Current Events > My brother cannot comprehend the Monty Hall problem.

(+) Yeah! (+)

tiornys posted...

If I flip a card off of a deck, what are the odds that it is red? If I flip a second card off the deck, what are the odds that it's red if the first card was black? What if the first card was red?

I already said percentages were never my strong suit. That's why I asked for simple terms.

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I think what confuses a lot of people is the idea that each goat is a distinct goat. Same thing that confuses a lot of people on a lot of probability problems.

So let's just pan them each out.

If you choose Goat A and switch, you get the car.
If you choose Goat B and switch, you get the car.
If you choose the car and switch, you get either Goat A or Goat B.
If you choose Goat A and stay, you get Goat A.
If you choose Goat B and stay, you get Goat B.
If you choose the car and stay, you get the car.

It's really pretty straight-forward.

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GATTJT posted...

So I think at this point I'm confident in saying you're just pulling our legs.

You'd be wrong and also the reason your brother doesn't understand since you'd rather give up than accept someone just not getting it.

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markconigliaro posted...

Then don't think of it as the doors being eliminated, think of it as the host picking all those doors as a single group, and the host wins if the prize is in any of those doors. You get 1 door, the host gets 999 doors. The host then offers to trade his 999 doors for your single door. Obviously his group has a better chance of winning, so you switch.

Same with 3 doors. You get 1 door, the host gets 2. He offers to trade your single door for his 2 doors, and if either of those two doors have the prize, you win.

But it's not a group once they've been eliminated! The choice at that point is 50/50. Maybe when all the doors were in play it was whatever the odds were originally but it's now 50/50 since only two choices are left.

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dave_is_slick posted...

You'd be wrong and also the reason your brother doesn't understand since you'd rather give up than accept someone just not getting it.

Simplest way I can think to explain it is just increasing the number of doors. If there are 10 doors, then your one door is highly unlikely to be a car, whereas the other 9 doors are highly likely to contain the car in one of them. Like someone else said, you choose a door, the host gets the other 9 but decides to reveal what 8 of his doors have. Then he offers to give you his set of doors. It is statistically likely he has the car, so you should switch. Yes, you can wind up getting a goat, but that would be just bad luck at that point. This holds true for any amount of doors. Except 2 of course, but then that is truly 50/50.

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dave_is_slick posted...

But it's not a group once they've been eliminated!

The opened doors are not eliminated, they're just opened so you can see what was behind them. Think of it like that.

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just because there are two outcomes does not mean it is automatically 50/50

you can only either win or lose at the lottery, that does not mean there is an equal chance of either happening

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50/50 either I'll wake up tomorrow morning or die in my sleep tonight. Actually now that I think about it, all probability might as well be 50/50, either it's true or false.

Think we just solved all the questions of the universe.

dave_is_slick posted...

I already said percentages were never my strong suit. That's why I asked for simple terms.

If I just pick a door randomly, there's only a 1 in 3 chance that it's the car. It's pretty decent, but that also means there's a 2 in 3 chance I DIDN'T pick the car. And if you were the host, you know you're not going to show me the door with the car behind it, so that just leaves the last one. It's your 2/3 against my 1/3.

dave_is_slick posted...

You'd be wrong and also the reason your brother doesn't understand since you'd rather give up than accept someone just not getting it.

you're not "not getting it". you're trying not to get it. you keep saying the same (wrong) thing over and over, even after you've been told that it's wrong, and shown why.

you want people to stop thinking you're trolling? read what's being told to you. learn from what's being told to you. going "nuh-uh" for 100 posts is only wasting everyone's time.

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Delirious_Beard posted...

you can only either win or lose at the lottery, that does not mean there is an equal chance of either happening

Nobody is eliminated in the lottery, everyone is in play. That's not a good example at all if you're trying to help me get this.

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Delirious_Beard posted...

just because there are two outcomes does not mean it is automatically 50/50

you can only either win or lose at the lottery, that does not mean there is an equal chance of either happening

This is only the illusion of having only two outcomes. Just like with the MH problem, there are multiple distinct lose scenarios. One where the winning numbers were 1, 1, 1, 1, 1, 1; one where the winning numbers were 1, 1, 1, 1, 1, 2; one where the winning numbers were 1, 1, 1, 1, 1, 3...

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dave_is_slick posted...

GATTJT posted...
Using your "right or wrong" logic, my chances of winning the lottery are 50/50, either I do or I don't.
I'm not. What I am doing is seeing that it ultimately comes to two choices at the end. That is not the case with the lottery

No, you are. Your possibilities in the lottery are either win or lose, but those hold different probabilities. This is exactly the same

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3 choices equals 33% chance. 1 choice is removed, and it becomes 50%. The notion of winning if you switch is predicated on the assumption you know which door the prize is likely behind. It's a guessing game, 50/50 that's what it is.

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Medussa posted...

you're not "not getting it". you're trying not to get it.

Bullshit. Why in the fuck would I do that? If you're not interested in trying to help, get the fuck out of here. That's how people learn, by getting what's "wrong" repeatedly explained in different ways until one just clicks. Nothing has clicked for me.

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How dumb is your brother?

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008Zulu posted...

3 choices equals 33% chance. 1 choice is removed, and it becomes 50%. The notion of winning if you switch is predicated on the assumption you know which door the prize is likely behind. It's a guessing game, 50/50 that's what it is.

It would be 50/50 if one door was removed before you chose a door at all, but that is not the case. You select a door while you have a 1/3 chance, then the host, and I need to reestablish that he knows what is behind every door, opens a door to reveal a goat. Your door is still at 1/3, the opened door's chances do not split equally into the two closed doors. Do you really think that, if there were instead 10 doors, and you choose one, that door's chances of having a car suddenly increase to 1/2 simply because 8 other doors were revealed to have goats?

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dave_is_slick posted...

Why in the fuck would I do that?

people do it all the fucking time. playing dumb on the internet to get people to bash their heads into the wall for hours of their lives.

dave_is_slick posted...

If you're not interested in trying to help, get the fuck out of here.

10 people have already tried to help you. reread through the topic. nothing I can say will add to the dozens of posts that already explained it to you.

dave_is_slick posted...

Nothing has clicked for me.

I don't believe you, but for shits and giggles, how about we try something different. you tell us why your reasoning is correct and we're all wrong. maybe we'll figure out where things are breaking down.

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dave_is_slick posted...

But it's not a group once they've been eliminated! The choice at that point is 50/50. Maybe when all the doors were in play it was whatever the odds were originally but it's now 50/50 since only two choices are left.

There are a bunch of responses since you last posted this but I think main focus should be on convincing you, using the 1000 door example, that the doors chosen by the host still matter for the probability. What this means is that your door is a 1/1000 chance of being right while one of the doors the host leaves has a 999/1000 chance of being right.
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dave_is_slick posted...

I already said percentages were never my strong suit. That's why I asked for simple terms.

Would you accept that the odds are different for the second card depending on what the first card was?

dave_is_slick posted...

Bulls***. Why in the f*** would I do that? If you're not interested in trying to help, get the f*** out of here. That's how people learn, by getting what's "wrong" repeatedly explained in different ways until one just clicks. Nothing has clicked for me.

Let's go back to the lottery example.

You get to pick either one set of lottery numbers or all but one set of lottery numbers.

Which do you go for? Obviously the "all but one" option.

If I then say "Hey, did you know that one of those lottery tickets you have is a loser?" that shouldn't impact anything. You already knew you had a loser. In fact you have tons of them, because only one of your millions of tickets is the right combination.

So now I tell you "Hey, all but one of your lottery tickets are guaranteed losers". Has anything changed?

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Would you rather pick 1 door out of a million, or 999,999 doors?

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DevsBro posted...

I think what confuses a lot of people is the idea that each goat is a distinct goat. Same thing that confuses a lot of people on a lot of probability problems.

So let's just pan them each out.

If you choose Goat A and switch, you get the car.
If you choose Goat B and switch, you get the car.
If you choose the car and switch, you get either Goat A or Goat B.
If you choose Goat A and stay, you get Goat A.
If you choose Goat B and stay, you get Goat B.
If you choose the car and stay, you get the car.

It's really pretty straight-forward.

If someone doesn't get it after reading this, they're never going to get it.
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If you dont understand this problem, try Gettier Problems/cases

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It didnt "Click" for me, until i heard a variation with more Animals. More duds (999 duds, 1 prize)

But be careful about going overly complicated on it.

I have no fucking clue how i would go around actually explaining it to someone else...

then again im... not good at explaining things.

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As others have stated, the problem confuses people because it's stated in an (intentionally or not) inherently misleading way.

The culprit verbiage is "Would you switch?" This leads people into thinking that they are being presented with a 50/50 scenario. They are not. A better way to phrase it is, "Now, as previously (in the first Round), you have the opportunity to select one of the three doors. This time (in the second Round), one of the three doors is open. Now, given you can select one of the three doors (including the door that has already been revealed), would you select the same door that you did the first time?"

If you state it like this, then people get that they had a 33% chance of selecting the right door in Round 1. However, now (in Round 2) with one of the doors open (they can pick the open door), it makes sense to choose a different door. This way, you increase your odds to a 66% chance of selecting the right door. Essentially, by selecting a different door/not re-selecting the same door, you're giving yourself two opportunities to select the right door instead of one.

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GATTJT posted...

The opened doors are not eliminated, they're just opened so you can see what was behind them. Think of it like that.

This is a very helpful way to think about it.

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Medussa posted...

people do it all the f***ing time. playing dumb on the internet to get people to bash their heads into the wall for hours of their lives.

Those people are fucking pathetic and if you could see post history you'd see I've consistently held that view.

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Medussa posted...

I don't believe you, but for s***s and giggles, how about we try something different. you tell us why your reasoning is correct and we're all wrong. maybe we'll figure out where things are breaking down.

Also, you don't believe me, yet if you read the topic like you keep telling me to, you'd see others have said they just weren't getting it until the right explanation for them was given. Are these people liars?

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I'm not going to argue it anymore.

explain how post #102 is incorrect.

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Wait, how is this even a debate? If someone walked in blind after all the other doors were opened and then picked one of the two remaining doors, that would be a 50/50

But not if you have the extra information gained from the start. Its not 50/50 any more. How do some people not understand this? It doesnt even have anything to do with the wording of the question.

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GATTJT posted...

It would be 50/50 if one door was removed before you chose a door at all, but that is not the case. You select a door while you have a 1/3 chance, then the host, and I need to reestablish that he knows what is behind every door, opens a door to reveal a goat. Your door is still at 1/3, the opened door's chances do not split equally into the two closed doors. Do you really think that, if there were instead 10 doors, and you choose one, that door's chances of having a car suddenly increase to 1/2 simply because 8 other doors were revealed to have goats?

If only two doors exist, it's 50/50. The act of elimination just increases yours odds that you picked the winner.

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Tag

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I get it but this kinda stuff if why I hate probability and stats. Most of it is just stupid and theoretically should be interesting but never is in practice

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I think the problem arises because people aren't pessimistic enough lol.

If you have 3 options at the start and only one has the car, then you only have a 1/3 chance of picking the car, and a 2/3 chance of picking a goat. In other words, you will probably pick a goat door on your first choice. Emphasis on probably. Don't be deceived by the "chance" of getting it right the first time. Having a chance isn't good enough. You could get lucky, but you probably won't. The chances are greater you will pick the "wrong" option your first guess. Remembering this is key.

The host isn't going to reveal the car. They will only reveal a goat. So after the host reveals another option as a goat, this changes nothing about your first choice. You only had a 1/3 chance of picking the car, so while you may have done that first try, you still probably picked a goat the first time. You essentially need to assume you will pick a goat the first time (since you would have been more likely to in terms of probability), and will then have another goat revealed to you. At this point the final door most likely has the car, (not definitely though) so switching makes the most sense in terms of probability.

It is easy to get confused if you make the mistake of conflating higher probability with being the "right" choice. There is no sure-fire way to guarantee you get the car. It's not impossible that you picked the car the first time, and could be switching to a goat. It's just less likely. Higher probability is never a guarantee.

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Maybe I can break some brains if I point out that the guy who says it's 50/50 at the second selection is technically right, and here's why. There's a 2/3 chance that switching in general will win you the car. There's a 1/3 chance that keeping your door in general will win the car. Choosing randomly, that means we'll switch half the time and keep half the time. That means that, by chance, we'll still switch doors and win 1/3 of the time and that, by chance, we'll keep doors and win 1/6 of the time, meaning we'd still expect to win 50% of the time choosing randomly.

Note, however, that a 50% probability of you winning isn't the same as saying it doesn't matter which one you pick. If I have a recording of this year's Super Bowl and play it out, the Chiefs will win 100% of the time. If I say the Chiefs will win, I'll be right 100% of the time. If I say the 49ers will win, I'll be wrong 100% of the time. If I flip a coin, I'll still be right half the time but it's not the same as saying it doesn't matter which one I pick.

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And if the refs were fair the 49ers win 100% of the time

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Serious Cat posted...

Maybe I can break some brains if I point out that the guy who says it's 50/50 at the second selection is technically right, and here's why. There's a 2/3 chance that switching in general will win you the car. There's a 1/3 chance that keeping your door in general will win the car. Choosing randomly, that means we'll switch half the time and keep half the time. That means that, by chance, we'll still switch doors and win 1/3 of the time and that, by chance, we'll keep doors and win 1/6 of the time, meaning we'd still expect to win 50% of the time choosing randomly.

Note, however, that a 50% probability of you winning isn't the same as saying it doesn't matter which one you pick. If I have a recording of this year's Super Bowl and play it out, the Chiefs will win 100% of the time. If I say the Chiefs will win, I'll be right 100% of the time. If I say the 49ers will win, I'll be wrong 100% of the time. If I flip a coin, I'll still be right half the time but it's not the same as saying it doesn't matter which one I pick.

What witchcraft is this?

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What if i want the goat?
That's some good BBQ right there

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ROBANN_88 posted...

What if i want the goat?
That's some good BBQ right there

the car can be exchanged for money, which can buy many goats.

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NeoShadowhen posted...

I understand the argument. Has it been tested? Seems like it would be wildly easy to program.

Indeed it's easy to program a test for this, there are many examples on github. Here's a simple one:
https://github.com/spcask/monty-hall-sim

And yes, the chances of winning actually DO increase when switching doors. It's not just theoretical talk.

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Of fucking course this topic is about to hit 150 posts
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This is something I can understand from a statistical point, but in practice it still doesn't click for me. Doesn't help there are so many different examples that change the context.

In practice, to me, there are two choices: 1/3, then 1/2. However the 1/3 choice is completely arbitrary, as that third door is simply thrown out. There may be two Goats (1 and A), but no matter what your first choice will not pick Goat A. You either picked Goat 1 or the Car. So when you have that second choice...has anything changed? Now it's Goat 1 or the Car, again. The odds never really changed in practice.

It's simply not intuitive for me, idk. That first choice doesn't seem real.

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so basically
first chance to get a car is 33%
but i have a 66% chance of getting a goat

and if i do get the goat on the first pick, which is 66% chance, and the other goat is removed, i have a better chance of taking the car by switching my pick?

am i getting this right?

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Blue_Dream87 posted...

This is something I can understand from a statistical point, but in practice it still doesn't click for me. Doesn't help there are so many different examples that change the context.

In practice, to me, there are two choices: 1/3, then 1/2. However the 1/3 choice is completely arbitrary, as that third door is simply thrown out. There may be two Goats (1 and A), but no matter what your first choice will not pick Goat A. You either picked Goat 1 or the Car. So when you have that second choice...has anything changed? Now it's Goat 1 or the Car, again. The odds never really changed in practice.

It's simply not intuitive for me, idk. That first choice doesn't seem real.

It hinges on what you know before you make your choice. You know that you only have a 1 in 3 chance of picking the car on your first pick, in which case there are goats behind each of the unselected doors. You know that you have a 2 in 3 chance of picking a goat on your first pick, in which case there are one of the goats and the car behind the remaining doors.

Notice that no matter what you pick, the host will be able to show you one of the goats. If you pick the car (which you will do 1/3 of the time), he will show you a goat. If you pick one of the goats (which you will do 2/3 of the time), he will show you the other goat. The host showing an arbitrary goat after the choice doesn't change the fact that you still only have a 1/3 chance of picking the car on your first choice and are always better off switching.

So yes, you have a 50-50 shot by picking either of the two remaining doors at random, but you also have a 50-50 shot at picking the winner of a boxing match at random with no information. The key is that knowing the initial probability gives us weighted odds. Showing the goat behind one of the doors is the equivalent of showing us that the underdog in a boxing match showed up, but doesn't change the odds that he wins.

The seeming paradox comes because you'll win 66.6% of the time switching, 33.3% of the time staying with your initial selection, and 50% of the time making a truly random selection. In cases like this, 50/50 isn't the same as not mattering because not all randomness is equally random.


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it's better to switch since you don't want to stick with your initial ass percentage. Out of 3, your pick is always 33%. Out of 10, your pick is always 10%. If other choices are eliminated, you can stick with your initial 10% or switch to 90%, what's left over. Even if the numbers are wrong, it gets the job done for me conceptually.

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Guys, play with it yourself if you want anectodal evidence:

http://www.shodor.org/interactivate/activities/SimpleMontyHall/

Try it a bunch of times, either constantly choosing the other door or constantly choosing the same door.

I just did it twenty times, choosing the different door each time, end result of doing so was winning the car 16 out of 20 times, which is due to the small sample size statistically higher than you'd expect. But the more you do it, the closer you'd get to seeing a 66% success rate switching choices versus only 33% if you stayed.

This isn't a debate, the people choosing to stay are wrong on a statistical standing.

Edit, up to 50 tests and at 37 wins versus 13 losses by switching all 50 times. I could keep going, but that's enough.

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I like how this topic is just people re-explaining the math for the hundredth time in slightly different ways. If people don't get it by now, they ain't getting it.
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Serious Cat posted...

It hinges on what you know before you make your choice. You know that you only have a 1 in 3 chance of picking the car on your first pick, in which case there are goats behind each of the unselected doors. You know that you have a 2 in 3 chance of picking a goat on your first pick, in which case there are one of the goats and the car behind the remaining doors.

Notice that no matter what you pick, the host will be able to show you one of the goats. If you pick the car (which you will do 1/3 of the time), he will show you a goat. If you pick one of the goats (which you will do 2/3 of the time), he will show you the other goat. The host showing an arbitrary goat after the choice doesn't change the fact that you still only have a 1/3 chance of picking the car on your first choice and are always better off switching.

So yes, you have a 50-50 shot by picking either of the two remaining doors at random, but you also have a 50-50 shot at picking the winner of a boxing match at random with no information. The key is that knowing the initial probability gives us weighted odds. Showing the goat behind one of the doors is the equivalent of showing us that the underdog in a boxing match showed up, but doesn't change the odds that he wins.

The seeming paradox comes because you'll win 66.6% of the time switching, 33.3% of the time staying with your initial selection, and 50% of the time making a truly random selection. In cases like this, 50/50 isn't the same as not mattering because not all randomness is equally random.

Thank you, it's starting to click together.

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jumi posted...

What witchcraft is this?

The average win rate assuming equally split take/keep door decisions is 50%


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