You are a participant on "Let's Make a Deal." Monty Hall shows you three closed doors.
decision/monty.hall Added material
Problem
You are a participant on "Let's Make a Deal." Monty Hall shows you three closed doors. He tells you that two of the closed doors have a goat behind them and that one of the doors has a new car behind it. You pick one door, but before you open it, Monty opens one of the two remaining doors and shows that it hides a goat. He then offers you a chance to switch doors with the remaining closed door. Is it to your advantage to do so?
Show solution
Under reasonable assumptions about Monty Hall's motivation, your chance of picking the car doubles when you switch. The problem is confusing for two reasons: first, there are hidden assumptions about Monty's motivation that cloud the issue; and second, novice probability students do not see how to handle the information that the opening of the door gave them. Monty can have one of three basic motives: 1. He randomly opens doors. 2. He always opens the door he knows contains nothing. 3. He only opens a door when the contestant has picked the car. In each case, the information from Monty opening the door is: 1. The probability of the car distributes evenly over both remaining doors. 2. The probability of the car in your door does not change. 3. The probability of the car in your door goes to 100%. These result in very different strategies: 1. No improvement when switching. 2. Double your chances by switching. 3. Don't switch! Most people think that (2) is the intended interpretation of Monty's motive. However, even given this interpretation, many people at this point still have trouble with this puzzle. Here are two other arguments that sometimes do the trick. 1. Increase the number of doors from three to 100. If there are 100 doors, and Monty shows that 98 of them are valueless, isn't it pretty clear that the chance the prize is behind the remaining door is 99/100? 2. All agree that you have a 1/3 chance of picking the car originally. That means that the other two doors represent a 2/3 chance. Now, after Monty opens the one of them that is a goat (which he must always be able to do assuming he knows which one it is), the remaining door must now represent the 2/3 chance. So you should pick it. In the real game show the situation was much more complex. Interviews with Monty Hall indicate that he sometimes lured the contestant who had picked the car with cash incentives to switch. However, if Monty always adopted this strategy, contestants would soon learn never to switch, so one presumes that Monty offered another door even when the contestant had picked a goat. At any rate, analyzing the problem when Monty was luring people away from the car is difficult, since it requires knowing something about Monty's probability of bluffing. See the paper by Fernandez and Piron for an analysis of this game. The original Monty Hall problem (and solution) appears to be due to Steve Selvin, and appears in American Statistician, "A Problem in Probability," Volume 29, Number 1 (February 1975), page 67. It should be of no surprise to readers of this group that he received several letters contesting the accuracy of his solution, so he responded two issues later in American Statistician, Volume 29, Number 3 (August 1975), page 134. However, the principles that underlie the problem date back at least to the fifties, and probably are timeless. See the references below for details. These references are selected from the voluminous literature on this subject because they are current and contain bibliographies: Leonard Gillman, "The Car and the Goats," American Mathematical Monthly, Volume 99, Number 1 (January 1992), page 3 Ed Barbeau, "The Problem of the Car and Goats," College Mathematics Journal, Volume 24, Number 2 (March 1993), page 149. Contains a list of equivalent or related problems. Luis Fernandez and Robert Piron, "Should She Switch? A Game-Theoretic Analysis of the Monty Hall Problem," Mathematics Magazine, Volume 72, Number 3 (June 1999), page 214
Additional material
[ Editor: postings from rec.puzzles and sci.math, 1994-1997. ]
In article <2khttp$dtb@cony.gsf.de> mue@gsf.de writes:
>The difference is in when you are offered a switch.
>First interpretation (Marilyn's): Monty _always_ offers a switch.
> => best strategy: always switch
>Second interpretation: Monty _only_ offers a switch if you chose
> the door with the price
> => best strategy: never switch
The analysis in the second interpretation here is quite obvious. I
actually had in mind a more general problem:
You're on a game show. The rules of the show are that the host may
not lie, and if he makes a promise to you, he must honor the promise
before the end of the game unless you agree to release him from it.
Also, the host may not harm you or take anything from you. Other
than that the host can do whatever he pleases.
Now the host shows you three doors and says there is a car behind
one of the doors, and no car behind the two others. He then says
that if you select a door, he will give you the aforementioned car
if it is behind the door you select, otherwise you get nothing.
So you select door X. But then the host opens door Y (Y != X),
revealing that there is no car behind it. He then promises that if
you release him from his first promise, he will give you the car if
it is behind the remaining unopened, unchosen door (door Z, Z != Y
and Z != X).
Now should you "switch" (you get the car if and only if it is behind
door Z) or "stay" (you get the car if and only if it is behind door
X)? Assume that you want to get the car, and you have no additional
information that gives you any clue as to the host's likelihood to
behave in any particular manner.
Incidentally, the archive entry gives three interpretations, none of
which I believe covers this interpretation. It also gives one of the
least satisfactory (in my opinion) arguments for the "2/3" answer in
the case where Monty always offers the switch.
-- David A. Karr <karr@cs.cornell.edu> (February 1994)
In article <barnett.768592555@cnn.eng.convex.com> barnett@convex.com
(Paul Barnett) writes:
>I'm sorry, but this doesn't wash. You are trying to imply a linkage
>between independent events.
>...
>In the Monte Hall problem, one has to presume that Monte knows which
>door has the prize, and he opens only the 'losing' door(s) until two
>doors remain: the one chosen by the contestant and one other door.
And this presumption means that things are no longer symmetrical. If
(as you suggested in the deleted part) Monte had opened doors at random
and happened not to have hit on the prize up to now, then indeed the
chances are even between your door and the remaining one: nothing has
made the remaining one more likely. But if Monte has been using his
knowledge to select losing doors to open (and is known always to do
so, then offer the chance to switch), he has also indirectly been
selecting the winning door to *not* open. This gives you some limited
information about which door is the winning door...
>At this point, the contestant has a choice: stay with their original
>door, or change to the remaining door. There are 2 possibilities.
>Therefore, the probability of choosing the correct door is 1/2.
And by the same argument, there are two possibilities: either I will
win a lottery tomorrow or I won't. Therefore the probability of each
happening is 1/2? Dream on...
>Keeping the original choice, or switching to the other has no effect
>on the probability, unless you have some insight into Monte Hall's
>behavior that I haven't noticed.
Yes. Let me tell you a hypothetical story: one day, Monte needed to get
away from the show quickly. So he said to the contestant: "You look
smart enough to realise that all the business of opening doors and
showing them to be empty is a waste of time. We both know that I must
be able to open all but one of the remaining doors and show them to be
empty. Also, we both know that if the prize is behind the door you've
chosen now, switching will lose you the prize at that point, while if
it is behind any of the others, I will be forced to open all but that
door and switching will win you the prize. So I'll cut the whole
process short: do you want to switch from getting what's behind the
door you chose to getting what's behind *all* the other doors?"
Assuming that Monte is making a fair offer and not trying to "psych"
you out of the prize, do you still believe the chances are 50:50?
(After all, either the prize is behind the door you originally chose or
it is behind the set of doors you're being allowed to switch to - just
two possibilities.)
If not, please identify why the probabilities are different for the
original method of offering you the switch and the new offer - i.e.
identify some situation in which switching in the original method would
have gained you the prize and switching in the new offer wouldn't, or
vice versa.
>... Predictably, the probability that #35
>will roll out as one of the first 33 balls is:
>
> 1/50 + 1/49 + ..... + 1/3
Here's a hint about probability calculations: if your formula gives an
answer which is outside the range 0 to 1, you've made a mistake! In
this case, 1/50 + 1/49 + ... + 1/3 is about 2.9992. The correct
formula is:
1/50 + 49/50*1/49 + 49/50*48/49*1/48 + ... + 49/50*48/49*...*4/5*3/4*1/3
= 1/50 + 1/50 + ... + 1/50
= 48/50 = 0.96
calculated as Prob(#35 on first throw) + Prob(no #35 on first
throw)*Prob(#35 on second throw, given no #35 on first throw) + ...
>Somehow, you managed to roll out 33 balls, and #35 is still one of the
>two remaining. You have certainly defied the odds, but the final
>rollout is AN INDEPENDENT EVENT. What has happened before is of no
>consequence, except that it brought you to this point.
No. The fact that you got there by defeating the odds is significant.
If you play your game 50 times, you expect to be eliminated before
reaching the "two balls left" stage 48 times, to get #35 on the next go
once and not to get #35 on the next go once. Given that you have
beaten the odds so far, only two of these games have survived and you
expect to "win" by getting #35 next time once - i.e. a 50% chance of
success.
On the other hand, if you play 50-door Monte fifty times, you expect
your initial guess to be right once and wrong 49 times. Monty can and
does show you 48 empty doors in *every* one of those games. So there
are fifty surviving games: you've got the correct door in one of them.
This gives a very different probability to there being two surviving
games and you've got the correct door in one of them.
-- David Seal <dseal@armltd.co.uk> (May 1994)
In article <1995Aug3.154849.3582@relay.nswc.navy.mil>
dchill@relay.nswc.navy.mil (oanews) points out:
>Reruns of LMAD are still on. Watch a few. Monty does not always show
>an empty door. No matter how much you would like it to be true, it is
>not.
From the rec.puzzles archive: ==> decision/monty.hall.p <==
Interviews with Monty Hall indicate that he did indeed try to lure
the contestant who had picked the car with cash incentives to
switch. However, if Monty always adopted this strategy,
contestants would soon learn never to switch, so one presumes that
occasionally Monty offered another door even when the contestant
had picked a goat. At any rate, analyzing the problem with this
strategy is difficult, since it requires knowing something about
Monty's probability of bluffing.
If one assumes Monty reveals a losing door and offers a switch
with probability = p when the initial guess is correct
with probability = q when the initial guess is wrong
Then it is easy to show that an always switch strategy will win the
prize with probability = 1/3 + (2q-p)/3, while a not switch strategy
wins with probability = 1/3. Therefore, if Monty wants to minimize the
cost of awarding prizes he can use any strategy with p >= 2q. In this
case there is no advantage to switching, but this won't be apparent to
the average contestant.
If this does reflect what went on in the actual show, it is interesting
to note that since Monty is less likely to show a door when the first
guess is wrong, Monty's failure is an additional clue to the
contestant. Therefore, once the contestant realizes that Monty does
not intend to show a door the contestant should ask to switch to one of
the other doors. If we assume that this request will be granted this
raises the game to a new level of complexity.
-- Richard Byrd <richard@spot.Colorado.EDU> (August 1995)
Bif wrote:
> If you increase it to 100 doors, where you pick one, Monty opens 98,
> and then offers you the choice, most people think that you _should_
> switch.
>
> Somehow, the logic that applies to 100 doors does not apply to 3
> doors:
>
> "Well, I only had a 1 in 100 chance to begin with, so it probably is
> behind the other door."
>
> "Well, I only had a 1 in 3 chance to begin with, but now I have a
> 50-50 chance, so I'll just stick with this one. Besides, Monty is
> going to trick me anyway."
I tried the 100 doors argument once with a pretty bright friend. Still
lost in the numbers, he agreed that 100 meant you should switch, but
that 3 didn't. He even argued that the cut-off point between where you
shouldn't switch and where you should was at about 4 or 5 doors.
I did much better by creating two fictional characters for him: Mr.
Always-Switch and Mr. Never-Switch and I asked him what the odds were
that each one would win a given game.
> But that's not nearly as frustrating as trying to explain where the
> "other dollar" went in the "Three men and a Bellboy" puzzle...
The difference between these two puzzles and a lot of puzzles is that
these two involve getting your "stupid" friend to guess the wrong
answer, then proving your superiority by proving him wrong. I'm pretty
sure that's the reason these puzzles cause so many arguments.
-- Dave Gates <davegates@davegates.com> (November 1997)