The full Monty

Yes, but try getting milk out of a car.

Monty Hall was the host of the US television game show Let’s Make a Deal. An idealized version of one of the games on the show is known as the Monty Hall Problem, and it reveals some aspects of basic probability theory that many people find counter-intuitive. In the game, there are three doors; behind the winning door is the top prize — e.g. a car — and behind each of the other two (losing) doors is a booby prize — traditionally, a goat. (We presume you’d rather win a car than a goat.) Monty Hall knows beforehand which door is the winner. As the contestant, you get to choose a door to open, and will receive whatever lies behind the door. However, to introduce some extra tension, Monty will offer you the chance to change your mind.

This is how it works. The game begins with you choosing a door. Monty opens one of the other two doors, revealing a losing door; of course, he can always do this because at least one of the doors not chosen initially by you is a loser, and Monty knows where the winner is. Monty says something like, “Are you sure you’re happy with the door that you’ve chosen? Or would you like to change to this one?”, indicating the other unopened door (i.e., neither the one you first chose nor the one that he revealed).

So… do you switch?

Try to work your strategy out before reading any further!

Solution

This problem uses only elementary probability, but contains some subtleties for the unwary. Marilyn vos Savant discussed the problem and its (correct) solution in her “Ask Marilyn” column in Parade magazine in 1990; many readers wrote to the magazine claiming that she was wrong.

It is beneficial to switch (you actually double your chances of winning). Hopefully it is clear that if you don’t switch, you have probability 1/3 of winning. That is,

\mathbb{P}\left[ \mbox{win if you don't switch} \right] = \mathbb{P}\left[ \mbox{choose the winning door initially} \right] = \displaystyle\frac{1}{3}.

Now, if you do switch, you lose if and only if you chose the winning door to start with (since if you did, you switch to a losing door, otherwise you switch to the winner). So

\mathbb{P}\left[ \mbox{win if you switch} \right] = \mathbb{P}\left[ \mbox{choose a losing door initially} \right] = \displaystyle\frac{2}{3}.

If this argument is a little too sneaky, think about how the game proceeds for a given configuration of winning and losing doors.

Describe a configuration by some list of the three outcomes (W, L, L), and, without loss of generality, suppose the first entry in the list is the door that you chose initially. Then there are three distinct possibilities, WLL, LWL, and LLW, and each of these is equally likely (probability 1/3). Then

\mathbb{P}\left[ \mbox{win if you don't switch} \right] = \mathbb{P}\left[ \mbox{WLL} \right] = \displaystyle\frac{1}{3}.

What happens if you do switch? If we are in situation WLL, then Monty reveals one of the two losing doors (it doesn’t matter which), and if you switch then you end up with the other losing door. On the other hand, if the situation is LWL, Monty will reveal the third door and switching will take you to the second door (the winner!). Similarly for LLW. So

\mathbb{P}\left[ \mbox{win if you switch} \right] = \mathbb{P}\left[ \mbox{LWL, LLW} \right] = \displaystyle\frac{2}{3}.

At this point, with any luck, some of you are thinking something like this: “But how can it make any difference? When I chose my door originally I knew there must be a goat behind at least one of the other ones; now Monty has confirmed there was a goat behind at least one of the other ones… which I knew already, thank you. Nothing’s changed, so how can it possibly benefit me to switch?”

The reason for the change in probability is the fact that Monty’s actions do impart additional information, because, with the knowledge that he possesses, he’s opened a particular door. (Think about some generalizations: what if there were 100 doors and Monty had revealed 98 losing doors?)

To see how Monty’s actions impart additional information, consider a variation of the original problem.

Further problem

Going back to the original (3-door) problem: suppose instead that Monty suddenly forgets where the winning door was, panics, and opens one of the two remaining doors randomly. By chance, he happens to pick a losing door to reveal, much to his relief, a goat. Does this change the probabilities?

Solution. We use some basic conditional probability. Using the same description as above for the possible configurations,

\mathbb{P}\left[ \mbox{Monty picks a losing door} \mid \mbox{WLL} \right] = 1,

since if you initially choose the winning door, Monty will always choose to open one of the other two (losing) doors. On the other hand, if you initially choose a losing door, Monty’s chances are only 50–50:

\mathbb{P}\left[ \mbox{Monty picks a losing door} \mid \mbox{LWL} \right] = \mathbb{P}\left[ \mbox{Monty picks a losing door} \mid \mbox{LLW} \right]=~\dfrac{1}{2}.

Now the only winning configuration for the strategy of not switching is still WLL. By Bayes’s rule,

\mathbb{P}\left[ \mbox{WLL} \mid \mbox{Monty picks a losing door} \right] = \displaystyle\frac{ \mathbb{P}\left[ \mbox{Monty picks a losing door} \mid \mbox{WLL} \right] \mathbb{P}\left[ \mbox{WLL} \right]}{ \mathbb{P}\left[ \mbox{Monty picks a losing door} \right] }.

Again using the fact that all three configurations are equally likely a priori, we obtain

\mathbb{P}\left[ \mbox{WLL} \mid \mbox{Monty picks a losing door} \right] =\displaystyle\frac{1\times\frac{1}{3}}{(1 \times \frac{1}{3} ) + (\frac{1}{2} \times \frac{1}{3} ) + (\frac{1}{2} \times \frac{1}{3})}  = \frac{1}{2}.

So the probability that either strategy wins, given that Monty happened to pick a losing door, is now 1/2: Monty’s incompetence has evened things up. Note that Monty’s actions now give different information: the fact that he has picked a losing door suggests that it is more likely that the configuration is WLL, so switching becomes less attractive.

A quick (perhaps too quick) way to see the new answer, avoiding calculations, is to order configurations so that, as before, the first is the door you pick initially and now the second is the door that Monty picks randomly. Again the three configurations WLL, LLW, LWL are equally likely, but the condition on Monty’s revealing a losing door restricts us to the first two (WLL, LLW) of which the first wins if you stick and the second if you switch.

Of course you (the contestant) might wonder whether your personal probabilities have changed, after all, since you presumably can’t know about Monty’s sudden lapse, but that’s a different (Bayesian) subject.

(Andrew Wade / DP)

Advertisements
This entry was posted in Articles. Bookmark the permalink.

2 Responses to The full Monty

  1. Pingback: Those two fun fallacies exploded | Degree of Freedom

  2. Pingback: A few farewells | Degree of Freedom

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s