Monty Hall Problem... For Newbies

[Dr. Keith]

One slight nit to pick:

Well, that's my bit of scientific investigation for the day, I'm off for a well earned cup of tea.

 

[jt512]

Regardless of whether it appears to be a 50/50 choice and outcome, the numbers show that swapping is always the better option to maximise your chance of winning, whether you choose after the door is opened or before.

Say what?! Are you claiming that if you randomly pick a door, then randomly change to one of the other two doors, and then Monty reveals where one of the goats is, that you have better than 1/3 probability of winning? Because that claim would obviously be false.

 

[Brian-M]

Say what?! Are you claiming that if you randomly pick a door, then randomly change to one of the other two doors, and then Monty reveals where the goat is, that you have 2/3 probability of winning? Because that is obviously not true.

I think he means that it doesn't matter whether your choice to switch or not switch to the door that isn't opened to reveal the goat is made before or after the goat is revealed.

 

[jt512]

I think he means that it doesn't matter whether your choice to switch or not switch to the door that isn't opened to reveal the goat is made before or after the goat is revealed.

As I understand his claim, he picks a door, then changes to another door before having any knowledge of what is behind any of the three doors, and that this procedure gives him a better than 1/3 probability of winning. What you described, above, seems to be this procedure.

 

[Paul C. Anagnostopoulos]

I think the clearest explanation is the one where switching is equivalent to getting both of the doors you didn't pick first. After all, Monty has opened one of them, so you "get" that one, and you also get the other unopened one.

Enumerating all the possibilities does the trick, too.

~~ Paul

 

[wollery]

Say what?! Are you claiming that if you randomly pick a door, then randomly change to one of the other two doors, and then Monty reveals where one of the goats is, that you have better than 1/3 probability of winning? Because that claim would obviously be false.

In the classic MH problem you decide to stick or switch after he reveals the goat. I'm saying that you can make that decision (whether to stick or switch after the goat is revealed) before the goat is revealed. In fact, you can make the decision (whether to stick or switch after the goat is revealed) before you even make your first selection. You can make the decision (whether to stick or switch after the goat is revealed) before you get to the studio if you want. Hell, your great great grandfather could have made that decision for you 100 years ago. At what point in time you make the decision to stick or switch (after the goat is revealed) makes no difference to the probability of winning the car. Switching after the goat is revealed gives you a 2/3 probability of winning, sticking with your original choice gives you a 1/3 probability of winning.

 

[jiggeryqua]

After all, Monty has opened one of them, so you "get" that one, and you also get the other unopened one.

By that reasoning, if you stick you get the one Monty has opened and the unopened door you chose - inasmuch as you never 'get' the goat (the inevitable, irrelevant goat). You're insisting that G+? is of more value than G+?.

 

[jt512]

In the classic MH problem you decide to stick or switch after he reveals the goat. I'm saying that you can make that decision (whether to stick or switch after the goat is revealed) before the goat is revealed. In fact, you can make the decision (whether to stick or switch after the goat is revealed) before you even make your first selection. You can make the decision (whether to stick or switch after the goat is revealed) before you get to the studio if you want. Hell, your great great grandfather could have made that decision for you 100 years ago. At what point in time you make the decision to stick or switch (after the goat is revealed) makes no difference to the probability of winning the car. Switching after the goat is revealed gives you a 2/3 probability of winning, sticking with your original choice gives you a 1/3 probability of winning.

In other words, the correct strategy is to switch.

 

[wollery]

In other words, the correct strategy is to switch.

Yes.

 

[s4zando]

By that reasoning, if you stick you get the one Monty has opened and the unopened door you chose - inasmuch as you never 'get' the goat (the inevitable, irrelevant goat). You're insisting that G+? is of more value than G+?.

The bolded is an interesting way to present the problem to favor 50/50. Each goat includes a door, either way.

But the door is not the prize, its what's behind the door that is represented by the (?). The two (?)'s are not equal, and we know this because at this point in the game, one (?)= a goat, and one (?)= a car. I think it would be more accurately expressed:

G+.33(car)<G+.67(car)

 

[jiggeryqua]

The bolded is an interesting way to present the problem to favor 50/50. Each goat includes a door, either way.

But the door is not the prize, its what's behind the door that is represented by the (?). The two (?)'s are not equal, and we know this because at this point in the game, one (?)= a goat, and one (?)= a car. I think it would be more accurately expressed:

G+.33(car)<G+.67(car)

And that's an interesting way to explain a problem by merely inserting the numbers you're trying to justify. My own formula (G+?=G+?) jokingly addressed an earlier attempt to explain the problem by reasoning that when a goat is shown, the 'other' door is now worth more than the other door. I made that deliberately confusing because really, so was the explanation. It didn't account for why two indistinguishable doors are made different by an unconnected, inevitable, irrelevant goat. It merely claims that whichever door you pick, it's more likely to be the other one. Explaining that 'six of one' is the same as 'half a dozen of the other' doesn't help the innumerate.

 

[Prometheus]

Yes you can.

Look at the 100 door variation and that is clear.

You can pick any door of the 100 and have a 1/100 shot at the prize.

MH then removes 98 of the doors you did not pick, showing goats behind all of them. As I said in the above post, seeing those goats not only gives you information about the door they were in, but about the remaining door you did not pick.

That 99th door (the remaining door that you did not pick) almost definitely has the prize.

Stick with your original door and it is still 1/100. Yet the other door is up to 99/100 (or is it 98/100?) even though originally it would have been 1/100.

Sure there is still a 1/100 chance that the prize was in the door you originally picked. But it is nowhere near 50/50.

Your chance of winning in that instance has clearly changed "partway through it".

Either you do not understand that, or you are perpetuating the marble fallacy.

I don't remember that exactly but it goes something like this: Say you have a bag with 99 black marbles and 1 white marble. Just because there are 2 colors does not mean you have a 50/50 chance of picking the white marble.

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.

If you buy a Powerball ticket, you have something like a 1 in 175 million chance of winning the jackpot. This remains true each and every time you buy a ticket. Nevertheless, something like 1 in 75,000 regular lottery players eventually win a jackpot at some time in their lives. Buy becoming a regular player and shelling out a buck or two per week, you markedly improve your odds of someday being a winner, even though each and every single time you play, you are blowing your money on a 1/175M shot.

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

 

[Modified]

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.

The thing you're missing is that Monty can not eliminate the door that you have chosen. Your analysis would be correct if you have to choose A or B ahead of time without knowing what they will be, then Monty eliminates 98 irrelevant doors, labels the remaining two A and B, and you can then choose to change your selection. But that isn't the game. He is eliminating doors from the set of 99/100 that you have not chosen. There is a 99/100 chance that the car is in that set. Unless he shows you every door, as far as you know it is still a 99/100 chance.

 

[Meed]

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.

Huh? No, you start with a 1/100 chance of having chosen correctly among the only two doors that Monty will not open, and you finish with exactly the same 1/100 chance (if you stick with your original pick).

How could you possibly have a 50/50 chance of picking the right door at the start when the car is behind only 1 out of 100 of the doors?

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.

It's incorrect to say that if 100 contestants played, 99 of them would win. Since chance is involved, it's any number of them could win or lose.

It would be correct to say that if an infinite number of groups of 100 played, that the average number of winners per group would be 99.

You've still got a 50/50 chance of winning your one trial. I understand that this seems counter-intuitive.

It seems counter-intuitive because it isn't right. But I don't understand what the reasoning is behind your conclusion. I think there must be some miscommunication going on somewhere.


ETA: If 100 people played once and each of them had a 50/50 chance of winning, then we would expect 50 out of that 100 to win on average, not 99.

 

[Daald]

Hey let me add the method that helped me get this in case it helps someone. The 100 door method is good but I would probably still have had a problem with it.

So here it is:

Since you always have the choice to switch you do this:

Pick 2 doors. Let's say A,B.

Say out loud "C".

Have Monty resolve between A and B for you

Switch and go with that.

So in essence you were allowed to pick twice from the original problem set...hence 2/3.

 

[Prometheus]

The thing you're missing is that Monty can not eliminate the door that you have chosen. Your analysis would be correct if you have to choose A or B ahead of time without knowing what they will be, then Monty eliminates 98 irrelevant doors, labels the remaining two A and B, and you can then choose to change your selection. But that isn't the game. He is eliminating doors from the set of 99/100 that you have not chosen. There is a 99/100 chance that the car is in that set. Unless he shows you every door, as far as you know it is still a 99/100 chance.

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.

Huh? No, you start with a 1/100 chance of having chosen correctly among the only two doors that Monty will not open, and you finish with exactly the same 1/100 chance (if you stick with your original pick).

How could you possibly have a 50/50 chance of picking the right door at the start when the car is behind only 1 out of 100 of the doors?

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." 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.

It's incorrect to say that if 100 contestants played, 99 of them would win. Since chance is involved, it's any number of them could win or lose.

Then it's a good thing I didn't say that. I said, "of a whole population of contestants, 99/100 of them would win by switching." Of course I am aware that the actual accuracy of that statement depends on the actual size of the population in question, so it's just silly to assume that I intended "whole population" to be read as, "a group of 100".

It seems counter-intuitive because it isn't right. But I don't understand what the reasoning is behind your conclusion. I think there must be some miscommunication going on somewhere.


ETA: If 100 people played once and each of them had a 50/50 chance of winning, then we would expect 50 out of that 100 to win on average, not 99.

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.

 

[Meed]

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."

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.

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.

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

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.

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.

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.

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.

 

[bruto]

I would think the 50/50 option exists only if you know which of the doors Monty will open. In the hundred door game, if you know Monty will always open all but your choice and number 63, say, then you know that the real game is between those two doors. Otherwise, the game is between your door and all the others.

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.

You choose one door, and then Monty says "will you stick, or will you switch your bet from that one to whichever of the other 99 holds the car. Of course only one of those 99 can possibly hold the car. Opening the 98 that don't tells you nothing that you did not know beforehand: that there's only one car.

How is it possible that for any individual who plays the game once, the chance is 50/50, but if a hundred people troop through the studio, each playing the game once, the odds of making a right choice are one in a hundred? Something does not add up.

 

[Meed]

I would think the 50/50 option exists only if you know which of the doors Monty will open. In the hundred door game, if you know Monty will always open all but your choice and number 63, say, then you know that the real game is between those two doors. Otherwise, the game is between your door and all the others.

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.

You choose one door, and then Monty says "will you stick, or will you switch your bet from that one to whichever of the other 99 holds the car. Of course only one of those 99 can possibly hold the car. Opening the 98 that don't tells you nothing that you did not know beforehand: that there's only one car.

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

How is it possible that for any individual who plays the game once, the chance is 50/50, but if a hundred people troop through the studio, each playing the game once, the odds of making a right choice are one in a hundred? Something does not add up.

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

 

[Modified]

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 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.