Monty Hall Problem... For Newbies

[wollery]

Prometheus, if the probability of winning for a whole population is 99/100 by switching then the probability of winning for each individual player is 99/100 by switching. The probability of the whole population is the same as the probability of any single member of that population.

Likewise, if the probability for any single player is 50/50 then the probability for the whole population is 50/50. You can't divorce individual probability and group probability, one defines the other.

If I throw a die 600 times then on average I'll throw each number 100 times. The whole population probability of getting a 6 is 1/6. My individual probability of getting a 6 on any individual throw is 1/6.

Okay, let's look at the MH problem with 100 doors.

My individual probability of picking the right door right off the bat is 1/100. I know that there are 99 goats and MH is going to reveal 98 of them, but that doesn't change the probability of my first choice being correct, because I don't know where the car is or which doors MH is going to open. That he opens 98 doors doesn't change the fact that my choice was a 1/100 probability of picking the car. The door I have chosen still has a 1/100 probability of being the right door, even after 98 of the remaining 99 are opened. There's no miraculous change in probability.

Now wait a second, If I toss a coin my individual probability of getting the car is 50/50 because the group probability becomes 50/50, as I explained in my earlier post (you're averaging out the 1/100 probability with the 99/100 probability), but my best bet is to switch rather than toss a coin, because then my individual probability when switching is the same as the whole population probability for switching, which is 99/100. Tossing the coin makes my probability worse. So why on Earth would I toss a coin to make that decision? I practically halve my chances of winning by tossing a coin. It makes no sense to do it that way. If I understand the maths (and I do) I'm going to switch doors. Yes, there's a 1/100 chance I'm wrong, but there's a 99/100 chance I'm right, and I'll take those odds over a 50/50 every time.

With the switching tactic 1/100 people will lose. I feel sorry for them, but 99/100 will drive away in a new car. Why wouldn't anyone switch?

Seriously, actually try this stuff for yourself, run a trial in real life. Do it by always switching, then do it by always sticking, then do it by tossing a coin. The numbers don't lie, your best bet is always to switch doors, even though you only have two doors to choose between.

 

[Rasmus]

On the off-chance that you're not simply trolling ...

No, I do know ahead of time, that no matter which I choose first, Monty will eliminate 98 irrelevant doors, and I will then be left with a choice of two.

Yes, but you do not know *which* doors are irrelevant.

Let's play another game with a million doors:

You pick door #35.830.

Now Monty starts opening doors, one after the other ... he knows where the car is, and you know that he knows. You also know that Monty will not open open the door with the car early on. The car will only be revealed at the last possible point - behind the door you picked, or the last door Monty decided to open.

Monty just moves from door door...

397124|715940|663220|538902|944130
114286|993475|292264|290414|720465
7722|220853|910631|259563|793970
382906|69092|511100|968919|305172
803515|160681|46549|497406|612624
585309|357179|401502|890387|163803
812587|646594|974064|30125|349982
398630|360321|467729|741717|141112
796162|608103|59407|250442|699147
483995|602927|905418|75646|788842
968005|349286|721784|6693|840995
359100|206705|964873|263532|498649
612525|305235|151161|926373|893340
768869|789036|117490|900420|447520
539506|942380|980992|747989|625911
90453|923812|94280|765549|328459
235418|503084|881346|536251|215987
363973|525853|759703|624016|211698
268605|220936|639567|758739|969109
807255|184541|563865|148061|866579

... and you sit there, wondering if it's even worth your time. You have a one-in-a-million shot of getting the car, and finding out that the car is behind door #7 or #68.7786 or #386.004 is going to take forever...

Now, there'S only a few doors left:

#484.194
#230.349
#442.115
#422.236
#28.488 #66.071
#860.185
#106.477
#822.079

at this stage, do you honestly think you have a 10% chance of winning the car?

Or is it much more likely that Monty had a reason to spare door #860.185 from being opened? (Or maybe he had a reason to not open #66.071?)

I didn't say you have a 50/50 chance of picking the correct door at the start. I said, "You start the game with a 50/50 chance of having chosen correctly among the only two doors that Monty will not open."

But that's saying the same thing!
If the chances end up being 50/50 then you must have had a 50/50 chance of doing it right from the beginning. Nobody is moving the car around, after all.

No matter what you do, there are just two doors that Monty won't open--the one you pick, and one which may or may not hide the car.

Yes, but out of all the doors, these have not equal chances of remaining unopened.

Let's go back to my game with the million doors. Monty had time to open a few more doors, and we're left with your original choice (#35.830) and the one door Monty hasn't opened yet: #230.349

Do you honestly think the car is equally likely to be behind the door you originally chose, and the one door out of 999.999 that Monty conciously decided to leave closed? Do you really think that door #230.349 being still closed is just a matter of chance, and that he could have just eas easily left closed #12?

Or do you think that *maybe* he had a reason to not open that particular door?

Once again,

for a whole population the standard analyisis is correct; on average 99/100 of them should win by a policy of always switching. That much is not in question. But an individual contestant cannot take advantage of that fact because, unlike the regular lottery player who can string together a bunch of 1/175M chances into a lifetime expectation of 1/75k, Monty only let's him play once.

That doesn't change his actual chances of winning.

If the entire population plays the game, and 99 in 100 win a car (or 7 in a million or whatever you have), than the chances of each individual are 99% (or 0.0007%)

In other words: There being 2 options, or possible scenarios does not imply that they are both equally likely.

What do you think my chances are of winning the lottery this week, if I play precisely one line? There are two possible scenarios: I win, or I lose.

 

[MetalPig]

I look at it like this.
You pick a door. Then you get the opportunity to choose to win everything that's behind the other doors, except for any goats that may be behind those doors. So switching is always the right choice.

Also:
In the case of 3 doors, the odds that the car is behind a door you didn't choose is 2/3.
The fact that MH opens a door with a goat does not change that, since opening that door does not give you any new information. All it does is tell you that behind at least one of the doors you didn't choose there is a goat, but you already knew that.

 

[Red Baron Farms]

No. Monty is omniscient and so will always eliminate 98 irrelevant doors. From the very outset, I know that no matter what, the car is either behind my first choice, or behind the door Monty will not eliminate. You start the game with a 50/50 chance of having chosen correctly among the only two doors that Monty will not open, and you finish with exactly the same 50/50 chance. That, of a whole population of contestants, 99/100 of them would win by switching just doesn't matter because you can only play once. You've still got a 50/50 chance of winning your one trial. I understand that this seems counter-intuitive.



You cannot similarly improve your odds with Monty, because no contestant ever gets more than one "ticket".

You are almost there. "Monty is omniscient and so will always eliminate 98 irrelevant doors." That part is correct.

"You start the game with a 50/50 chance of having chosen correctly among the only two doors that Monty will not open, and you finish with exactly the same 50/50 chance." That part is close. It is true that the odds stay the same. It is not true that they are 50/50. The odds you pick the right door, in a 100 door 1 car scenario are 1/99. That stays the same. You will always either be on the 1 side or the 99 side. 1 if you stay and 99 if you change.

The reason is you pick first, before Monty has a chance to help you by removing 98 wrong doors. After he helps you, then you can either stay, and refuse his help. Or change and take advantage of his omniscient help. He basically gave you 98 free picks by telling you what on that 99 side you should avoid.

Remember, there is always just 2 sides, your pick and all the others you didn't pick. But luckily Monty helps you cheat by letting you in on his omniscience for 98 of those on the other side you didn't pick. This way you can't mess up. You get an effective 99 picks with just 1 pick. Except you don't get to keep the 98 goats, only the car 99% chance, or the last goat 1% chance.

 

[Alferd_Packer]

I think the problem is that people think of this as a Monte Carlo situation where both sets of choices are unconnected with each other.

 

[Didactylos]

Always enjoyed this puzzle. I heard of it first time many years ago, AFAIR it was thinking of the problem as an offer to switch from the first box you choose to both of the other boxes that made me accept that switching was best.

I'm not so fond of the hypotetical Quiz Shows with 100 or 1000 boxes. I suppose they are introduced to illustrate the absurdity of the 50/50 fallacy, but obviously that's not always working. Introducing umpteen boxes also introduces more ways to twist up your thinking, if you know what I mean.

I thought of a way to reformulate the problem a la the first paragraph:

Let the contestant divide the boxes into two groups, one of two boxes and another with just one box. From the group with two boxes Monty opens a box with a goat. Now let the contestant open either the remaining box from the group where Monty have culled one dud box or select the box in the group Monty have left alone.

More wordy, so again more room for misunderstanding. But its more explicit that the first choice matters. Also bypasses our reluctance to second guess ourselves. Having won and then thrown the price away hurts much more than just never having won.


Some keep insisting that since there are two choices in round two, the chance for each must be 50/50, anything else is just sophistry. I thought of a way to make the fallacious thinking in this clear:

Hopefully all can agree on these two setups:

- Choose 1 out of 3, Monty shows a goat and offers reselect. Supposedly its now 1 out of 2 if you stay.

- Choose 1 out of 3, Monty doesn't open anything or offers another choice. Clearly 1 chance of 3.

Then somewhere in this progression the odds must change from 1/2 to 1/3. If you can't figure out where, maybe you should consider that its 1/3 all the time.

- Choose 1 out of 3, goat exposed, reselect offered but withdrawn before you can react.

- Choose 1 out of 3, goat exposed, no reselect offered.

- Choose 1 out of 3, the remaining boxes are removed and a goat exposed without your knowledge but before you are allowed to open the chosen box.


Note that I have only skimmed the last couple of pages, the above is a reply to the OP and may not apply in any way to the present discussion.

 

[bruto]

There are two ways that game could work. One is that the car is always behind door 63 and the other is that Monty is willing to open the door with the car behind it. Either way, the odds of this version of the game would not be 50/50.

But if you don't know which door Monty will open, then all you know is that the car is behind any one of the hundred.

My scenario was that Monty will always open all the doors except your choice and #63. This means that the only possible locations for the car are your choice and #63, and I would have thought that would make the odds 50/50. No other door could possibly contain a car.

It does tell you something that you did not know beforehand. It tells you that the car isn't behind those particular doors.

yes, but the informaton is of no use. You always knew that 98 of the doors contained goats. Knowing which 98 they are is of no use in the game.

Huh? What game are you talking about here and how did you come up with those odds?

I am responding specifically to Prometheus's 100 door scenario, in which he claims an individual player who plays the game once has a 50/50 chance, even though 99 of a hundred players get the car.

 

[Dr. Keith]

That defies probability and common sense. If you have a choice between two lottery tickets, one with a 999999/1000000 chance of winning $1 million, and one with a 1/1000000 chance of winning $1 million, according to you it doesn't matter which you choose.

This. I would like to see some reply to this.

 

[JohnnyG]

My scenario was that Monty will always open all the doors except your choice and #63. This means that the only possible locations for the car are your choice and #63, and I would have thought that would make the odds 50/50.
<snip>

Wouldn't those odds actually be 0/100? Monty cannot know ahead of time which door you are going to choose, so the only door he can place the car to be sure he doesn't reveal it would be door 63.

 

[JohnnyG]

To me the original problem is very simple. We know that there are only two options after the goat is revealed - stay or switch. Therefore, the two probabilities must sum to 1.

I think everyone agrees that the probability of guessing the right door in the beginning are 1/3. That choice was made without knowing which door Monty is going to reveal, so that information does not alter the odds of winning if you stay. Therefore the odds of winning if you stay remain 1/3 and the odds if you switch have to be 1-1/3 = 2/3.

Odds of 50/50 would require that the information obtained by revealing the door with the goat is used when selecting the door in the first place. That can only happen if the goat is revealed prior to selecting the door in the first place - but it isn't, therefore that information cannot alter the original 1/3 odds.

 

[jiggeryqua]

Along with the Irrelevant Goat, it's worth considering the Inevitable (Irrelevant) Car. Of course, when it's time to open the doors the car becomes very relevant, but when the problem is posed it's as inevitable and irrelevant as the goat. There is a car - we have no more idea where it is than we do the location of the Inevitable Goat, but it's just as inevitable. It's also irrelevant - because the odds never change on account of it.

You pick a door. Is the car more likely to be somewhere else? Yes.

Monty shows a goat. Is the car more likely to be somewhere else? Still yes. Nothing has changed.

No numbers necessary .

 

[Meed]

My scenario was that Monty will always open all the doors except your choice and #63. This means that the only possible locations for the car are your choice and #63, and I would have thought that would make the odds 50/50. No other door could possibly contain a car.

As I said, the only way to execute the scenario you describe is for the car to always be behind door 63.

Think about this game from Monty's perspective. Imagine the car is behind some door other than 63. Let's say it's behind door 3. Now the contestant selects door 44. If you now open every door except for the contestant's choice (44) and 63, you will be opening door 3, the one with the car behind it. So you have to put it behind 63 to avoid this. If it's always behind door 63 then it's not 50/50. Switching always results in a win.

yes, but the informaton is of no use. You always knew that 98 of the doors contained goats. Knowing which 98 they are is of no use in the game.I am responding specifically to Prometheus's 100 door scenario, in which he claims an individual player who plays the game once has a 50/50 chance, even though 99 of a hundred players get the car.

It is of use. It tells you which doors not to pick.

 

[bruto]

Wouldn't those odds actually be 0/100? Monty cannot know ahead of time which door you are going to choose, so the only door he can place the car to be sure he doesn't reveal it would be door 63.

I guess so, now that you say it that way. I was thinking erroneously that the door chosen was somehow different. Brain cramp.

I still say that the opening of the doors has no bearing on the choice, since it's done after the choice is made. It tells you which doors not to pick, but there's no option of picking them anyway. The game is to stick with your first choice or switch to whichever of the other 99 might have the car - essentially all of them. You always knew there would be at least 98 with no prize, so opening 98 does not alter the choice.

I agree that the problem is simple, assuming that it is played fairly. The premise of a fair game is that the car is placed randomly behind one of the possible doors, and that Monty, though he cannot control where the car goes, knows where it is, and never can or will reveal the car.

 

[Prometheus]

If you aren't painstakingly clear about what you mean then I'm afraid I'm likely to remain completely lost as to what you're trying to say.

I assumed that when you said "you start the game with a 50/50 chance of having chosen correctly" that you were referring to the start of the game as the point in which you make the decision to switch or stick and the past tense "having chosen" referring to the initial pick. That is the only way I can make sense of that sentence. And based on that interpretation, it seems you are saying that no matter what door you pick initially, there is a 50/50 chance that it will have been the right pick. If that were the case then it would follow that your initial pick has a 50% chance of being correct.

Yes. From before you make your initial pick, we know that the car will wind up either behind the door you choose, or the one Monty doesn't open. 50/50

That is not in dispute. If there is an implied "therefore" associated with that statement, then you'll have to explain.

I agree that that is not in dispute. I repeated it only because it seemed as though you thought I was disputing it.

That is only true because each individual contestant has a 99% chance of winning. If each individual contestant had a 50% chance of winning by switching, then an average of 50/100 would win by switching in the population.

No. Each individual contestant only gets to play once. They don't get 100 trials each so you cannot say that an individual has 99 chances out of 100. Each individual, is only ever going to have 2 doors to choose from, their initial pick, and the one Monty offers them a chance to switch to.

People increasing their odds in the lottery by playing it multiple times has nothing to do with the problem. We're talking about a population of people playing once, not a population of people playing multiple times. If an individual plays the 100 door Monty once (using the switch strategy), their odds are 99/100. If they play five times, then their odds of winning at least once would increase to 9,999,999,999/10,000,000,000.

You seem to be arguing that the initial odds are 50/50 and you can't improve those odds without playing multiple times. But I don't understand why you think the initial odds are 50/50. I also don't understand why if you think that, that you aren't also claiming that a population of people playing the game once would have a 50% win/loss ration.

For the same reason that people who play the lottery once don't have a 1/75k chance of winning, but people who play every week do.

My point is that Monty's game is not like the lottery.

 

[Drs_Res]

The only way to consider this as two different games, one with a 1/3 chance and one with a 50/50 chance is:

Game 1: You pick one door out of three. (1/3 chance)

Intermission:

• Monty opens a goat door.
• Monty randomly shuffles the remaining doors. (1 goat and 1 car)

Game 2: Pick one of the two doors. (50/50 chance)

If this was the case, then the game should have just started off with 2 doors in the first place.

 

[Roboramma]

No. Each individual contestant only gets to play once. They don't get 100 trials each so you cannot say that an individual has 99 chances out of 100. Each individual, is only ever going to have 2 doors to choose from, their initial pick, and the one Monty offers them a chance to switch to.

He didn't say that "an individual has 99 chances out of 100". He said each individual has 99/100 chance of getting the car. This can be seen from the fact that summed over many individuals an average of 99 out of 100 win the car. I am at a loss for how you can not understand this.

The fact that there are two choices doesn't make the odds 50/50.

 

[Tanalia]

Yes. From before you make your initial pick, we know that the car will wind up either behind the door you choose, or the one Monty doesn't open. 50/50

This is your error -- they chances are not equal because Monty's choice depends on the initial pick.

 

[Skeptic Ginger]

Sorry, I never read the instructions. I answered the answer I already knew. Feel free to subtract my answer from your results. Considering it's 60:25ish in favor of an answer that should be more equal if the persons answering don't know the answer, I'm clearly not the only one that didn't follow the OP rule not to vote if you knew the answer.

 

[Delvo]

This is your error -- they chances are not equal because Monty's choice depends on the initial pick.

In other words, your 2-way choice is not between two doors. It's between one door and the combination of the other two doors.

But the more important goofs in the 50/50 argument are these:

1. One individual's odds being different from the odds for a large group. By definition, they're exactly the same thing, just with two different labels; they can't possibly, in any way, in any situation, anywhere, at all, ever, be two different things. Saying otherwise isn't really even saying two things are separate from each other; it's saying one thing is separate from itself. It's impossible gibberish.

2. Coming up with theories to try to beat reality. The trials have been done many times. The results are in. Saying it should be 50/50 isn't arguing one theory against another; it's arguing against the FACT that it's actually one-third/two-thirds according to the results of actually DOING it.

 

[ynot]

The fact that there are two choices doesn't make the odds 50/50.

If there was 100 choice options to begins with . . .

The odds for the first choice are 99 - 1 against being correct and those odds for whatever is chosen don’t change for the second part of the game.

That the 99 alternate choices have been reduced to a single alternate choice for the second part of the game means the alternate choice option has odds of 99 - 1 for being correct.

Do you want to stay with 99 - 1 against or change to 99 - 1 for?

In other words - Do you think your 1 choice is more likely to be correct than any of the 99 alternate choices?

ETA - Monty might as well ask at the beginning of the game – “Do you want to open 1 or 2 doors to find the car?”.