It WAS a 33.33 percent chance and still would be if they didn’t show the goat behind on of the doors. These people seem to be ignoring the fact that this is not the same math problems as the original. It is now a 50 percent chance. They are trying to be cool by over thinking the issue. This is math, not psychology.
I think I’ve wrapped my head around it.
If you picked a goat in first place (when you had three choices) switching would always result in a car. The odds of you picking a goat, initially, are 2/3. Picking to switch is basically betting that you had a goat in the first place.
Only if they don’t show you the goat behind one of the doors.
I gave 2 answers because you did not specify which door is chosen by who. That actually affects the outcome due to conditional probability. Depending on which is chosen, it changes the answer from 1/2 to 2/3. People assume the probability of winning is spread equally among all possible choices.
I also find the prior knowledge of the host as a confound to the whole situation as well.
Well, no. The reasoning behind changing your door being correct 66% of the time would be that you now have knowledge your outcome can either be car or goat, not goat, goat, car.
Initially you had a 66% chance of getting a goat, right? With only two outcomes (goat or car) you now know that if you had a goat in the first place (which is most likely the case) that you should change.
So then, if they showed the car and let you pick between only the 2 goat doors you would still only have a 66 percent chance of getting a goat and 33 percent chances that it’s a car?
What if there were two cars and 1 Goat?
Ok, I think I get ya. Right, I think the part that confounds people most is the issue of conditional probability as it applies here. The prior knowledge of the host does impact the problem - in other words, the problem as originally stated (I just paraphrased the problem, I didn’t design it or come up with it) says the gameshow host reveals to the contestant a door with a goat behind it. It doesn’t really say that there would be a chance that the host would reveal a door with the car behind it. Then again, it wouldn’t make sense for the host to ever present the question as stated if there was any chance that he’d reveal the car in the first place. Thus, the selection for which door is opened by the host is not at random. I suspect that many people without having covered any type of stats theory wouldn’t quite get how to put the fact that the door selected by the host isn’t really random into a probability equation. Most people just look at the problem as “well, there are only two doors left. There’s a goat behind one, a car behind the other. Must be a 50/50 chance with either door, so it doesn’t matter.” Which as you stated, is false if the host’s selection is not entirely at random. Nice work.
“I’ll take what’s in the Boxxy, Wayne!”
I also want to bet on Ben Richards.
Brock
phantom50
No. Then you’d have lost as there was only one car.
The situation you’re given is that you have three options. You make a choice and someone tells you that one of your other options is unavailable to you now- that you have the ability to chose between the two remaining options (one of which must be correct).
That’s the state of affairs: the question is do you change doors or not.
If the odds don’t change when a goat is revealed, why should they when are car is revealed? These numbers are magical and can change a 50 percent chance into a 33 percent chance. I’m sure they could turn one of the goats into a 2nd car cause apparently in this world, anything can be anything.
Ok, try and follow me on this one.
If you only had the one door or the other situation, as a standalone, then yeah, it’s 50/50. I’m inclined to agree those odds are simpler to understand and explain.
However: the situation is more complex than that. You are not being given the choice amongst three options anymore- you now actually have had your odds compounded. Think of it in terms of wanting to get it wrong in the first place. You had a good chance of getting it wrong in the first place, as you had only one car behind those three doors. But- with the knowledge that either your door or the other door has the car now you have been given a better chance and picking the right door after changing your choice.
The math is simply 1:3 divided into 1:2. You’re playing off the chance you got it wrong, not right, by changing your door at the end. This second chance opportunity lets you bet against yourself, initially. Which you couldn’t do otherwise.
The odds of you getting it wrong in the beginning were 2/3. Make any more sense?
Yeah, I know that. That’s not math, that’s assumption.
No its probability.
Probability that is derived from an assumption. It still doesn’t change the fact that by revealing the goat it went from a 1/3 chance to a 1/2 chance. Acting like that doesn’t change the “probability” is silly.
In fact your chances improved that you picked the right door in the first place.
Let me put it this way, after the goat is revealed he gives you the chance to choose another door. How many doors are there to choose from?
It’s only a 1/2 chance that you picked the car the first time around. Now the chance of picking the car after a goat is revealed is 2/3 due to the knowledge that one of the doors ISN’T a car. Check the wiki article on the problem, it’s got a good visual that explains it.
This is so easy to understand. I don’t know why it is so hard.
You have a 2/3 chance of picking the goat. Therefore you are most likely to pick the goat. When the other goat is revealed, you then know that you don’t have to worry about what is behind that door. Your odds improve to 2/3 instead of 1/2 because there are two goats to initially choose from. By switching doors you flip those odds around putting you at favor of winning the car… because you probably choose the goat at first..
If you didn’t choose the goat at first, you will loose. But your odds of choosing the car at first were only 1/3..
Switch everytime and you will win 2/3 of the time.
2/3rds of the time.. it works.. everytime.
Here’s a stab at another visual (doesn’t scale well for large numbers, but for the size of this problem it gets the point across)… again realize the host had prior knowledge of which doors had a goat.
G=goat, C=car
These are the possible permutations. I labeled the goats with numbers to distinguish the redundant scenarios. You can quickly see it could be reduced down to only 3, as delineated by the dotted lines. I’m using column one as the contestant’s choices.
Columns 2 and 3 represent the host’s possible selections after the contestant has chosen. The host knows which door had the goat before he revealed it.
The goats with a strikethrough represent what the host revealed. In situations 1-4, there is 100% chance that the host will only be able to pick from one door (bc the other one has the car). In situations 5 and 6, the host has his choice between 2 goats, it doesn’t really matter. In essence, 2/3 of the time you know absolutely that the host has intentionally revealed a door because the other door must have the car. The other 1/3 of the time it’s a tossup.
Going back to the contestant now, the contestant has a choice between 2 doors - the original door (column 1) and the remaining unopened door. Look at the possible choices if the contestant were to switch. 2/3 of the choices reveal a car, only 1/3 reveals a goat. Thus, 2/3 of the time switching will reveal you a car.