I Didn't Want to Get Involved, but... (Monty Hall)

Great Debates
1

The Monty Hall Dilemma has reared its ugly head again. I have trouble with it because it’s like one of those two-way pictures in which what it looks like depends on how you’re squinting. Here’s the problem, for our Martian friends:

In the game Let’s Make a Deal, a contestant is asked to pick one of three doors. If he picks the right one, he wins a car. If he picks one of the wrong doors he wins crap. The contestant picks a door, but before it is opened the host, Monty Hall, opens up one of the doors that had crap behind them. Now there are two unopened doors, and the contestant is asked whether he will take what’s behind the door he originally picked, or if he’ll switch to the other unopened door.

Which should he choose?

The gut assumption is that it doesn’t matter, that both doors are equally likely to be hiding the car. But a lot of people have argued that the probability is actually that switching will get you the car 66.7% of the time.

A friend of mine wrote a program to demostrate the problem, which is a practical approach. The problem seems to be that the answer still depends on how you do the algebra. Since I didn’t speak C, I wrote my own program, but I got the opposite results. His program ran random scenarios and got the 66.7/33.3 results. Mine ran through all the finite number of dilemmas and got the 50/50 result.

Maybe somebody who speaks Basic can tell me how I’m setting the problem up wrong, but it looks to me that of the 12 possible outcomes, it comes down even whether switching will get you the car.

100 s=0
105 REM The number of winning switches
110 ns=0
115 REM The number of winning non-switches
120 t=0
125 REM The total number of dilemas possible
130 for x=1 to 3
135 REM x is the door the prize is behind
140 for y=1 to 3
145 REM y is the contestant’s initial pick
150 for z=1 to 3
155 REM z is Monty’s elimination
160 if z=x then goto 210
165 REM Monty doesn’t eliminate the door the prize is behind
170 if z=y then goto 210
175 REM Monty doesn’t eliminate the door the contestant has already picked
180 t=t+1
185 REM The scenarios that pass Monty’s Choice are counted
190 if y=x then ns=ns+1
195 REM No-switch winners are counted
200 if y<>x then s=s+1
205 REM Switch winners are counted
210 next z
220 next y
230 next x
240 print “Switch:”, s/t100;"%"
250 print “No switch:”, ns/t100;"%"
260 print “Total dilemas:”, t

In case you don’t happen to have a basic interpreter handy, try pasting the
code into Cocoa.

2

The problem is you’re treating Monty’s choice as an independent variable. It’s not the case. It’s dependent on the player’s choice.

If the player picks the prize, Monty has two choice. If the player does not pick the prize, Monty has one choice.

If Monty has two choice, you have to count the choices at 50% probability. If he has one choice, you count that choice at 100% probability.

What happens is that There are 12 “real” cases overall. In 6 you win by staying, in 6 you lose. However, in the 6 you win, you have to divide by two (because Monty has a choice of doors), whereas in the 6 you lose, you don’t divide because Monty’s choice is not probabilistic.

You have to be very careful in counting cases in probability, especially when one choice set is coerced or dependedent on a previous choice.

Time flies like an arrow. Fruit flies like a banana.

3

I still don’t get it. The way I figure it is this:

  1. There are three doors the prize could be behind
  2. There are three doors the contestant could choose initially
  3. There are three doors Monty could choose.
  4. That leaves 27 possible scenarios, except,
  5. Only 2/3s of those 27 will occur, because Monty will not eliminate the door the prize is behind. That puts us down to 18 scenarios.
  6. Only 2/3s of those 18 will occur, because Monty will not eliminate the door the contestant has chosen. That leaves us with 12 scenarios.

So, I’ve already taken into account the limitations of Monty’s choice from both variables. That’s where the 12 possible outcomes come from. But then you want to factor in Monty’s choice again? I don’t see why that’s methodologically correct.

4

I think we should rename this the Dracula problem.

Anyway, the short answer is that while there are twelve possible scenarios, not all of the scenarios are equally likely to happen. The six where switching is the winning strategy are twice as likely to occur as the six where holding is the winning strategy.

5

You’re going at it backwards. You’re still assuming Monty’s choice is independent of which choice you make. But that’s not so, and it’s causing you to count the winners twice. There are really only three winners and six losers.

Work it forwards.
To simplify, first assume that the prize is behind door #1 (but you don’t know which is door #1)

I pick #1 and Monty can pick #2 or #3. That’s two cases right (P1M2 and P1M3)?

I pick #2 and Monty must pick #3. That’s one case right (P2M3)?

NO! The sum of P1M2 + P1M3 = P2M3!

Look at it another way. Let’s look at four cases and distribute the cases equally.

I pick 1, Monty picks 2
I pick 1, Monty picks 3
I pick 2, Monty must pick 3
I pick 2, Monty must pick 3

Yet another way

If I pick door #1, Monty has a 50% chance of picking door #2. So that’s 1/3 * 1/2 = 1/6.

If I pick door #2, Monty has a 100% chance of picking door #3. So that’s 1/3 * 1 = 1/3.

Yet another way:

(P = player’s pick, Z = Prize Door, M = Monty’s door)

P Z M Outcome by NOT switching
1 1 2 Win
1 1 3 Win
1 2 3 Lose
1 3 2 Lose
2 1 3 Lose
2 2 1 Win
2 2 3 Win
2 3 1 Lose
3 1 2 Lose
3 2 1 Lose
3 3 1 Win
3 3 2 Win

So not switching has a 50% of winning, right? NO! Look at the cases again… You’ve calculated that you have a 50% chance of picking the correct door on the first try! That’s not true: You have a 33.3% chance of picking the correctly on the first try, thus you should see that you’ve set up the problem wrong.

Time flies like an arrow. Fruit flies like a banana.

6

Little Nemo wrote:

Obviously, by possible scenarios I didn’t necessarily mean equally possible scenarios then my count would have been two – win or lose. When I count twelve of them, I mean the following:

P is the door that the car is behind
C is the door the Contestant first picks
M is the door Monty eliminates

P C M
1 1 3 - 1
1 2 3 - 2
2 1 3 - 3
2 2 3 - 4
1 1 2 - 5
1 3 2 - 6
3 1 2 - 7
3 3 2 - 8
2 2 1 - 9
2 3 1 - 10
3 2 1 - 11
3 3 1 - 12

SingleDad wrote:

No, I’m not. The fact that I’m not is documented in the REM statements of the Basic code in my first post above and in my follow explanation up afterward. I’ve eliminated the choices that Monty would not make because of the choice the contestant has made, and the choice he would not make because of his knowledge of which door the car is behind. Please read my post, and if you still think I don’t account for the limitations of Monty’s choice, explain to me how.

Funny, Little Nemo thinks I should count them twice. You think I have already counted them twice.

Little Nemo – which of the twelve scenarios above is twice as likely to happen?
SingleDad – which two of the equally likely scenarios above are repititions of eachother?

No. I’ve calculated that at the point of dilemma, you have a 50% chance of having already picked the right one, not at the point of trilemma. You can’t equate the two without ignoring Monty’s choice. Because one wrong answer is going to be eliminated, the choice was destined to be 50/50 to begin with. You either get it right, or you get it wrong, and if you get it wrong you can get it wrong two ways, so that the probability really splits up more like this:

P is the door that the car is behind
C is the door the Contestant first picks
M is the door Monty eliminates

P C M
1 1 3 - 1 Car
1 2 3 - 2 Goat
2 1 3 - 3 Chicken shit
2 2 3 - 4 Car
1 1 2 - 5 Car
1 3 2 - 6 Goat
3 1 2 - 7 Chicken shit
3 3 2 - 8 Car
2 2 1 - 9 Car
2 3 1 - 10 Goat
3 2 1 - 11 Chicken shit
3 3 1 - 12 Car

Which is a shortcut representation. A full table would actually have columns for which doors had which other prizes behind them. But mathematically, it would work out the same. Half of the losers would be one thing, and the other half the others. So the probabilities actually work out to:

50% Car
25% Goat
25% Sack of chicken shit

I’m not sure what you mean, SingleDad, that I’m working it backwards, except that you think I think I’ve calculated the probability at the trilemma when I’ve actually calculated the probability at the dilemma.

7

I wrote:

That should read:

Obviously, if by possible scenarios I didn’t necessarily mean equally possible scenarios then my count would have been two – win or lose.

8

P is the door that the car is behind
C is the door the Contestant first picks
M is the door Monty eliminates

**P C M **
1 1 3 - 1<-Contestant is correct initially
1 2 3 - 2
2 1 3 - 3
2 2 3 - 4<-Contestant is correct initially
1 1 2 - 5<-Contestant is correct initially
1 3 2 - 6
3 1 2 - 7
3 3 2 - 8<-Contestant is correct initially
2 2 1 - 9<-Contestant is correct initially
2 3 1 - 10
3 2 1 - 11
3 3 1 - 12<-Contestant is correct initially

As you can see, you have the contestant being correct in their initial guess 6 out of twelve times. This cannot be correct. Each of the twelve does not have an equal chance of occuring. Let’s look at the situations before Monty open a door:

**P C M **
1 1 - - 1
1 2 - - 2
2 1 - - 3
2 2 - - 4
1 1 - - 5
1 3 - - 6
3 1 - - 7
3 3 - - 8
2 2 - - 9
2 3 - - 10
3 2 - - 11
3 3 - - 12

This is what you had. Since each possibility is equally likely, we need to remove the duplicates.

**P C M **
1 1 - - (a)
1 2 - - (b)
2 1 - - ©
2 2 - - (d)
1 3 - - (e)
3 1 - - (f)
3 3 - - (g)
2 3 - - (h)
3 2 - - (i)

Each of these is equally likely. Just because Monty has more options at this point in cases (a), (d), and (g) doesn’t make them twice as likely to have already happened.

This post will probably make more sense if you go from bottom to top, as it’s currently in reverse chronological order.

9

Johhny Angel, try this code:
<BLOCKQUOTE><font size=“1” face=“Verdana, Arial”>code:</font><HR><pre>
100 randomize
110 rem s is the number of times staying wins
120 s = 0
130 rem t is number of games we’ll play
140 t = 100
150 for i = 1 to t
160 rem x is door prize is behind
170 x = int(rnd(3)+1)
180 rem y is contestant’s initial pick
190 y = int(rnd(3)+1)
200 rem z is Monty’s elimination.
210 on x-1 goto 220, 260, 300
220 if y = 1 then z = 2
230 if y = 2 then z = 3
240 if y = 3 then z = 2
250 goto 330
260 if y = 1 then z = 3
270 if y = 2 then z = 1
280 if y = 3 then z = 1
290 goto 330
300 if y = 1 then z = 2
310 if y = 2 then z = 1
320 if y = 3 then z = 2
330 print "Initial Pick = "; y, "Win = "; x, "Eliminate = "; z,
340 if y = x then print “win = stay” : s = s + 1
350 if y <> x then print “win = switch”
360 next i
370 print "Win by Staying = “; s; " (”; s*100/t; “%)”
380 print "Win by Switching = “; t-s; " (”; (t-s)*100/t; “%)”
999 end




I tried to use an array to make the elimination process a little clearer, but I couldn't make arrays work on cocoa. Also, I'm not impressed with cocoa's random numbers. If you have a better interpreter, try using it, and for better flavor, try changing t in line 140 to something like 1000 or better.
10

I think you’re all over analyzing the original problem and adding assumptions. The problem, as it was originally stated was this:

There are three doors. Behind one is a prize. You are asked to choose a door. One of the remaining doors is opened to reveal no prize, leaving two doors in the equation. Should you switch doors?

The answer is YES because switching gives you the same odds as choosing BOTH of the other doors in the first place. Alternatively, you can think of it like this: Switching doors gives you the same odds of winning as trying to choose a loosing door to discard.

It’s important to realize that the assumptions are (1) One of the doors you didn’t choose first will be opened (2) Only a losing door will be opened. These assumptions could be restated as: A losing door that was not chosen will be eliminated from the competition. This has the effect of giving the remaining unchosen door a probability of 2/3 of having the prize while your original choice still has a probability of 1/3.

I’ve got a perl version of the 3 doors simulation, if anyone is interested…

11

JoeyBlades: Yes. But the interesting question here is why is the 12 case analysis incorrect? The answer is that it incorrectly fails to propagate in the second clause is incorrect.

We have a complex (two clauses) question here.

In the first clause, a contestant picks a door. It’s obvious that the contestant has a 1/3 chance of initially picking the correct door. To simplify, let’s assume that the doors are labelled on the inside, and the contestant cannot see the number. Furthermore, let’s assume the prize is behind door #1. As omniscient observers, we know what door the contestant picks.

Therefore there are three cases (C is the contestant’s choice:

Table 1
C
1
2
3

Now we have defined a property of our case set: C=1 == 1/3, C=2 == 1/3, C=3 == 1/3. We must propagate this property throughout our analysis.

Now Monty picks a door (M). If the contestant picks #1, Monty’s choice is free: He may pick either #2 or #3. If the contestant picks #2 or #3, Monty’s choice is coerced: He *must[/] pick #3 or #2, respectively. We can also determine the optimal strategy in each case(O=(S)witch or O=(H)old)

Table 2
C M O
1 2 H
1 3 H
2 3 S
3 2 S

Therefore H=2 and S=2 making them equally optimal, right?

But wait! We have screwed up. C=1 == 1/2, C=2 == 1/4, C=3 == 1/4! We have failed to ]propagate the property we defined in table 1! There are two ways we can restore this property. Either halve the probability of C1M2 and C1M3 (Table 2a), or duplicate the probabilities of C2M3 and C3M2 (Table 2b):

Table 2a
C M F O
1 2 .5 H
1 3 .5 H
2 3 1 S
2 3 1 S

C=1F == 1/3 C=2F == 1/3, C=3*F == 1/3, good! Thus H=.5+.5/sum (f) = 1/3, S = 1 + 1/sum (f) = 2/3

Table 2b
C M F O
1 2 1 H
1 3 1 H
2 3 1 S
2 3 1 S
3 2 1 S
3 3 1 S

C=1 == 1/3, C=2 == 1/3, C=3 == 1/3, good! Thus H = 1 + 1 / sum (f) = 2/6 = 1/3, and S = 1 + 1 + 1 + 1 / sum (f) = 4/6 = 2/3.

One of these techniques is required to preserve the C=x == 1/3 property defined in Table 1.

Either way, we see that the H=1/3, S=2/3 for either analysis, therefore switching has the better chance of success.

Time flies like an arrow. Fruit flies like a banana.

12

Right on, Little Nemo, I agree with you that this should be called the Dracula problem.

Recently discussed in this same forum.

The Monte (Door # X) Dilema

Cecil’s column:

On “Let’s Make a Deal,” you pick Door #1. Monty opens Door #2–no prize. Do you stay with Door #1 or switch to #3? (02-Nov-1990)

13

Here is the question, the answer, and an applet.

Monty Hall Question

14

waterj2 wrote:

So the problem with my methodology is that it is not sufficient just to eliminate the cases that won’t happen, but to also eliminate cases that are effectively equivalent to ones already counted, even if they involve two different actual possibilities. But I still don’t understand how you make the decision to do that without begging the question. What justification is there for this move independent of its ability to produce the expected results?

billehunt wrote:

The problem wasn’t that I didn’t believe the random results my friend was getting. It’s that I didn’t see why counting all possible cases shouldn’t yield the same results.

SingleDad wrote:

Your ad absurdum is well taken, but it has to mean that the twelve possibilities I calculated are not twelve equally likely possibilities. My question is, how is that so? Why is it that the three cases in which the contestant picks the car the first time, Monty’s two choices don’t count as two equally likely possibilites?

15

OK, we start out with the 12 possibilities you mentioned. In 6 of these, Monty had a choice of which door to open. Say, for example, you pick door 1, and the car is behind door 1. There’s a 1/9 chance of this happening. Half of the time, Monty opens door 2, half of the time he opens door 3–a 1/18 chance in each case.

This is different from the case when Monty doesn’t have a choice. Say you pick door 1 and the car is behind door 2. there’s a 1/9 chance of this happening, also. But this time, it doesn’t get split in half like before, Monty is forced to open door 3. This event has a 1/9 chance happening.

So we have 12 events, 6 each have a 1/18 chance of happening, the other 6 each have a 1/9 chance of happening. The first 6 correspond to when you switch, you lose–6/18 chance. The latter 6 correspond to when you switch, you win–6/9 chance.

…ebius sig. This is a moebius sig. This is a mo…
(sig line courtesy of WallyM7)

16

Johnny, it’s late and I’m not going to read all of the responces posted her. But I didn’t say that you should count some scenarios twice. I said some scenarios are twice as likely to occur as others.

Say there’s a six ided die with the following numbers on its faces: 1,1,2,3,3,4. You and I are going to bet. You give me a dollar every time everytime an odd number is rolled. I give you a dollar everytime an even number is rolled. By the logic of your OP, it’s a fair game: there are four possible results, I win on two of them and you win on two of them.

17

You guys have really have done a great job turning a simple problem into something ugly and hard to follow. …twelve possible scenerios, but half are twice as likely… are any of you in management?

Anyway, since some of you seem to have difficulty understanding how to applying basic mathematics above a primary school ‘how many blue marbles are left in the bag’ level, lets do this the simple primary school way using as little maths as possible (ie. I’ll define the problem and then solve it using simple logic):

Question:
What’s my chance of winning if I SWITCH (where I say to Monty Hall ‘I wanna switch’). We’ll just focus on the problem now, not every possibility as this just seems to confuse your little brains

Fact:
There is a 1 in 3 (33.3%) chance of picking the right door and a 2 in 3 (66.7%) chance of picking the wrong door in our initial guess.

If you pick the right one first off (33.3% chance) and switch you always LOSE. (If I’ve lost you already, give up now… you’ll never understand it, just get in your pick-up truck and head to your local bar and have some more beer).

Now, and this is the important bit, if you pick the wrong one first off (66.6% chance) and switch you will always WIN.

What…huh…Why? your little brain asks, well I go through it slowly.
[list=1][li]You pick the wrong door (2 in 3 chance or 66.7% probability)[/li][li]Monty eliminates the other wrong door[/li][li]you switch to the only door left, the PRIZE![/list=1][/li]
Therefore, when you switch you have a 66.7% change of winning and a 33.3% change of losing.
(If you go the stay path your chance of winning is only 33.3%…and no, if you can’t figure that one out I’m not going to explain it to you).

Rift
It takes effort to be this condesending :smiley:

18

You guys have really have done a great job turning a simple problem into something ugly and hard to follow. …twelve possible scenerios, but half are twice as likely… are any of you in management?

Anyway, since some of you seem to have difficulty understanding how to applying basic mathematics above a primary school ‘how many blue marbles are left in the bag’ level, lets do this the simple primary school way using as little maths as possible (ie. I’ll define the problem and then solve it using simple logic):

Question:
What’s my chance of winning if I SWITCH (where I say to Monty Hall ‘I wanna switch’). We’ll just focus on the problem now, not every possibility as this just seems to confuse your little brains

Fact:
There is a 1 in 3 (33.3%) chance of picking the right door and a 2 in 3 (66.7%) chance of picking the wrong door in our initial guess.

If you pick the right one first off (33.3% chance) and switch you always LOSE. (If I’ve lost you already, give up now… you’ll never understand it, just get in your pick-up truck and head to your local bar and have some more beer).

Now, and this is the important bit, if you pick the wrong one first off (66.6% chance) and switch you will always WIN.

What…huh…Why? your little brain asks, well I go through it slowly.
[list=1][li]You pick the wrong door (2 in 3 chance or 66.7% probability)[/li][li]Monty eliminates the other wrong door[/li][li]you switch to the only door left, the PRIZE![/list=1][/li]
Therefore, when you switch you have a 66.7% change of winning and a 33.3% change of losing.
(If you go the stay path your chance of winning is only 33.3%…and no, if you can’t figure that one out I’m not going to explain it to you).

Rift
It takes effort to be this condesending :smiley:

19

Sorry to beat this dead horse some more, but I think I can clear up some of the confusion for those who still think the chances of winning are 50/50 once Monty opens the goat-door. (Rift did explain it pretty well, but I want to show why people are confused by this)

I think the majority of the confusion comes from failing to realize the effect that Monty’s knowledge of the prize location has on the game. True, if Monty randomly picked one of the other doors, that you did not pick, and it just happened to reveal a goat, the chances that your door was correct vs. the remaining door would indeed be 50/50. However, since Monty systematically reveals the wrong door, he is indeed doing the same thing as offering you two doors (2/3 chance) vs. your original game (1/3 chance).

Many of you 50/50’ers have used the raffle ticket scenario as a defense. If 1000 people buy tickets, and 998 scratch off their tickets and they are not winners, you and one other dude are left holding tickets. Should you switch? In this scenario, the answer is a resounding NO. Your chances are 50/50 and it doesn’t matter if you switch or not. So how does this jive with the Monty Hall solution? Well, the 998 people had no idea who had the winning ticket, they were choosing at random. If someone who knew where the winning ticket was instructed 998 of the raffle ticket holders to scratch off their tickets, leaving just you and one other raffle-ticket-holding dude, now should you switch? YES YES YES Your odds have just gone from 1/1000 if you don’t switch to 999/1000 if you do. Is it clear now?

20

why doesn’t somebody, in the name of science, spend several days watching the show non stop, on say, the game show network, and tally the results?

not it.