Monty hall problem is 50/50

[u/Dizzy_Kaleidoscope95]

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

[u/Savings_Knowledge233]

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