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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)

#81

Post by saalweachter »

btilly wrote: 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…

_Let's Make A Deal_ was on the air for twenty (thirty?) years at the time of this controversy.

There was sufficient evidence for Savant's Monty -- the car was revealed 0% of the time, not 33 or 66% of the time.

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

#82

Post by rmacqueen »

kergonath wrote: post #68
rmacqueen wrote: post #62

I swear this problem only ever causes confusion because the original wording is ambiguous about what the host's motivations are. Once you realize the host is trying to get you to lose and knows what's behind each door and must open a door, then it's clearer that you should switch.

The host’s motives don’t matter at all. If the first choice is a goat, then the host does not have any choice about the door to open, so their intent is irrelevant. If the first choice is a goat, then the outcome is the same regardless of the door the host chooses. The host does not really have any agency here. The rules are intended to confuse people and trick them into thinking both remaining doors have the same pr…

I'm not sure if you would call it 'agency' but the host is following it's own specific door-picking policy - a policy which will never result in a car being revealed. It is precisely this policy and no other which makes the contestant switch the correct move. But the original wording doesn't say that - it just says the host opens a door and it happens to have a goat behind it.

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

#83

Post by HumblyTossed »

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?

The way I think about it, in general, is that Monty, by purposely not opening the correct door, is giving you a huge hint and thus changing the game which changes the odds. Think of it similarly to those people who play the same lotto #s all the time because they think eventually it has to hit. Well, no. It's a new game every time. If the lotto commission retired each combination that won each week, then sure. And this is essentially what Monty is doing. He's retiring some doors and changing the odds.

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

#84

Post by orblivion »

btilly wrote: 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…

So long as you describe a Malicious Monty that might not have shown you the door, you could also have a Benevolent Monty who won't show you the door if you've gotten it right.

Though I'm not sure I've ever considered the Malicious Monty, only the other two.

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

#85

Post by orblivion »

saalweachter wrote: post #81
btilly wrote: 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…

_Let's Make A Deal_ was on the air for twenty (thirty?) years at the time of this controversy. There was sufficient evidence for Savant's Monty -- the car was revealed 0% of the time, not 33 or 66% of the time.

I've never seen the show; I'm curious, did he always open one of the doors?

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

#86

Post by niel »

tromp wrote: post #66
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?

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?

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

#87

Post by hinkley »

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?

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.

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

#89

Post by mihaic »

dynm wrote: post #75
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?

I actually like to back up even further and start with this problem: There are 1000 doors. You pick one. There's a 1 in 1000 chance you get it right. Now, do you want what's behind your door, or what's behind the other 999 doors? https://dynomight.net/2020/09/17/making-the-monty-hall-probl...

I think this is the clearest explanation I've heard of the principle, kudos!

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

#90

Post by shadowgovt »

Arguably the most fascinating thing about the Monty Hall problem is that although this solution is extremely widely known in academic circles, the game is still played today on Let's Make a Deal! in the exact same way: three doors, pick one, a Zonk is revealed behind another, switch or keep?

(Although there is one variable I don't know the answer to: I don't know if they always give you the offer to switch. Sometimes I believe it's a different offer, "Keep the door or take this money out of my hand". The host being able to choose which offer is made shifts the odds a bit, because the host may be choosing the offer based on whether you've picked the right door initially, and may be making that choice in a biased or randomized way. There's hidden data there we don't have as observers).

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