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

Sorry, with two doors left it's 50-50 whether you stick with door #1 or switch to door #2. At this point there are only two doors left. Door #3 is now out of the equation and it's a new deal so to speak. She may be smart but she's not a gambler. If someone wants to lay me 6-5 we can bet on this all day long, and they get to pick which door they want. I would bust Marilyn Vos Savant at this game!

You go broke. They have run simulations to death on this. It’s not even debated anymore.

The only way you win is bet with enough odds to overcome the 66%.

It’s no random or 50% once someone who knows what’s behind the doors gets rid of one after you’ve picked a door first.
 
Sorry, with two doors left it's 50-50 whether you stick with door #1 or switch to door #2. At this point there are only two doors left. Door #3 is now out of the equation and it's a new deal so to speak. She may be smart but she's not a gambler. If someone wants to lay me 6-5 we can bet on this all day long, and they get to pick which door they want. I would bust Marilyn Vos Savant at this game!

1. You're wrong.

2. You're so wrong that even if you were right you'd still be wrong. You can't expect to bust anyone at truly even odds.
 
You can even do an at home simulation with cards.

Take two deuces and an ace.

Mix them up face down. Then set one to the side.

Flip the two others over and anytime there is a 2 and an Ace as those two cards, mark a win in the “switch” column.

This would be the same as assuming you always switch. Do it 100, or 500, or 1000 times. See how many you win.

Then do the same amount of times just picking a random card and turning it over. If your one card picked is the Ace, that’s a win.

See which one wins more. You’ll be surprised.
 
You’ll see the only time you “lose” when swapping doors is when both of the doors you didn’t choose are goats.
 
You can even do an at home simulation with cards.

Take two deuces and an ace.

Mix them up face down. Then set one to the side.

Flip the two others over and anytime there is a 2 and an Ace as those two cards, mark a win in the “switch” column.

This would be the same as assuming you always switch. Do it 100, or 500, or 1000 times. See how many you win.

Then do the same amount of times just picking a random card and turning it over. If your one card picked is the Ace, that’s a win.

See which one wins more. You’ll be surprised.

It works well with a penny under one of 3 cups too.

Once you do it a few of times, you quickly realize that the answer always comes down to whether or not you picked the correct cup the very first time.

From there you understand that the odds are 1 in 3 of picking the correct cup initially, and consequently the same if you don’t change cups.
 
Wait a minute, the solution shown in post #10 continues the assumption that door #3 is a possible winner, but we know it isn't.
So all possibilities that include #3 as a choice or #3 as the prize door are false choices.
We know #3 is a loser so we aren't going to make that choice and we also know that any option showing #3 as the prize door are also false and can be eliminated.
There can only be one prize door and it isn't #3. If you pick #1 and stick with it knowing #3 is a loser, you are 50/50 to win.
If you switch to #2 you are still 50/50 since it is either #1 or #2.
The confusion comes in when they reintroduce #3 as a possibility when we know it isn't.
 
You go broke. They have run simulations to death on this. It’s not even debated anymore.

The only way you win is bet with enough odds to overcome the 66%.

It’s no random or 50% once someone who knows what’s behind the doors gets rid of one after you’ve picked a door first.
Yes BUT. The deal is one time. One flip for a car or the goat. There is no chance for the 66% to manifest. Further the door Monty spots you is a revision of the proposition and in a true proposition, the proposer would be playing you with that extra information.


Sorry, with two doors left it's 50-50 whether you stick with door #1 or switch to door #2. At this point there are only two doors left. Door #3 is now out of the equation and it's a new deal so to speak. She may be smart but she's not a gambler. If someone wants to lay me 6-5 we can bet on this all day long, and they get to pick which door they want. I would bust Marilyn Vos Savant at this game!
Gotta agree with this although I wouldn't push it past the one bet :D
 
Yes BUT. The deal is one time. One flip for a car or the goat. There is no chance for the 66% to manifest. Further the door Monty spots you is a revision of the proposition and in a true proposition, the proposer would be playing you with that extra information.



Gotta agree with this although I wouldn't push it past the one bet :D

There is no “but”.

Your odds are 66% if you switch. One time or 1,000 times.

If you don’t switch, your odds are 33%. One time or 1,000 times.
 
Wait a minute, the solution shown in post #10 continues the assumption that door #3 is a possible winner, but we know it isn't.
So all possibilities that include #3 as a choice or #3 as the prize door are false choices.
We know #3 is a loser so we aren't going to make that choice and we also know that any option showing #3 as the prize door are also false and can be eliminated.
There can only be one prize door and it isn't #3. If you pick #1 and stick with it knowing #3 is a loser, you are 50/50 to win.
If you switch to #2 you are still 50/50 since it is either #1 or #2.
The confusion comes in when they reintroduce #3 as a possibility when we know it isn't.

Let’s say there are 3 doors.

You pick door 1. But you can’t see behind it.

I *know* what is behind door 2 and 3. And I’m going to open one and show you. It will *always* be a goat. Because I can’t open the car. That’s the catch and what makes this different.

So, 66% of the time, the car *will* be behind one of my doors.

I’m not allowed to open the car. So, 66% of the time, I’m going to tell you exactly where the car is.

Make sense?
 
You’ll see the only time you “lose” when swapping doors is when both of the doors you didn’t choose are goats.
The fun thing about this as a prop bet, is that even when people are told the correct solution, and have it explain to them in relatively simple terms, they don’t believe it.

At some point it’s not robbing somebody. It becomes educating. An act of kindness, if you will… 😁
 
Let’s say there are 3 doors.

You pick door 1. But you can’t see behind it.

I *know* what is behind door 2 and 3. And I’m going to open one and show you. It will *always* be a goat. Because I can’t open the car. That’s the catch and what makes this different.

So, 66% of the time, the car *will* be behind one of my doors.

I’m not allowed to open the car. So, 66% of the time, I’m going to tell you exactly where the car is.

Make sense?
No, 100% of the time the car will be behind one of the doors. The odds are 1 in three of winning. You eliminate 1/3 of the choices by opening door three. Now the car is behind #1 or #2, #3 is no longer a player. My choice is #1 or #2 and I'm 50/50.
 
There is no “but”.

Your odds are 66% if you switch. One time or 1,000 times.

If you don’t switch, your odds are 33%. One time or 1,000 times.
The switch one time doesn't mean heads OR tails but it does mean you're getting played win or lose. And what about the part where, by switching, you doubled your time guessing thereby increasing Monty's chances by 100% ?
 
No, 100% of the time the car will be behind one of the doors. The odds are 1 in three of winning. You eliminate 1/3 of the choices by opening door three. Now the car is behind #1 or #2, #3 is no longer a player. My choice is #1 or #2 and I'm 50/50.

That’s the wrong answer and why it’s a brain teaser.

It’s conditional probability.
 
There is no “but”.

Your odds are 66% if you switch. One time or 1,000 times.

If you don’t switch, your odds are 33%. One time or 1,000 times.
Wrong. Your odds are 1 in three initially. But Monty eliminates #3 so we can eliminate #3 from any further consideration, it's a goat.
That means we are now 1 in 2, 50/50. Twist all you want, there are now two choices, not three, one car and one goat. 50/50 switch or don't switch.
 
The switch one time doesn't mean heads OR tails but it does mean you're getting played win or lose. And what about the part where, by switching, you doubled your time guessing thereby increasing Monty's chances by 100% ?

Let’s say you have a Texas Holdem hand where you see the flop.

You get all your money in and you’re 33% to win.

Would you ever say it’s 50/50 because it’s a one time thing and you win or lose?

Of course not. It’s 33%
 
Wrong. Your odds are 1 in three initially. But Monty eliminates #3 so we can eliminate #3 from any further consideration, it's a goat.
That means we are now 1 in 2, 50/50. Twist all you want, there are now two choices, not three, one car and one goat. 50/50 switch or don't switch.

Lol….no.

That only works if he does *NOT* know what’s behind the doors.

The “conditional” part of the conditional probability is that he *knows* what’s behind the doors.
 
How about we watch the TV show and see the outcome of their decision each time. Or we could go to the archives and watch a dozen or so past shows and see what happens when there are two doors left and someone keeps their original pick or changes it. Without knowing what has happened on any of these past shows I will take 6-5 on a $100 (your 120 to my 100) for each show if someone can pull up the archive of the most recent fifteen to twenty shows. Deal or no deal? No cheating!
Quit while you're still even Jay.
 
That’s the wrong answer and why it’s a brain teaser.

It’s conditional probability.
The reason it's a brain teaser is that they try to reintroduce #3 as a possibility after it has been eliminated. Tricking you into thinking there are three possibilities when there are only two, you know #3 is a loser.
 
Guys, there’s literally a Wikipedia page for this.

It’s not a new question. This is ages old.

The only reason it’s not a casino game is that it’s too much of a sucker’s bet for enough people to fall for.
 
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