Let's Make A Deal

1. Mar 25, 2004

JohnDubYa

I hope this problem isn't old hat here. I know the answer, but I would be curious to see your explanations.

Suppose you are on a game show, where you are presented three curtains. Donkeys reside behind two of the curtains, and the big prize (tickets to Metallica) are behind the third curtain.

Suppose you pick curtain #1. Rather than show you what is behind curtain #1, the game show host opens curtain #2 to show you a donkey. Obviously the Metallica tickets must be behind curtain #1 (your original choice) or curtain #3.

Before opening curtain #1, the game show host decides to let you change your mind and pick #3 instead if you wish.

Should you switch to curtain #3, or stick by your original choice?

Or does it matter?

And why?

2. Mar 25, 2004

Damned charming :)

A note to everyone When I searched for this topic on the internet google searching "Monty Hall" turned up good results. It pointed out all though the question assumes Monty is not trying to trick the player in reality sometimes Monty did because he was giving to many cars away.

I think I have a better way of exaplaining why you should switch doors. First of all pretend their are 99 donkeys and one good prize rather than 2 donkeys and one good prize. Now imagine you have picked one door and the host and the lovely assistant are blocking the door you picked and one other door. when you are deciding wether to pick the door you realise that you can see the shadow of donkey hooves under 98 doors, this information would have been great to use before you picked and if you switch you are actually still making use of this information.
I still find this explaining this question also perplexs me, I fully understand how to calculate the actual probability but the probability the final probability seems dodgy unless . I find I am quite good at explaing probability questions (i Have tutored uni probability for 5 years. And I understand why some probabilities soudn strange to students. For example there is a simple reason why people get the following question wrong : "what is the chance that none of the students in a class of twenty has a common birthday". People interept the question as "what is the chance that I personally would have a common birthday with at least one person in a group of 20 "

3. Mar 25, 2004

HallsofIvy

Staff Emeritus
Are you absolutely certain that "Metallica" tickets are a better prize than donkeys?

The crucial point here is that the game show host knows which curtain the tickets are behind: he has knowlege you don't. As far as you know to start with, the tickets could be behind any one of the three curtains: your chances are 1 in 3.

Once the game show host opens the curtain revealing the donkeys (and, of course, he wouldn't open the curtain the tickets are behind), you now have knowledge you did not have before. There was a 1/3 chance that you chose the right curtain before. In that case, the host could open either of the two remaining curtains. There was a 2/3 chance that you chose a wrong curtain, in which case the host MUST open a wrong curtain, leaving the "right" curtain closed. Since there is a 2/3 chance of that happening, as opposed to a 1/3 chance of the other way, yes, you can improve your chances of winning from 1/3 to 1/2 by changing your choice.

"Marilyn Vos Savant" gave a very clear version of this several years ago: Suppose there were one million curtains with donkeys behind them and 1 curtain with the prize. You chose one of the 1000001 curtains, then the host removes all but your choice and one other! Since you had a 1 in 1000001 chance of choosing the correct curtain to start with isn't it clear that you had better change to the one left?

By the way, it is an interesting calculation to see what happens if Monty Hall (oops, "the host") did not know which curtain the prize was behind.

Suppose there are three curtains- you choose one. The host chooses one of the remaining two to open, at random but it happens not to be the prize. You can show that, in this case, using the conditional probability that host will not choose the correct curtain assuming you also did not, that whether you change or not, your probability of winning remains 1/3. No additional knowledge, no improved chances!

4. Mar 29, 2004

rattis

You start with a 1/3 chance of success. IWhen one curtain is revealed to be a failure the other 2 curtains have 1/2 cance of success, so it does not matter weather you change your selection. This is providing that you have no other knowledge such as hearing the donkey!

5. Mar 29, 2004

JohnDubYa

RE: "
Suppose there are three curtains- you choose one. The host chooses one of the remaining two to open, at random but it happens not to be the prize. You can show that, in this case, using the conditional probability that host will not choose the correct curtain assuming you also did not, that whether you change or not, your probability of winning remains 1/3. No additional knowledge, no improved chances!"

I don't agree. The game show host may not have any knowledge of the donkeys, but you do now that the curtain has been opened. I don't think it matters if the game show host knew which donkey the curtain was behind.

6. Mar 29, 2004

JohnDubYa

RE: "IWhen one curtain is revealed to be a failure the other 2 curtains have 1/2 cance of success, so it does not matter weather you change your selection."

I definitely don't agree, and I think this can be disproven with a simple computer algorithm. In fact, this would be a fun programming problem at the introductory level.

7. Mar 29, 2004

JohnDubYa

RE: "Are you absolutely certain that "Metallica" tickets are a better prize than donkeys?"

Yes, unless you are Napster.

8. Mar 29, 2004

Staff: Mentor

You actually improve your chances from 1/3 to 2/3 by changing your choice.

If you don't switch, your chances of winning remain at 1/3. If you flip a coin to decide which remaining curtain to choose, your odds rise to 1/2. But if you take full advantage of the new information given you--and always change your choice--you will increase your odds to 2/3.

Using Marilyn's example of 1,000,001 curtains, the odds are overwhelming that your first pick is wrong (1/1,000,001) and that the other choice is right (1,000,000/1,000,001). In this case, changing your choice is almost a sure thing!

9. Mar 29, 2004

TheStatutoryApe

If the Host chooses a curtain at random I'd say then that as long as it's a donkey he exposes then you have only changed the odds to 50/50. I actually had an arguement with a friend of mine over this particular question a while back. It comes down to the added information that the Host will not reveal the prize because he has foreknowledge of it's location. Due to that then by changing your choice your probability of having chosen the proper curtain increases to 2/3 instead of 1/3. And my friend actually did write a program that illustrated the point.

10. Mar 30, 2004

HallsofIvy

Staff Emeritus
You choose one of the three curtains- there is an a-priori probability of 1/3 that you have choose the right curtain.
Suppose the game show host does NOT know (or has forgotten) which curtain has the prize but opens one of the remaining two "at random" (the two being equally likely). Of course, if he opens the curtain that DOES have the prize, all bets are 0ff- and next week the show has a new host- so we must look at the conditional probability given that he DOES choose the curtain that does not have the prize.

IF the curtain you chose was the one with the prize (there was a 1/3 prob of that) then it doesn't matter which curtain the host chooses, it will NOT have the prize behind it and if you switch, you lose.

IF the curtain you chose was not the one with the prize (there was a 2/3 probability of that) then there is also a 1/2 probability that the host will also choose a curtain without the prize so that the show CAN "go on": there is a (1/2)(2/3)= 1/3 probability that this will happen. Now, if you switch, you win.

Notice that there is a 1/3 probability that switching will win and a 1/3 probability that switching will lose: your probability is exactly the same as if you ignored the host and kept the same curtain. (Of course, now there is also that 1/3 probability that the host will open a curtain that reveals the prize and ends the show!)

That is: IF the host knows which curtain the prize is behind and intentionally opens a curtain that does NOT have the prize (the way the show is supposed to work) then switching curtains will raise your probability of winning from 1/3 to 1/2. IF the host does NOT know which curtain the prize is behind and by sheer luck opens a curtain that does NOT have the prize, swiching curtains DOES not change your probability of winning.

(Personally, I think donkeys would sound better than metallica!)

Last edited: Mar 30, 2004
11. Mar 30, 2004

Staff: Mentor

No, the probability of winning goes from 1/3 to 2/3. It doubles.
Sure it does. I don't care what the host knows or doesn't know. If he opens a curtain that does not have the prize (whether by blind luck or on purpose) you are being given new information. Switch curtains and you double your chances of winning.
I'm with you on that one!

12. Mar 31, 2004

Integral

Staff Emeritus
A good way to look at this.

Pick a door any door (or is it a curtian? does it matter? Sure, door is easier to spell!)
You have a 1/3 chance of picking the winning door. That means that there is a 2/3 chance that the prize is behind one of the remaining 2 doors. The host shows you the door that it is NOT behind. By doing this he has not changed the odds, but he has eliminated the uncertainty. The odds remain 2/3 that the prize is behind the remaining door.

When I first encountered this problem I was skeptical, so I wrote a simple VB program to guess a 1000 times. The results were exactly 2/3.

13. May 21, 2004

mikesvenson

it seems to me that 2/3 chance would mean that 2 out of 3 doors contained the prize. since the prize is only behind one door out of two(at the middle of the game) then why wouldnt the chances be 1/2? It seems to me that at first there is a 1/3 chance of picking the right door, then after a losing door is removed leaving only two choices (one being the prize), then that would convert the odds to 1/2. No matter which door you chose to begin with, you are still making a definate choice to either stick with the original door or switch, both of witch seem to me to have a 1/2 chance of winning after one of the losing doors has been removed. This is how it makes sense to me, but as i have read above i might not know what im talking about.

14. May 21, 2004

matt grime

you need to know about conditional probability, and to read the question carefully. it is important whether or not the host knows in advance where the prize is, and is thus guaranteed to open a non-prize door. if so then you must change, if not then it doesn't matter

15. Jun 6, 2004

O Great One

Could someone explain why it matters whether the host knows in advance or not? If the host picks a door that doesn't have the prize, 2/3 of the time you will select the prize by switching. It seems that it doesn't matter whether the host knows where the prize is.

16. Jun 7, 2004

JohnDubYa

Agreed. Every logical analysis of this game that I have seen is unaffected by whether or not Monty knows which curtain has the Metallica tickets.

17. Jun 7, 2004

Staff: Mentor

Monty's ignorance will cost you

Of course it matters whether Monty knows where the prize is!

If Monty doesn't know, then the game will work a bit differently. 1/3 of the time he would reveal the prize! Well, no point in switching then--you know you lost. 2/3 of the time, he would--again, by chance alone--reveal an empty curtain. In that case, you'd better switch! Your odds improve from 1/3 to 2/3 for that case where an empty curtain is revealed.

In the original game, in which Monty always reveals an empty curtain (because he knows where everything is), your chances of winning are 2/3 by using the strategy of always switching to the remaining curtain.

In this new variation, in which Monty knows nothing, your odds of winning are 2/3 * 2/3 = 4/9 if you use the strategy of always switching to the remaining curtain whenever Monty reveals an empty curtain.

18. Jun 7, 2004

JohnDubYa

I think you have a fault in your logic. When you multiplied the two 2/3's, you incorporated both cases where Monty picks the curtain that has the Metallica tickets, as well as the case where Monty picks the curtain with the donkey. But we can throw out the former, because the situation only calls for those instances where Monty reveals an empty curtain.

Suppose Monty doesn't know. You pick curtain #1, and he decides (on a lark) to open curtain #3, showing a donkey.

Do you switch?

To me, what matters is the information YOU gain, no matter how that information was gained. You were 2/3 likely to be wrong with your initial guess, and that hasn't changed.

19. Jun 7, 2004

Staff: Mentor

But I do throw out the former. That's why it's 2/3 * 2/3 and not 3/3 * 2/3.
Absolutely! Once I get to that point, I can make my odds of winning 2/3 by switching. (Of course, I only have a 2/3 chance of getting to that point.)
Right. And you gain useable information only when Monty reveals a donkey. Which doesn't happen quite as often in your new (and quite different!) scenario.

20. Oct 25, 2011

Graeioume

Everyone who believes there is only a 1/2 chance needs to read (or preferably listen to the audiobook of):

The Curious Incident of the Dog in the Night-Time

Not only is it an excellent book, it references this paradox