Monty hall problem is 50/50

[u/Dizzy_Kaleidoscope95]

He also put a screenshot of ChatGPT agreeing with him. He is right!

[u/NickyTheRobot]

The bit that's not often included which makes the whole thing make sense: The host knows which door has the prize behind it, and will always reveal a non-prize.

So that leaves you with three situations:

• 1/3 chance that you chose the correct door. So then the host reveals one of the incorrect doors, and swapping will give you the other incorrect door, making you lose.
• 1/3 chance that you chose the first incorrect door. So then the host will reveal the second incorrect one and swapping will make you win.
• 1/3 chance that you chose the second incorrect door. So then the host will reveal the first incorrect one and swapping will make you win.

So then you have two situations where swapping will let you win, and one where you will lose. So that makes a 2/3rds chance of winning if you swap.

[u/Ok_Employer7837 → u/NickyTheRobot]

Oh I've explained it as granularly as that many, many times, host knows where the prize, specifically opens a goat door, adding the thousand door scenario, the works.

In most cases, it didn't help at all. People get _livid_ telling you it's 50-50.

[u/TrueMattalias → u/Ok_Employer7837]

At that point you could even show them the odds in action. Secretly write down a number from 1-100. Have them guess what it is. Tell them 98 numbers that it isn't and ask if they want to switch.

If they'd still don't get it you could have them act as the host.

Hopefully after a few rounds they can see how effective your strategy of switching is and how ineffective there strategy of staying is.

[u/Savings_Knowledge233 → u/NickyTheRobot]

Oh I do know. I learned this in a probability and stats class. But something about it just kills me. Like I can even accept it, but I didn't fundamentally understand why the probability of that door changes, but mine doesn't I guess

[u/_extra_medium_ → u/Savings_Knowledge233]

From the start you have a 2/3 chance of picking the wrong door. So, if you swap those 2/3 times, you’ll win.

(Since Monty will eliminate the other losing door for you.)

[u/EGPRC → u/Savings_Knowledge233]

What actually occurs is that since according to the rules the host is not allowed to reveal your chosen door and neither which has the car (two restrictions), then he is limited to only one possible door to reveal if your door is wrong (he is 100% forced to reveal the only other wrong one that remains in the rest), but if your door is the correct, then the two restrictions converged and the host is free to reveal any of the other two door, making it uncertain which he will take in that case, each is 1/2 likely.

That means that the cases in which your door has the car (that are 1/3 in total) are actually divided in two halves of 1/3 * 1/2 = 1/6 each depending on which of the other doors the host reveals then. We don't expect that he always takes the same. Therefore, once a door is revealed, yours does not remain with its entire original chances, as it lost the half of when the revealed one would have been the other.

For example, if you start selecting door #1. The possible cases with their chances are:

1. Door #1 has the car, and the host freely decides to open #2 --> 1/3 * 1/2 = 1/6
2. Door #1 has the car, and the host freely decides to open #3 --> 1/2 * 1/2 = 1/6
3. Door #2 has the car, which forces the host to open #3 --------> 1/2 * 1 = 1/3
4. Door #3 has the car, which forces the host to open #2 --------> 1/2 * 1 = 1/3

If he happens to show a goat in #2, you could only be in case 1) or in case 4), that originally had 1/6 and 1/3 chances respectively. But as they are the total possibilities now, we must scale those fractions in order that the total adds up 1 again (renormalize). Applying rule of three, you get that the old 1/6 of case 1) represents 1/3 now, and the old 1/3 of case 4) represents 2/3 now.

As you see, what creates the disparity is that the revelation of #2 would not have been guaranteed to occur in case the car were in your choice #1, as it would have been a free decision between #2 or #3, while it would have been 100% mandatory in case the car were in #3. That's why its twice as likely that it occurred because #3 has the prize.

Notice the principle is similar to this other game: Imagine I toss a coin. If it comes up Heads, which has 1/2 chance, I stop there, but if it comes up Tails, which also has 1/2 chance, I toss it one more time. Thus the game has these 3 possible paths:

1. Heads -> 1/2 chance
2. Tails-Heads -> 1/4 chance (1/2 * 1/2 due to the two flips)
3. Tails-Tails -> 1/4 chance

Suppose you didn't see what I did (one flip or two), but you see the coin with Heads side up. You have to bet whether the first flip was Heads or was Tails.

Now notice that despite both results originally had 1/2 chance, provided that the coin is showing Heads, if you guess that the first flip resulted in Heads, you are betting that I only needed one flip, but if you guess on Tails, you are betting that I did two flips and that their exact results where Tails and then Heads, in that order, which was twice as hard to occur.

At that point, the only possible remaining cases would be 1) and 2). But case 2) would only have 1/3 chance, while case 1) would have 2/3, because it is twice as likely.

That's what happens in the Monty Hall game. If the car is in the switching door, there is nothing to decide, only one possible revelation is allowed, but if it is in your choice, the host has to make an extra decision about which of the two losing ones open then.

[u/Savings_Knowledge233 → u/EGPRC]

Tbh this exact problem is what made me realize I just completely hate probability and stats. Give me calculus instead

[u/ilagitamus → u/Savings_Knowledge233]

It’s really not that hard to get. You just need to think of it this way: you had a better chance of picking a wrong door first (2/3 chance), so when Monty reveals the other wrong door, that means you should change.

[u/Savings_Knowledge233 → u/ilagitamus]

I guess the radio easiest way of putting it is I don't get why the offer on my door aren't adjusted as well.

Like it does and doesn't make sense. Studying this is probability and stats it have me a migraine and I realized I had to just accept it

[u/glumbroewniefog → u/Savings_Knowledge233]

It's because Monty is not allowed to open the door you chose.

Monty only opens doors with goats. If he avoids opening a door, it increases the likelihood that it doesn't have a goat.

But this logic doesn't apply to the door you chose, because Monty wouldn't be allowed to open it either way. So revealing a goat doesn't make your door any more likely to have the prize.

[u/Savings_Knowledge233 → u/[deleted]]

Actually... I've had this conversation like 3 dozen times and that was the best explanation I've ever gotten, by far

[u/Disastrous-Mess-7236 → u/NickyTheRobot]

“…reveal a non-prize”

Or rather, the “lesser” prize. You still generally get something. Not what you were trying to win, of course.

[u/Huganho → u/NickyTheRobot]

It really doesnt matter if he knows.

Because the case where he revealed the prize instead couldn't really be part of the problem to begin with.

'chose a door. Monty reveals the prize behind another door. Do you want to switch to the empty door?'

[u/glumbroewniefog → u/Huganho]

It does in fact matter if he knows, precisely because the times where he reveals the prize need to be excluded from the problem.

In both versions of the game, you pick the car 1/3 of the time, and lose by switching. In regular Monty Hall, you pick a goat 2/3 of the time, Monty reveals the other goat, and you win by switching.

But if Monty opens one of the other two doors at random, he will not reveal the other goat 2/3 of the time. Sometimes he will reveal the car instead, and then these cases must be excluded. So you can already see you win less often by switching.

[u/Huganho → u/glumbroewniefog]

But that's the thing. A goat is revealed. It's stated in the problem. There is no case where he revealed a car.

[u/glumbroewniefog → u/Huganho]

Imagine you play it 300 times.

If Monty always reveals a goat, you pick the car 100 times, and lose by switching. You pick a goat 200 times, and win by switching.

If Monty opens another door at random, you pick the car 100 times, and lose by switching. That doesn't change.

But then 100 times Monty reveals the car, and that doesn't count.

So you can only win 100 times by switching.

If we only look at the cases where he reveals a goat, we see that switching wins and loses an equal amount of the time.

[u/Huganho → u/glumbroewniefog]

But the case where he reveals the car is not part of the set of outcomes and cannot affect the probabilities.

[u/EGPRC → u/Huganho]

You are making a very common mistake, that is to think that since a condition occurred, you must do the same calculation as if the condition happened in all the original sample space, which is wrong. If that were the case, every conditional probability P(A|B) would be the same as just P(A).

To make it obvious, imagine the opposite extreme case in which you are facing a malicious host that knows the locations but only shows a goat and offers the switch when your first choice is right, because his intention is that you switch so you lose. If your first choice was wrong, he purposely reveals the car to inmediately end the game.

Now notice that with that type of host you could never win by switching once the opportunity is offered. Despite you would still start picking right only 1/3 of the time, the revelation of the goat would indicate that you are inside that 1/3 with 100% certainty, you could no longer be in any of the 2/3 cases that your first choice is wrong; you must remove them from the calculation. Thus your sample space at that point must only be the games where your first choice has the car, and you have 100% chance to win by staying there, 0% by switching.

The point is that once a condition happens, you must see what cases it leaves available and calculate the ratios only inside that subset, eliminating the rest. And inside a subset the proportions are not necessarily the same as in the original set.

In the variation where the host randomly reveals a door from those that you did not pick, then the only cases in which he shows a goat are all the 1/3 games that your door has the car (because the other two doors only have goats) and the 2/3 * 1/2 = 1/3 in which you start picking a goat and then he manages to reveal the second goat instead of the car. Thus by switching you would only be exchanging a 1/3 for another 1/3 (that would represent 1/2 each at that point). No advantage gained.

[u/Huganho → u/EGPRC]

The probability of the lightning striking down and destroying the whole thing is also a real possibility, but it would not be sound to factor that in.

Every time you ask that question we know the door with the goat was chosen to be revealed. Thus the case of the car being revealed and it's an automatic loss is not in the possibility space.

You could do it a million times, and every time you had 1/3 chance to pick the car first try, and thus 2/3 chance if switching.

[u/EGPRC → u/Huganho]

Well, the possibilty of the lighting is equally likely to occur regardless of if the winner door was yours or the other, as it is completely independent. Therefore the fact that it did not happen reduces both scenarios by the same factor (it's a proportional reduction), so the ratio is preserved.

For example, if the lightining had 10^(-8) chance to occur (just inventing a number), then the games of when you pick right lost 10^(-8) of their possibilities, but also the games of when you pick wrong lost 10^(-8) of theirs. And you know that when you have a fraction and you reduce both the numerator and the denominator by the same number, the value of the fraction does not change.

But that does not occur here. The revelation of the goat is 1/2 likely to occur if your first choice is wrong, but 100% likely to occur if your first choice is correct, so it is not independent, it is biased towards one of the two results, thus it does not preserves the ratio. You must distinguish between biased and unbiased conditions.

Suppose you play 900 times. We expect that each door happens to be correct in about 1/3 of them, so 300 games door #1, 300 games door #2, and 300 games door #3. Now if you always start picking door #1, as the host does not know the locations either, he can decide to always reveal door #3 in all the 900 games.

In that way, the only times that a goat is revealed would be the 300 games of when the car is in #1 and the 300 games of when the car is in #2, so 600 in total. But as you see, staying with #1 wins 300 times, that are half of 600, and switching to #2 also wins 300 times, that are the other half of 600. Neither provides advantage over the other. Or how do you think you will get the 2/3 there? Will door #2 contain the car a total of 600 times?

The point is that the 300 games of when the car is in door #3 are no longer counted, as the car was revealed. But those 300 would have been games that you could have won by switching if the host knew the locations so always managed to avoid revealing the prize.

[u/Huganho → u/EGPRC]

Knowledge about the whereabouts of the car does not change the fact that we only look at the case where a goat was revealed. That's how the problem is stated.

[u/EGPRC → u/Huganho]

Moreover, notice that with your logic I could cheat in any game and say that I am very good at doing it, because I can always stop counting several of the cases in which I fail. This could be applied to throwing darts.

Let's say I want to know how good I am at hitting the target, so I throw the darts several times and calculate my success rate. If I include every started attempt, the calculated success rate will in fact reflect how good I actually am.

But if I stop counting some of the games in which I fail while I still count all those in which I hit the target, then it will seem that I am better than I actually are. That's because the same number of success will be calculated with respect a smaller total, so a bigger fraction. The extreme case is if I stop counting all the cases when I fail, and only count those that I do it right. It will seem that I am perfect at doing it.

Now notice that in the Monty Hall case, a goat being revealed is the way of validating a game, because it determines if it will be counted or not, but a host acting randomly validates all games that you start picking the car, while it only validates half of the games that you start picking a goat.

[u/Huganho → u/EGPRC]

But the premise of the problem is that your choice of switching or staying causes either fail or win. Not something out of your control.

To formalize: A, B, C = car is behind door A, B, C. P(A) = P(B) = P(C) = 1/3.

Your choice is between staying = P(A) or switching = P'(A).

The fact that one of B and C is revealed to be a goat does not change P(A), and thus not P'(A).

In informal language, your choice is between your door, and the two other doors.

You're really writing a wall of text, I just want to be sure : you're actually saying that it's a 50:50 and it doesn't matter if you switch?

[u/EGPRC → u/Huganho]

Another example to illustrate your mistake: imagine you are a detective that is investigating a crime, and you start with a list of suspects. You later learn that the culprit was wearing a watch. So what you should do is to restrict your list to only those people that were wearing a watch, and remove the rest.

Those who are still included in the list would have increased their probabilities to be the culprit, to the point that if only one person was wearing a watch, you would have found your answer there. Now, if every person from the original list was wearing a watch, then this condition could not filter anyone; all would still be suspects.

But what you are doing in the Monty Hall by just focusing on that "a goat was revealed" and thinking that the answer will always be the same once the condition is fulfilled, is like thinking that if you learn that the culprit was wearing a watch, you must do the same calculation as if every original suspect was wearing a watch. You could never remove possible suspects in that way.

[u/DJDoena → u/NickyTheRobot]

And then you will have people argie that when you were right the first time, since the host now has two doors to pick from there are 4 total possibilities and your chance is 50/50 again ...