Famous brain teaser

[Justin Passing]

I think it's still 50.50 on the remaining doors. All opening a don't win door does is reduce the odds to 50.50.

All opening the don't win does is shows that the car is behind either one of the two remaining doors is a win. The 3 doors were initially set up randomly. That nocks a hole in her argument.

Oh dear! Logic is about as good as the spelling...

 

[Trazor]

I think it's still 50.50 on the remaining doors. All opening a don't win door does is reduce the odds to 50.50.

If it was a 2 door puzzle you would be correct, but it's a 3 door puzzle.

Your first pick is a one in 3 chance, the odds of the car being in the other 2 doors is a 2 in 3 chance.
When the the host shows you a Zonk in the remaining 2 doors, it' still a 2 in 3 chance, the odds don't change.

The simple way to prove that you have twice as much chance, is to make some cards up and play the game.
You will find after let's say 100 games you will win the car on average 66 times.

Again here is an online simulation.

Monty Hall Problem Online. Run the monty hall game and simulation , over and over to understand the probability of this problem.

Monty Hall Problem --a free graphical game and simulation to understand this probability problem.

www.mathwarehouse.com

 

[ajohn]

Oh dear! Logic is about as good as the spelling..

if you can't work out why the idea doesn't work it's you with the problem.

It's pretty simple really. Pick a door. 2/3 chance of not winning 1/3 chance of winning. There are 2 doors that loose so another can always be opened even if the player chooses the other one. Doing that does not change the initial odds at all. They are still as they were but you have a 50/50 chance if you swap or stick with the initial choice.

 

[Trazor]

It's pretty simple really. Pick a door. 2/3 chance of not winning 1/3 chance of winning. There are 2 doors that loose so another can always be opened even if the player opens the other one. Doing that does not change the initial odds at all. They are still as they were but you have a 50/50 chance if you swap or stick with the initial choice.

Sorry your logic is awful, just play the game for real using cards, get someone else to run the game.

Do you think this online simulator is lying to you........

Monty Hall Problem Online. Run the monty hall game and simulation , over and over to understand the probability of this problem.

Monty Hall Problem --a free graphical game and simulation to understand this probability problem.

www.mathwarehouse.com

 

[transam]

Dunno

Don’t get it tbh

 

[JonathanM]

It's pretty simple really. Pick a door. 2/3 chance of not winning 1/3 chance of winning. There are 2 doors that loose so another can always be opened even if the player chooses the other one. Doing that does not change the initial odds at all. They are still as they were but you have a 50/50 chance if you swap or stick with the initial choice.

The key is in bold. This is what finally clicked for me, but it took me hours and hours. The odds never change, even after the first door is opened. There is always a 2/3 chance that the car is behind one of the doors you didn't choose originally, and only a 1/3 that your original choice was right. So, by swapping to the other unopened door, you have a 2/3 chance, but only 1/3 if you stay with your choice.

 

[IT Minion]

if you can't work out why the idea doesn't work it's you with the problem.

It's pretty simple really. Pick a door. 2/3 chance of not winning 1/3 chance of winning. There are 2 doors that loose so another can always be opened even if the player chooses the other one. Doing that does not change the initial odds at all. They are still as they were but you have a 50/50 chance if you swap or stick with the initial choice.

let's try your logic with a million doors.

Pick a door, there is a 1/1,000,000 chance of winning. There are 999,999 doors that lose so the host opens 999,998 of them. There are now only two doors.

Do you think the odds are 50:50 here? Or do you think that the odds you picked a winner the first time are tiny, and that switching to the other door is more sensible?

 

[Trazor]

let's try your logic with a million doors.

Pick a door, there is a 1/1,000,000 chance of winning. There are 999,999 doors that lose so the host opens 999,998 of them. There are now only two doors.

Do you think the odds are 50:50 here? Or do you think that the odds you picked a winner the first time are tiny, and that switching to the other door is more sensible?

The most easily understood explanation, so far.

 

[ajohn]

Have to give myself a slap on the wrist for thinking maths rather than the effect of opening the door.

You pick a door. 1 in 3 chance of picking the correct one. There is a 2 in 3 chance that correct one is in the other 2. Opening a no win door shows where the win one should be most often. So probability if some one sticks with the same door is 1/3. If they switch 2/3.

 

[IT Minion]

The most easily understood explanation, so far.

Yep, it's the one that explained it to me.

 

[Winglet]

let's try your logic with a million doors.

Pick a door, there is a 1/1,000,000 chance of winning. There are 999,999 doors that lose so the host opens 999,998 of them. There are now only two doors.

Do you think the odds are 50:50 here? Or do you think that the odds you picked a winner the first time are tiny, and that switching to the other door is more sensible?

So reguardless of how many doors there were before, there are now two doors. Lets call them A) and B).
So the prize is behind one of those doors, so you have 50% chance of finding the prize behind door A). If it isnt there then it is behind door B) (The other 50%)

So it is 50/50

 

[IT Minion]

So reguardless of how many doors there were before, there are now two doors. Lets call them A) and B).
So the prize is behind one of those doors, so you have 50% chance of finding the prize behind door A). If it isnt there then it is behind door B) (The other 50%)

So it is 50/50

That assumes nothing you saw earlier was relevant and that the prizes have been switched around. But they aren't.

One of the bits that people miss is that they forget the game show host knows which doors are which. You're not betting on a fair dice roll.

 

[Winglet]

Whatever you saw before is no longer relevant surely?
The situation is now that there are two doors.

How is it helpful me knowing that the gameshow host knows what is behind the doors? He has now made the process easier for me so it is no longer one in 3 so now at the present time I only have two options left.

 

[Pat ex]

How about a similar situation.
You're asked to pick an Ace of Hearts from a pack of 52 cards, face down, so you have no way of knowing, which it is, but the dealer does.

You choose a card, leaving it face down, so you don't know if you're correct or not.
You must agree that the chances of you choosing the right card, from the 52, is highly remote.
Now the dealer reveals 50 of the remaining cards, leaving just the two face down. The dealer knows which is the Ace of Hearts.

Are you better sticking with your original choice, or swapping to the other card?
The original odds remain, 1 in 52 of choosing the right card, it's just that you now know the other 50 are not the Ace of Hearts.
If you had been originally presented with a choice from 2 cards, it would be 50/50. You weren't, it was a choice from 52.

 

[IT Minion]

Whatever you saw before is no longer relevant surely?
The situation is now that there are two doors.

How is it helpful me knowing that the gameshow host knows what is behind the doors? He has now made the process easier for me so it is no longer one in 3 so now at the present time I only have two options left.

But the prizes haven't moved.

You either landed on the right door out of a million (you didn't) or the prize is behind the only other door left after the host removed 999,998 wrong guesses.