re: Let's discuss the Monty Hall problem (probabilities, odds)

Say you were in the business of giving away your own money.

Would you offer people to switch off of a goat and onto your own money?

The math part makes sense in theory. Using a gameshow to explain this is bunk.

Let's keep it easy and always pick door 1... keep in mind he can't show you door 1 (your pick) and he can't show you the car

1 2 3 stay switch
C G G W L
G C G L W
G G C L W


2/3 odds if switch
1/3 odds if stay

I've always struggled with this problem, especially since my probs teacher was a Chinaman. I still don't fully understand how to work it out mathematically, but this video gives the best logical description I've seen (jump to 2:10):
LINK



ETA: Here's another good one with more math (obviously presented by an Asian).
LINK

Switching only loses if you made the 1/3 choice and picked the door with the prize. He makes a great and simple point: you make your first choice, being 1/3. Switching allows you to bet against that 1/3 chance.

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Here's another good one with more math (obviously presented by an Asian).

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Someone help me here. OK I get 1/3 on initial pick door 1.

Goat shown door 3. So now its 1/2 with 2 remaining doors. You know one has a goat one has the prize.
Shouldn't matter. Coin flip at least the only way I can see it.

I think it's because the math is factored in if you were to switch each time your odds are better given the information given you didnt have before

Ie, he's never going to open the door with your prize, just the goat regardless which door you choose which is key.



the doors are 123. The car is behind door 1.

Scenario 1: you pick door 1
since the car is behind your door, it doesn't matter which other door is opened. Lets say its 2. You are now given the choice between 1 and 3.
You get the car if you stay

Scenario 2: you pick door 2
since the car is behind door 1, they must open door 3. Its now a choice between door 1 and door 2.
You get the car if you switch

Scenario 3: you pick door 3.
since the car is behind door 1, they must open door 2. Its now a choice between door 1 and door 3
You get the car if you switch.


Each of these 3 scenarios is equaly likely if you pick a random door. in 2 of them, you get the car if you switch. Hence, 2/3

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Someone help me here. OK I get 1/3 on initial pick door 1.

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I struggled reading the rest of your post, but this is all that matters. You have a 1/3 chance of making the correct choice. That means that the remaining choices have a 2/3 chance of being correct. Now the host eliminates one of those two choices. That means that switching (which gives you the remaining choice of those two)= 2/3 chance of winning.

Put simply, switching only loses if you make the CORRECT choice (1/3 chance) on your original pick.

An interesting anecdote I remember reading about the Monty Hall Problem:

In 1990, Marilyn Vos Savant had a column in Parade magazine where people wrote her for advice and often gave her brain twisters to solve for fun. Someone asked her about the Monty Hall Problem and she responded that you should switch doors. Her answer got a lot of derision from math professors who wrote in claiming she was wrong and that there is no advantage to switching. She saved those letters from all those PhD's and has since put them on her website. LINK In any case, Marilyn was right, switching is the optimal strategy.

In case you're wondering, Vos Savant has no expertise in anything other than having an astronomically high IQ (she was tested at an IQ of 228 at age 10 which put her in the Guinness Book of World Records as having the highest known recorded IQ). However, it is well known that childhood scores that use "mental age" are not accurate when the child becomes an adult. Nonetheless, her adult IQ is still estimated to be easily one-in-a-million (over 180). For comparison, Mensa is 1 in 50 (or about 130).

Her Wiki Page

Always switch. Doubles your odds.

quote:

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1 2 3 stay switch
C G G W L
G C G L W
G G C L W

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That is one way to look at it but the scenario is that he shows you door 3, which means the goat is definitely behind door 1 or 2. That means option three, in this scenario, has already been eliminated as a possibility. At the point in which we enter the event, their is no statistical advantage in switching or keeping the original selection (we may want to draw a conclusion based on the general psychology of the game, but that is not a mathematical decision).

For me, it is the equivalent of selecting the next coin flip in a series of coin flips. If you flipped heads 99 times in a row, the odds of you flipping heads on the next coin toss are still 50% even though the odds of you flipping heads 100 times in a row are 1 in 1267650600228229401496703205376.

That's just my thinking on the subject. Hit me up in 4-5 years when my son complete his Math degree with a concentration in statistic.

ETA: I have come to the switch side.

I agree. To me this advantage to switching only holds if you view the game as a single period event and calculate the probability as such. Once making this a multi-period game, which it is given the game has changed after revealing door 3, then the probabilities are 50-50.