Ignore Bait: Highest IQ, Many Questions, Odds makers invited...

If we follow the $, I think Monte did better than the contestants looking at the big picture.

Fatboy<——-still wants a goat
 
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It’s like the martingale system gambling. Always loses.

...
When young (and dumb) I made a ton of money one afternoon in Vegas expermenting with Martingale.
Thought I found the holy grail.
Colored up an swore they shorted me a few chips. No problem, I'll win that back right quick.
Lost the whole roll in a couple of minutes.

First and last time for that.
 
Here's what makes the problem both interesting and confusing. The choices are not "always switch" or "always keep". There is a third choice that makes your chances of winning 50/50. Let your choice be determined by a flip of a coin. If you initially chose right (1/3) half of that time you will give away a winner while half of the rest of the time you will keep a loser. That's 1/6 + 1/3 = 1/2 of the time you will lose. That's why people who assume the decision to keep/switch is 50/50 say there's a 50/50 chance of winning, and if they choose to make that decision 50/50 or close to it (by listening to the audience, for example) they're right. They make it 50/50 by assuming it's 50/50 and introducing a 50/50 event into their decision.
 
When young (and dumb) I made a ton of money one afternoon in Vegas expermenting with Martingale.
Thought I found the holy grail.
Colored up an swore they shorted me a few chips. No problem, I'll win that back right quick.
Lost the whole roll in a couple of minutes.

First and last time for that.
Same thing for me in Reno😂😂
 
Here's what makes the problem both interesting and confusing. The choices are not "always switch" or "always keep". There is a third choice that makes your chances of winning 50/50. Let your choice be determined by a flip of a coin. If you initially chose right (1/3) half of that time you will give away a winner while half of the rest of the time you will keep a loser. That's 1/6 + 1/3 = 1/2 of the time you will lose. That's why people who assume the decision to keep/switch is 50/50 say there's a 50/50 chance of winning, and if they choose to make that decision 50/50 or close to it (by listening to the audience, for example) they're right. They make it 50/50 by assuming it's 50/50 and introducing a 50/50 event into their decision.
You lost me on this logic - you are overthinking this.
 
You lost me on this logic - you are overthinking this.

It’s actually a fairly interesting observation if we think about it.

Let’s say we have 3 contestants. A, B, C.

A: always swaps
B: flips a coin
C: always stays

They each play 100 games. And we assume the 1/3 and 2/3 stats hold up perfectly for ease of illustration.

A: wins 67 games
B: (see analysis below)
C: wins 33 games

Player B, decides in his opinion it’s 50/50 and flips a coin to decide every time. For ease of analysis we will say the 50% choice holds up.

So Player B will hold 50 games and switch 50 games.

Hold = 1/3 of 50 = 17 games
Stay = 2/3 of 50 = 33 games

So, now Player B wins 50 games.


So, what he’s saying is that if a player decides incorrectly in their mind that it’s 50/50 and truly does pick a different strategy 50% of the time, we now introduce a win rate of 50% over the long haul.


That doesn’t mean the odds are 50%. Just that it’s a better strategy than always staying and a worse strategy than always switching.

Math is fun.
 
Basically, if a person takes a completely random approach they effectively killed part of the information that Monty provided them.

They just exclude a door that was opened and turn the odds into a true 50/50 via bad (but less terrible than just staying) decision making.
 
Instead of doors imagine it a dating scenario. Your potential spouse is behind one door and the other two are short term relationships.

If you pick right the first try there is no need for switching. Picking the spouse in a group of three vs picking the spouse in a group of two what difference does it make? Its exactly the dating to wedding scenario.

How often do you hear about people being left at the marital ceremony?

The argument about math and modeling only works academically or for standardized tests. This thread feels worthy for technical job interviews.
 
Depends on the scenario. The exact scenario says you pick a door. Then the host opens a door showing a goat and gives you the option.

If you knew going in that he was going to open the door and show a goat, you should have already made the decision to switch.

If you didn’t, know that, then he opens a door and reveals a goat, even if it was random and he could have opened the car, you now have additional information you didn’t have. Then he offers the switch.

It would be the same as two players in a hold ‘em game. One had a big decision and the other turns over one of their cards. There was the decision before that that was proper for the scenario and there was a decision after that is proper with the added information.


The only reason I say he has to open the goat or bring up that he knows what’s behind the doors is to highlight the odds to the non believers.

You pick a door and it’s a 33% chance.

If Monty doesn’t know what’s behind the cards, it’s also 33% that he picks a goat or the car. But if he knows where the car is, and is not allowed to pick it, then it’s 66%.

But, the actual scenario is that you pick a door. Then you’re shown an open door with a goat. Then you are given a choice of keeping or swapping. Is it to your advantage to swap?

That sounds like semantics, however in game theory and such, it’s extremely important to have the exact order of options and information if one is to take the example and extrapolate it to be useful decision making.

I'm little surprised by this response. My response (and all of vapoolplayer's responses) was to the following scenario:
1. There are three doors. One has a car and two have a goat.
2. You you pick a door.
3. After you pick, Monty opens one of the two doors that you didn't pick. Monty knows which door has the car and will only reveal a door that has a goat. If the car is behind one of the two doors you didn't choose, Monty will open the other one. If both doors have a goat, then Monty picks one at random.
4. Monty offers you a choice. Stick with your original selection or switch to the door that Monty didn't reveal.

As has been stated here many times, you should switch because the car will be behind the door that Monty didn't open with 2/3 probability. My only point is that, given that Monty will never open the door with the car, seeing a goat behind the door he opens does not tell you anything new.

Suppose you pick door #1.
If door #2 has the car, Monty will open door #3.
If door #3 has the car, Monty will open door #2.
if door #1 has the car, Monty pick one at random.

So if Monty opens door #2, you should switch to door #3. If he opens door #3, you should switch to door #2. The point of my post is that regardless of which door Monty opens, you should switch from your initial choice.

vapoolplayer's post makes my point. He said: "If you knew going in that he was going to open the door and show a goat, you should have already made the decision to switch." In other words, when Monty opens a door with a goat behind it, the only thing you "learn" is that one of the two doors had a goat behind it. But you already knew that. Monty opens one of the two doors you didn't pick, but only one can have a car behind it and he will never reveal the car. So when Monty opens a door with a goat, there is no new relevant information.

I agree with vapoolplayer that a different game would require a different decision rule. I was just surprised that he started talking about a different game than the one he had posted on multiple times.
 
This might be one of my favorite threads because of how wrong some people are and they dont have a clue about it.

I am not trying to convince anyone of anything here. I will just explain how I see it.

I choose a door. I know that at least one of the doors I didnt choose is a goat. Now lets say for arguments sake that after I have chosen my door nothing is revealed and I am given the option of staying with my door and trading for the two remaining doors. The decision with the best odds is to trade for the other door even though I know that one of them is a goat. The host revealing the goat doesnt change the odds. I am at either one third (my first choice) or two thirds (trading). The fact that the goat is revealed does not change a thing. I already knew it was there.
I agree. See my post on page 15 (#289) and my most recent post.
 
Instead of doors imagine it a dating scenario. Your potential spouse is behind one door and the other two are short term relationships.

If you pick right the first try there is no need for switching. Picking the spouse in a group of three vs picking the spouse in a group of two what difference does it make? Its exactly the dating to wedding scenario.

How often do you hear about people being left at the marital ceremony?

The argument about math and modeling only works academically or for standardized tests. This thread feels worthy for technical job interviews.

I literally said “what the fuck” out loud.

I wish this was a troll post. It would be amazing.
 
True, but you said you know where the car is 66% of the time when it's actually 100%

When I say me, I’m speaking from the contestant POV. I might have not been clear on that.

If we are saying the contestant knows 100%, then I’ll have to hear explanation to see what I’m missing.
 
I only read the first 10 pages or so. Did Major Miscue ever come around? :D

He finally admitted that he did in another thread, to counter somebody who mentioned his lack of understanding as it related to another topic.

He never admitted it here.

When people finally come to understand that they were wrong about something, why is it so hard for them to just man up and say so?

How refreshing would that be?
 
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