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Started By
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re: Let's discuss the Monty Hall problem (probabilities, odds)
Posted on 8/12/17 at 12:53 pm to TheArrogantCorndog
Posted on 8/12/17 at 12:53 pm to TheArrogantCorndog
quote:
You have a 1/3 chance of being right from the outset... just because he showed you a goat, doesn't change those odds, you still have a 1/3 chance of being right... 3 doors, and you picked one... the goat increases the odds of the last door to 2/3 by negating the goat's door to zero
He adds a value input to the remaining door, by stripping away the value of the goat's door... your initial door is still at 1/3 because you picked it with no added value input
You have a 1/3 chance of being right from the start, and a 2/3 chance of being right if you switch
I get all of this.
Here is my question, though. There are two goats, so no matter what door you pick, he's going to be able to show you a door with a goat behind it.
When he asks you if you want to switch, if you ignore the door you initially picked, you have a 50/50 chance of picking the door with prize behind it at that point. Is that incorrect?
1
Posted on 8/12/17 at 12:54 pm to TigrrrDad
quote:You are wrong even if you consider how it's phrased in the OP.
Tell me where my scenario is wrong then.
On the show, Monty knows where the goats are, so the odds are like so:
Odds of choosing right initially: 1/3
Odds of winning if you switch after he reveals a goat: 2/3
Now, the way it's phrased in the OP, the host doesn't necessarily know what's behind the door he opens, right? With that assumption, he could reveal the car, in which case you lose immediately. Your initial 1/3 odds did not change, because you didn't have a chance to switch. Now, since he revealed a goat, your odds do not magically increase to 50/50 by sticking with your initial choice. They remain 1/3, and switching STILL increases your odds to 2/3.
The ONLY way your odds could ever be 50/50 is if the host reveals a goat BEFORE you make your initial selection.
Posted on 8/12/17 at 12:55 pm to ShortyRob
quote:
How well are your fingers fitting in your ears
LOL
That's precisely what you are doing, because you won't show where my scenario is wrong.
Posted on 8/12/17 at 12:55 pm to Korkstand
quote:
On the show, Monty knows where the goats are,
He's intentionally ignoring this detail to cover up for the fact he's too stupid to understand basic probability.
Posted on 8/12/17 at 12:56 pm to Korkstand
I choose door # 1. #3 is opened and reveals a goat.
The car is behind #1:
I switch, I lose.
I stay, I win.
The car is behind door #2:
I switch, I win.
I stay, I lose.
Show me how this is not 50/50.
This post was edited on 8/12 at 12:04 pm
The car is behind #1:
I switch, I lose.
I stay, I win.
The car is behind door #2:
I switch, I win.
I stay, I lose.
Show me how this is not 50/50.
This post was edited on 8/12 at 12:04 pm
This post was edited on 8/12/17 at 12:57 pm
Posted on 8/12/17 at 12:56 pm to WestCoastAg
quote:
you have a 1/3 chance at choosing the car when you initially pick say door 1. the game show host is going to open door three knowing that a goat is behind there. so now, you know that there is actually only one remaining goat for the three doors. so now your odds of choosing the car have gone up to 2/3 if you change your choice to door
You know he's going to show you a goat no matter what. Just ignore your initial pick.
You are going to have a choice between two doors in the end no matter what. One will have the prize, one won't. 50/50. I've never understood this.
Posted on 8/12/17 at 12:58 pm to BayouBengals03
quote:Yes, it is incorrect, because your initial selection influences Monty's selection.
When he asks you if you want to switch, if you ignore the door you initially picked, you have a 50/50 chance of picking the door with prize behind it at that point. Is that incorrect?
Posted on 8/12/17 at 1:00 pm to Korkstand
Not according to the OP - in the OP you always pick #1 and he always picks #3, which always has a goat. The probability scenarios being discussed here vary from the OP.
Posted on 8/12/17 at 1:01 pm to TigrrrDad
quote:Again? I don't know what it will take for you to get it.
Tell me where my scenario is wrong then.
quote:
I choose door # 1. #3 is opened and reveals a goat.
That is the problem stated in the OP.
The car is behind #1:
I switch, I lose.
I stay, I win.
The car is behind door #2:
I switch, I win.
I stay, I lose.
50/50
You started the game by choosing between 3 doors. Your odds are definitely 1/3 there, right?
OK. At that point, no matter what happens, your odds of being right with that selection CANNOT change. Consider if there were 100 doors, and you chose 1. If Monty opens 98 other doors, and leaves you with yours and one other door, are your odds 50/50 at that point if you switch? Obviously not, because there is an extremely high chance that the other door has a car if you've seen 98 goats so far.
But if you stick with your first choice, do you think it's just as likely to be correct as the final door that Monty hasn't yet opened?
Posted on 8/12/17 at 1:03 pm to Korkstand
He's going to show me a goat no matter what, though, right? Or am I misunderstanding how the game show works.
I understand whole probability of it. It is likely I picked a goat (66.6%). So, if he shows me the other goat, and it's likely that I picked a goat, switching makes total sense.
I understand whole probability of it. It is likely I picked a goat (66.6%). So, if he shows me the other goat, and it's likely that I picked a goat, switching makes total sense.
Posted on 8/12/17 at 1:03 pm to TigrrrDad
quote:That interpretation of the OP is nonsensical. If you interpret the problem this way, then there is no sense having 3 doors to begin with, because the contestant doesn't have the option to choose at first.
Not according to the OP - in the OP you always pick #1 and he always picks #3, which always has a goat. The probability scenarios being discussed here vary from the OP.
Posted on 8/12/17 at 1:04 pm to BayouBengals03
quote:Have you read the thread? Several posters have explained why the odds are never 50/50.
You are going to have a choice between two doors in the end no matter what. One will have the prize, one won't. 50/50. I've never understood this.
Posted on 8/12/17 at 1:04 pm to TigrrrDad
quote:That would make sense if a goat was revealed first before you picked a door.
No, you won't - because I laid out every option in the OP scenario.
Tell me where my scenario is wrong then.
I choose door # 1. #3 is opened and reveals a goat.
That is the problem stated in the OP.
The car is behind #1:
I switch, I lose.
I stay, I win.
The car is behind door #2:
I switch, I win.
I stay, I lose.
50/50
But it wasn't. You picked door #1 first. Therefore you had a 33% chance of choosing right since you picked first.
This post was edited on 8/12/17 at 1:06 pm
Posted on 8/12/17 at 1:05 pm to Korkstand
I don't really care about this example, tbh.
I used to argue about the TV show Deal or No Deal all the time, and how you gain no probability advantage by switching your case at the end, because Howie Mandel and nobody else has any idea what is in either case.
I used to argue about the TV show Deal or No Deal all the time, and how you gain no probability advantage by switching your case at the end, because Howie Mandel and nobody else has any idea what is in either case.
This post was edited on 8/12/17 at 1:08 pm
Posted on 8/12/17 at 1:05 pm to Korkstand
Just like Vos Savant, they are straying away from the OP scenario. In Vos Savant's scenario, she even gave the option of the car being behind door #3 when running through all possibilities - wheras in the original scenario she proposed she already said a goat was behind door #3.
This post was edited on 8/12/17 at 1:06 pm
Posted on 8/12/17 at 1:07 pm to BayouBengals03
quote:Yes, he will always show you a goat. However, it does not matter as far as the probabilities are concerned AFTER he has revealed a goat, whether intentionally or not.
He's going to show me a goat no matter what, though, right? Or am I misunderstanding how the game show works.
If Monty opens randomly, then he may reveal a car, and your initial 2/3 chance of losing plays out without the opportunity to increase your odds.
BUT
Even if he does open randomly, ONCE HE REVEALS A GOAT, THE REMAINING DOOR ALWAYS HAS 2/3 CHANCE OF BEING A CAR.
I know it can be hard to wrap your mind around, but that's the way it is.
EDIT: I am thinking more about my last assertion here.
This post was edited on 8/12/17 at 1:10 pm
Posted on 8/12/17 at 1:07 pm to Korkstand
quote:
The ONLY way your odds could ever be 50/50 is if the host reveals a goat BEFORE you make your initial selection.
That's wrong. If the host is an equal player, i.e. he's picking at random, and he reveals goats until one door is left, the odds of him doing that unlikely thing are the same as the odds that you picked the winning door in the first place, and that makes switching 50:50.
Posted on 8/12/17 at 1:07 pm to TigrrrDad
quote:
Yes, I grew up watching Let's Make A Deal.
And my response to what I quoted still stands.
I choose door # 1. #3 is opened and reveals a goat.
The car is behind #1:
I switch, I lose.
I stay, I win.
The car is behind door #2:
I switch, I win.
I stay, I lose.
50/50
while that is correct, you are not considering all factors of the problem.
Posted on 8/12/17 at 1:09 pm to TigrrrDad
quote:
What is incorrect about the scenario I posted?
Nothing technically, but you are ignoring the door which the host revealed.
The host is allowing you to essentially switch to the option of 2 of the 3 doors as opposed to keeping your single door.
It is true and in a "single selection" you have a 50/50 shot of being correct or incorrect. But that isn't really the case here as the car was placed within one of the three doors and has not been moved so it isn't a 2 door problem.
This post was edited on 8/12/17 at 1:20 pm
Posted on 8/12/17 at 1:10 pm to Spock's Eyebrow
quote:Not quite, because he doesn't have the option of opening your chosen door.
That's wrong. If the host is an equal player, i.e. he's picking at random, and he reveals goats until one door is left, the odds of him doing that unlikely thing are the same as the odds that you picked the winning door in the first place, and that makes switching 50:50.
Your odds of being correct can never increase from 1/3 if you don't switch.
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