Suppose there are 3 doors, only one of which is the winning door. Once you choose one of the doors, I pick one of the two remaining doors and tell you (truthfully) that it is not the winning door. Should you change your choice to the remaining door? The answer is yes because changing your choice gives you 66.7% chance of getting the winning door. Why is the chance 66.7% instead of 50%? Recently I got into a discussion of this probability problem with my sister and no matter how I explain, she says it makes no sense. Does my explanation make sense to you? Suppose I phrase the question this way: Once you choose one of the doors, I tell you that of the two remaining doors, one of them is definitely not a winning door. Do you want to abandon the door you chose and instead choose BOTH of the other doors (one of which is definitely not a winning door)? Realistically, there is no difference between the original question vs how I rephrased it. My sister insists that I'm speaking nonsense.
Your sister is correct. Let’s say the doors are A,B and C. I chose door A and you said door B is not correct. Now I am left with door A and door C and one of them is correct. Therefore there is a 50% chance I am correct, whichever one I choose. This is due to you eliminating one of the options, namely door B from the choices I can choose from.
You essentially get to choose two doors if you switch after the reveal. If it's truly random, then the chance of the prize being behind any one door is approximately 33% for each door, i.e., once you choose any door at this instant, you get only 33% chance of being correct. Now you divide the doors into two groups on the basis of your choice: group 1 containing the one door you have chosen with 33% probability of getting right and group 2 containing two doors you have not chosen with (33*2)%= 66% probability. When out of the second group, one of the doors is shown to not contain the prize (note that the host here is deliberately opening the door not containing the prize), that would mean the 66% chance shifts to the only remaining unchosen door as the sum of probability will remain unchanged for the second group even after one door gets eliminated. P.S. It's horribly counterintuitive.
The monty hall is a bit of stastical illusion since at first it doesn't seem like ot makes sense. You're correct but it's a bit of a pickle for a lot of people to wrap their head around at first. Might want to show a table with the stats or a video to help put it into perspective
I get it but that’s because I already knew it, though maybe it would be hard to understand any explanation immediately... maybe make it visual? I recently learned that during a presentation, 67% of what people remember is the visual part! And only 5% was auditory... the other 27%.... I forgot... maybe you only remember 72% of a presentation? Average. This isn’t relevant at all. I’ll mute myself
Strictly speaking it isn't that you don't make sense because I've heard of this before and it didn't make sense either. I mean, I understand that it is true, but the explanation still doesn't make sense to me.
You're mostly correct, but there are unspoken assumptions in your original question. The probability of the non-winning door being revealed matters. If there's a 100% chance that you'll reveal one of the remaining doors as non-winning, then it's true that changing gives a 66.7% chance to win. But if the chance isn't 100%, and you just happen to reveal a non-winning door, that changes things. For example, lets say you pick a door, and then I pick any of the 3 doors at random to open. If I pick a winning door or the door you chose, we start over, if I pick a losing door that you didn't choose, we continue. In that scenario, even if on the first try I happen to reveal a non-winning door you didn't choose (so you can't tell the difference between my scenario and your original one), changing your choice only gives a 50% chance to win. It's easier to see why this is if you list out all the possible outcomes. Spoiler You pick A, I open A, winning door is A: Start over because I opened your door. You pick A, I open A, winning door is B: Start over because I opened your door. You pick A, I open A, winning door is C: Start over because I opened your door. You pick A, I open B, winning door is A: Continue You pick A, I open B, winning door is B: Start over because I opened the winning door. You pick A, I open B, winning door is C: Continue You pick A, I open C, winning door is A: Continue You pick A, I open C, winning door is B: Continue You pick A, I open C, winning door is C: Start over because I opened the winning door. You pick B, I open A, winning door is A: Start over because I opened the winning door. You pick B, I open A, winning door is B: Continue You pick B, I open A, winning door is C: Continue You pick B, I open B, winning door is A: Start over because I opened your door. You pick B, I open B, winning door is B: Start over because I opened your door. You pick B, I open B, winning door is C: Start over because I opened your door. You pick B, I open C, winning door is A: Continue You pick B, I open C, winning door is B: Continue You pick B, I open C, winning door is C: Start over because I opened the winning door. You pick C, I open A, winning door is A: Start over because I opened the winning door. You pick C, I open A, winning door is B: Continue You pick C, I open A, winning door is C: Continue You pick C, I open B, winning door is A: Continue You pick C, I open B, winning door is B: Start over because I opened the winning door. You pick C, I open B, winning door is C: Continue You pick C, I open C, winning door is A: Start over because I opened your door. You pick C, I open C, winning door is B: Start over because I opened your door. You pick C, I open C, winning door is C: Start over because I opened your door. The "continue" and "start over" possibilities aren't even. When you look at just the possibilities where you keep playing, there's a 50% chance that you end up in a situation where changing will result in a win.
His sister is dead wrong, the Monty Hall problem has been discussed to death for decades. If you're struggling to understand why, it helps for some people if the scope of the experiment is expanded while keeping the rules the same: You have 100 doors, 1 of the doors has a prize behind it and 99 do not. You choose a door, let's say the first door. The host opens 98 of the doors, all of them except for the first door and door number 37. The probability that you picked the door correctly on the first guess was 1%, 1 in 100. Therefore, the probability that the prize was behind one of the other doors is 99%, or 99 in 100. When you eliminate 98 of those other doors, the probability that it was behind one of them is not reduced, it simply condenses the probability onto that remaining door. If you're still struggling, write down on a piece of paper or in a word document or wherever you like all 9 possible variations that you can have, along with if you would win or lose with each of those possibilities. You will find that in 6 of them you win by switching and 3 you win by sticking. There's no arguing against that.
My friend explained it to me and I understood. Basically at First whichever you choose its 1/3, 33%. Moving Forward with the Help of the Person you get a 50% Chance because its 1/2 but, If He tells you that the door which Sure doesnt have the prize isnt the door you Chose, its still 1/2 since its one of the two Doors which can have the prize
Your spoiler is very poorly laid out. 1: Pick door A, door B or C is eliminated, prize is behind door A. 2: Pick door A, door B is eliminated, prize is behind door C. 3: Pick door A, door C is eliminated, prize is behind door B. 4: Pick door B, door A or C is eliminated, prize is behind door B. 5: Pick door B, door A is eliminated, prize is behind door C. 6: Pick door B, door C is eliminated, prize is behind door A. 7: Pick door C, door A or B is eliminated, prize is behind door C. 8: Pick door C, door B is eliminated, prize is behind door A. 9: Pick door C, door A is eliminated, prize is behind door B. These are the only 9 possibilities that have any statistical bearing upon the situation, with the classic Monty Hall problem. 3 you win by sticking, 6 you win by switching. One, four and seven are functionally identical and are counted as one possibility each. When you initially picked a door, the probability of the prize being behind a door you didn't pick is 2/3 or 66.6%. If one of those options is eliminated, it does not change the fact that there was a 2 in 3 chance of it being behind one of the other two doors - all it does is make it so that *if* it was behind one of the other doors, then it *has* to be behind the one remaining door that was left unopened, and thus there is a 2 in 3 chance of winning if you switch to it.
I heard this after I played Zero Escape: Zero Time Dilemma. Had to look up a video on Youtube to understand, also, the explanation is much easier to understand with bigger numbers.
Is this some "careful, sometimes probability is skewed" kind of tale for students? Cuz even if conceptually the probability of the wrong door moves over to the other door, IRL there's no difference between the two unopened doors?
No. I'm describing a variation (a very well studied and debated one) of the classic Monty Hall problem that examines the unspoken assumptions in the original. You can't simply combine possibilities without adding their probabilities together. "You pick A, door B is eliminated" and "You pick A, door C is eliminated" both have the same probability. If you combine them like you have there, you have 9 options but those 9 options aren't all equally likely. Instead of listing the possibilities, you can do a quick simulation. Set the winning door to a random integer from 1 to 3 inclusive, set the opened door to a random integer from 1 to 3 inclusive. The player's chosen door can be set randomly or they can always pick the first door or whatever else, it makes no difference (you can try it both ways to confirm). If the opened door is the same as the winning door or the player's door, then start over with new random doors. Otherwise, check whether the player would win by changing doors. You can try it out yourself here, just click "Run code". I just ran it and changing doors won 500,132 times out of 1,000,000.
Your entire post is changing the basic rules of the Monty Hall problem to suit your own argument, when your rule variation was not discussed or asked for.
It's pointing out a flaw in the typical formulation of (and this thread's description of) the premise. The conclusion of "changing wins 66.7% of the time" is only true if you add further assumptions, so by very slightly changing the circumstances (possibly even without the player's knowledge) while still following the described rules, that conclusion can become wrong. It also demonstrates a flaw in the explanation of "choose both of the other doors", because that explanation sounds good but doesn't quite describe what's happening, as you can see in this example where "choosing both other doors" doesn't increase your chance of winning.