host opens the door with the ordinary goat is 500. (Because the host cannot tell between goats, and we assume that the host choses a goat at random.)
The expected number of games in which your initial choice is the (door with the) ordinary goat is 1,000. Then the expected number of times the host opens the door with the ordinary goat is 0.
The expected number of games win which your initial choice is the (door with the) golden goat is 1,000. Then the expected number of times the host opens the door with the ordinary goat is 1000.
So in 3,000 trials, the host will open the door with the ordinary goat 1500 times. If you stick you will win in 1000 trials, a win probability of 2/3, and if you switch you will win in 500 trials, a win probability of 1/3. It is better to stick.
I hope you enjoyed this puzzle. I’ll be back in two weeks.
Credit: The earliest reference seems to be here. Thanks to Henk Tijms for alerting it to me.
I’ve been setting a puzzle here on alternate Mondays since 2015. I’m always on the look-out for great puzzles. If you would like to suggest one, email me.
