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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)

#201

Post by tshaddox »

Karellen wrote: post #50
gerad wrote: post #39

The intuitive way for me to understand the Monte Hall problem is to pretended there are 1000 doors. You pick 1. There’s a 1 in 1000 chance you get it right. The host then opens 998 doors that don’t have the prize. Do you keep your original or do you switch? Are the odds 50:50?

What improvement does your 1000-door variant have over the 100-door variant described in the article?

The most common unintended interpretation seems to be that Monty Hall is only coincidentally opening a door with a goat behind it. I don't quite understand how people arrive at that interpretation, but the 1,000-door variant should make it even more of a stretch to interpret it as Monty Hall coincidentally doing something so extremely improbable.

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

#202

Post by calderknight »

Lendal wrote: post #184
lamontcg wrote: post #167

The asymmetry is that if you pick the one with the car first (only 1/3rd chance) then Monty Hall can show you either of the other doors. If you pick one without a car first (2/3rd chance) then Monty Hall must show you the door that doesn't have a car and the remaining door always has the car. So 2/3rd of the time you have a 100% chance of a car if you switch, 1/3rd of the time you have 0% chance of a car if you switc…

Aha! This is the first explanation that reveals the most common error. The fact that the reveal isn't just a random door. The host must not reveal the car. That changes everything. I believe that's what most people are missing. But if you actually stated that explicitly, you would give away the answer and the problem wouldn't be fun anymore. The fun is in not explicitly stating every unconscious assumption you have a…

That's clear. Otherwise the question would just be about speculating on the host's personality.

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

#203

Post by tshaddox »

jncfhnb wrote: post #115
KrazyButTrue wrote: post #100

Earlier quoted context omitted.

> 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.

Doesn’t matter. If you don’t have a guarantee that he will always open a goat door, his opening of the goat door doesn’t give you information.

The statement of the problem is the guarantee, just like the statement of the problem is the guarantee that there is always 1 car and 2 goats, that you always get to initially choose 1 door, etc.

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

#204

Post by topaz0 »

empyrrhicist wrote: post #157
topaz0 wrote: post #95

The explanation in the article is fine as far as it goes, but I think it's much more helpful to emphasize one key fact, which is that MH has to show a goat, and can't open your first choice door, even if it's a goat. That means that if you did pick a goat the first time around, he can only have revealed the other goat, in which case switching will necessarily give you the car. That happens with probability 2/3. This…

nvm

Actually it does matter. If you lose automatically when a car is shown, that will happen in 1/3 of games: none of the times your first pick was a car (0 * 1/3 = 0), half of the times your first pick was a goat (1/2 * 2/3 = 1/3). The remaining 2/3 are equally split between the times you chose a car first (which are guaranteed to complete) and the times you chose a goat first and MH randomly chose the other goat rather than the car.

(Note: this analysis does depend on the constraint that MH doesn't open your first pick door! Edit: actually the answer also becomes equal odds if he can pick any door, regardless of whether it was your first pick or the car. In that case, 1/3 of games will not complete, and the rest will be 50/50 between the unopened doors.)

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

#205

Post by calderknight »

dekhn wrote: post #112
ghaff wrote: post #101

Earlier quoted context omitted.

I think this is the original "Ask Marilyn": Suppose you’re on a game show, and you’re given the choice of three doors. Behind one door is a car, behind the others, goats. You pick a door, say #1, and the host, who knows what’s behind the doors, opens another door, say #3, which has a goat. He says to you, "Do you want to pick door #2?" Is it to your advantage to switch your choice of doors? Craig F. Whitaker Columbia…

Yes, I'm literally talking about this sentence: "host, who knows what’s behind the doors, opens another door, say #3, which has a goat." I parsed that as "50% of the time, monty opens another door and it has a car and you win immediately, and 50% of the time, monty opens another door and it has a goat". In retrospect I think my brain just sort of pictured that and proceeded to assume there was no reason to switch, an…

I don't think your interpretation of the sentence is sensible. The sentence mentions that the host knows what's behind the doors. So, if he is allowed to open the door with the car, the problem would become insoluble and would just be about speculating on the host's personality. And it definitely doesn't support the conclusion that the probabilities become 50/50.

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

#206

Post by andrewla »

The reason the scenario is kind of twisted is that in the iterated game of "you pick door, he shows goat, switch or no", vos Savant's answer is clearly correct.

But in the iterated game of "you pick door, and host chooses whether to show a goat or to let you open the door you initially chose" I don't think you (or he) can get better than 50-50. If you chose the correct door initially, then he can try to trick you into switching by opening a goat-door. But if you chose a goat-door, then he can let you open your selected door (or offer some other deal -- $1,000 cash or the door you chose). If he only shows the goat-door when you have selected the car-door, then you should never switch, so his optimal strategy is to show a goat-door when you choose a goat-door 1/3 of the time, so you basically get no information from his actions.

The "always switch" solution imagines this as a one-player game rather than a two-player game, with Monty Hall actively trying to sabotage.

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

#207

Post by dwpdwpdwpdwpdwp »

I taught this problem to a group of high schoolers as part of a statistics curriculum. It was engaging for two reasons:

The topic of the vitriolic responses to Marilyn is not boring.

It’s a great problem for illustrating the importance of simulation in statistics. By breaking off into pairs, half of whom played the switch strategy and half of whom played the stay strategy, we did hundreds of trials and found empirical evidence in favor of Marilyn’s answer. Furthermore, by actually playing the game, whatever any ambiguity in the wording of the problem was quickly resolved.

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

#208

Post by bemusedthrow75 »

Swizec wrote: post #162
empyrrhicist wrote: post #154

Earlier quoted context omitted.

Your re-framing is actually not quite right either (or at least, not complete). Key point: if the host picks the car, the game essentially re-sets. Imagine the extreme scenario with 999 goats and 1 car. You select the first door, the host opens 998 doors at random, leaving your selected door and one other. Two scenarios are now possible: 1. There was a car in the 998 doors that got opened. Tough luck, you lose the ga…

Also you're picking between Reward and Bigger Reward. Positive EV no matter your choice so eh don't worry about it too much :) A nice goat fetches up to $1000 to according to a quick google.

I am suddenly reminded of the clever, morally educational joke about the dead horse raffle.

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

#209

Post by andrewla »

gerad wrote: post #39

The intuitive way for me to understand the Monte Hall problem is to pretended there are 1000 doors. You pick 1. There’s a 1 in 1000 chance you get it right. The host then opens 998 doors that don’t have the prize. Do you keep your original or do you switch? Are the odds 50:50?

Did the host have to open 998 doors, or could he have chosen to just let you open your selected door? If he's "out to get you" then you get very little information here, because he could have just let you open your door. The only reason he'd open the other doors is to try to get you to switch, which means that it's pretty close to 50-50.

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

#210

Post by kragen »

calderknight wrote: post #205
dekhn wrote: post #112

Earlier quoted context omitted.

Yes, I'm literally talking about this sentence: "host, who knows what’s behind the doors, opens another door, say #3, which has a goat." I parsed that as "50% of the time, monty opens another door and it has a car and you win immediately, and 50% of the time, monty opens another door and it has a goat". In retrospect I think my brain just sort of pictured that and proceeded to assume there was no reason to switch, an…

I don't think your interpretation of the sentence is sensible. The sentence mentions that the host knows what's behind the doors. So, if he is allowed to open the door with the car, the problem would become insoluble and would just be about speculating on the host's personality. And it definitely doesn't support the conclusion that the probabilities become 50/50.

it's pretty counterintuitive that his personality enters into it, isn't it?
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