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Marilyn vos Savant and the Monty Hall Problem (2015)

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Re: Marilyn vos Savant and the Monty Hall Problem (2015)

#131
post #60

As stated, the answer is indeterminate. To demonstrate, consider these scenarios. - Savant's Monty: The host knows where the car is, and wants to prolong the game. Switching wins 2/3 of the time. - Ignorant Monty: The host has no idea where the car is. Revealing the goat was luck and your odds remain 50-50. - Malicious Monty: The host knows where the car is, and wants you to lose. The fact that the host didn't reveal…

Ignorant Monty can't exist. Monty is the host of a game show, not a contestant. He's not trying to win a prize, and if he opens a prize door, it's a production failure. Whether he knows or not, somebody told him which door to open, and that door does not have the prize behind it. Malicious Monty, however, can exist. He only ever opens a second door when you've picked correctly, but when you haven't simply accepts you…

Ignorant Monty absolutely can exist. The person standing in front of the audience is not necessarily the person who arranged the prize. And if he's not, he knows nothing more than the audience.

I was, however, wrong that Monty was ignorant. By his own statement, he always knew the answer. But whether or not he was on your side depended on his mood. And he applied various forms of pressure to get you to make the right, or wrong, decision.

Therefore the problem was not stated correctly. A correct statement needs to not just state what you witness, but the counterfactuals about possibilities that you didn't witness.

Re: Marilyn vos Savant and the Monty Hall Problem (2015)

#132

Earlier quoted context omitted.

The problem as described does not specify his behavior. Only what happened. If all you know is that he opened a goat door you have learned nothing. It could be that he always opens a goat door meaning you should switch. It could be that he only opens a goat door if you picked the car and you should not. It could be he only opens a goat door if you picked a goat and you should. Probability cannot determine which of th…

"If all you know is that he opened a goat door you have learned nothing." This is precisely where you are wrong.

Ok let’s play!

We play this game 1000 times.

33% of the time you pick the car and I show you a goat. You switch and lose 100% of the time.

66% of the time you pick a goat and I don’t show you a goat. Assuming you don’t switch because you believe this means you have an equal odd on your current door, you lose 100% of the time.

Congrats. You have lost every single round.

Re: Marilyn vos Savant and the Monty Hall Problem (2015)

#133
post #60

As stated, the answer is indeterminate. To demonstrate, consider these scenarios. - Savant's Monty: The host knows where the car is, and wants to prolong the game. Switching wins 2/3 of the time. - Ignorant Monty: The host has no idea where the car is. Revealing the goat was luck and your odds remain 50-50. - Malicious Monty: The host knows where the car is, and wants you to lose. The fact that the host didn't reveal…

> Nowhere in the problem are Monty's knowledge and motivation stated. "and the host, who knows what’s behind the doors, opens another door [..] which has a goat." The question is clear: The host (1) knows what is behind each door and (2) always shows a goat. It's clearly a determinate problem.

(Rereads the article.)

You're right. The problem is stated multiple times in the article. As is usual, most of the statements do not address knowledge. But Marilyn's own statement did. However she did not address motivation.

Monty himself claims that his motivation varied depending on his mood. The actual game had more complications. And his statement about the real game was, "My only advice is, if you can get me to offer you $5,000 not to open the door, take the money and go home."

You can find the article I got that quote from at https://www.nytimes.com/1991/07/21/us/behind-monty-hall-s-do....

Re: Marilyn vos Savant and the Monty Hall Problem (2015)

#134

Earlier quoted context omitted.

> His purposeful behavior could also be random. It doesn't matter. Because the problem explains what happens: he opens a door and reveals a goat. That is crucial information. There's now only one goat and one car left. But the choices have not been shuffled: you're definitely still pointing at your original choice. What were the odds that your original choice is a goat? Two in three. If you picked a goat, what is gua…

Incorrect. You have no way to know if he would only show you a goat if you picked the car. If that were the case then switching is a guaranteed loss. Simply knowing that he opened a door and showed a goat does not mean he would have done this regardless of your choice. Not a probability problem without this info.

> You have no way to know if he would only show you a goat if you picked the car.

What do you mean?

It's specified in the scenario. He shows you a goat. It's right there. This isn't a variable. It's a fact.

Your only job is to work out whether, given the scenario described, it makes sense to switch. Given all your possible choices.

Re: Marilyn vos Savant and the Monty Hall Problem (2015)

#135
post #87

Earlier quoted context omitted.

No explanation worked for me. I did a truth table, and said, "...huh." You don't need to model all three door choices. Just say you pick Door A, not Door 1. The first door you didn't pick is B, and C is the other door. Then map the prize behind each of 3 doors, and whether you stay or change your answer. Then count the number of successes for change versus stay. Spoilers: It's 3/6 vs 2/6.

My experience was I coded a simulation in Python and got 50/50. ... this revealed a major flaw in my intuition that helped me to grasp it. I had subconsciously been assuming Monty sometimes opens the door revealing the prize instead of always opening a goat door. When I coded that into the simulation, of course it broke (because the player was now playing foolishly, looking at the prize and intentionally choosing the…

This is an even better learning experience than coding it right the first time, and I am so happy you had it!

Re: Marilyn vos Savant and the Monty Hall Problem (2015)

#136
post #86
post #66

Earlier quoted context omitted.

Your formulation is once again ambiguous. Instead of clearly stating that Monty had to open 998 empty doors, you state that he did that. Which could mean he happened to do that. Had to vs happened to makes a big difference. Had he opened them at random and by (extremely small) chance they happened to be empty, the odds are quite different.

I'm trying to understand why the odds would be different if Monty opened empty doors by chance versus on purpose. Does it really change anything?

The difference is if Monty has a chance to accidentally open the door with the car.

Re: Marilyn vos Savant and the Monty Hall Problem (2015)

#137
post #86
post #66

Earlier quoted context omitted.

Your formulation is once again ambiguous. Instead of clearly stating that Monty had to open 998 empty doors, you state that he did that. Which could mean he happened to do that. Had to vs happened to makes a big difference. Had he opened them at random and by (extremely small) chance they happened to be empty, the odds are quite different.

I'm trying to understand why the odds would be different if Monty opened empty doors by chance versus on purpose. Does it really change anything?

It changes. Monty's behaviour influenced by his knowledge were the car is. He is leaking information. It is a probabilistic leak: if you first picked by a chance the only door with a car, then Monty is free to open any of remaining doors.

But if you had chosen a door with a goat, then Monty has no choice at all, he must open the only door with a goat that you didn't pick. It is a leak.

From other hand if Monty picked a door by random, he would not leak his secret, but he might open a door with the car accidentally.

Re: Marilyn vos Savant and the Monty Hall Problem (2015)

#138
post #61

Earlier quoted context omitted.

The remaining doors are not random. The one you picked still has a 1/3 chance of winning, that doesn't change with opening another door. However switching doors is essentially the same as picking all the other doors at once, since the door with a goat is not opened, thus the 2/3. Imagine the thing with 100 doors and imagine the host isn't opening any doors, but just telling you what's behind them and it becomes prett…

>The one you picked still has a 1/3 chance of winning If there are only two choices remaining, I guess I don't understand the assertion that there is still a 1/3 chance of winning. Suppose you were new to the contest and were allowed to put money down on the choice at the point that there were two doors left. Then you repeated the bet every new contestant. Would one door would be more profitable than the other?

> Suppose you were new to the contest and were allowed to put money down on the choice at the point that there were two doors left.

That would be a completely different game. The goat doesn't get reshuffled. The doors that remain are not random and most importantly, you know which door you picked in the first round. It doesn't become a 50:50 chance just because there are two doors, as the goat isn't distributed over those two doors, but across all three.

Imagine 1000 doors. You pick one. In round two you are asked if you want to stay with that one door or pick all the other 999 doors at once. What do you do?

Re: Marilyn vos Savant and the Monty Hall Problem (2015)

#139

Earlier quoted context omitted.

Incorrect. You have no way to know if he would only show you a goat if you picked the car. If that were the case then switching is a guaranteed loss. Simply knowing that he opened a door and showed a goat does not mean he would have done this regardless of your choice. Not a probability problem without this info.

> You have no way to know if he would only show you a goat if you picked the car. What do you mean? It's specified in the scenario. He shows you a goat. It's right there. This isn't a variable. It's a fact. Your only job is to work out whether, given the scenario described, it makes sense to switch. Given all your possible choices.

You cannot assume that your context would apply from all starting conditions!

That’s why this problem is kind of bad. It does not describe the behavior of the host. It describes the perspective of a contestant halfway through the game.

Dumb example. Host flips a coin 9 times in a row and lands heads each time. If you believe this to have happened by chance then it’s probably rigged because that’s insane and it’ll likely be heads again.

If it’s ALWAYS heads then you haven’t learned anything.

Re: Marilyn vos Savant and the Monty Hall Problem (2015)

#140

Earlier quoted context omitted.

"If all you know is that he opened a goat door you have learned nothing." This is precisely where you are wrong.

Ok let’s play! We play this game 1000 times. 33% of the time you pick the car and I show you a goat. You switch and lose 100% of the time. 66% of the time you pick a goat and I don’t show you a goat. Assuming you don’t switch because you believe this means you have an equal odd on your current door, you lose 100% of the time. Congrats. You have lost every single round.

You've restated the problem (incorrectly) -- changed it.

There's still a goat you could show me. And it is a fact that you show me a goat. Nowhere in the problem does it suggest there is a chance you show me a goat.

I do honestly admire your dogged commitment, and I think the way you are committed shows up an important point about the article and the history of the problem.

Which is that one can quite clearly fairly argue the point, as you are doing, without resorting to misogynistic or patronising rudeness as so many did at the time!

But you're still wrong. :-)

And I'm going to leave it here.

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