Thread: Brain teaser.

I would stick with the original decision because I would rather have a goat than a car.

You ALWAYS switch for you best chance.

If you stick you have a 1/3 chance.
If you swap you get 2/3 chance, double.

Didn't mythbusters do something on this?

Called the Monty Hall problem

Indeed and often referred too as the Monty Hall Problem.

The popular reasoning goes like this:

after opening a door to reveal a goat, Host leaves you with two possibilities, so the odds must be 50/50 and therefore there is nothing to be gained by switching.
You could switch, but there is no reason to do so.
The popular reasoning is wrong.


As strange as it may seem, you should switch doors – it doubles your chances of winning!
Without switching, you would only win a third of the time.
By switching doors, you will win two thirds of the time.


How is that possible?

Perhaps the easiest way to see that switching is the proper strategy is to work through an example.

Let’s assume you choose Door 1.
If the car is behind Door 2, then Host (Monty) has no choice but to reveal Door 3 (a goat).

Switching from Door 1 to Door 2 wins for you.

Likewise, if the car is behind Door 3 then Host (Monty) is forced to reveal Door 2.

Switching from 1 to 3 wins for you.

The only time you lose by switching is if you happened to choose the right door from the outset.

But we know that only happens one third of the time.

The other two thirds of the time switching is the correct strategy.


If you’re still not convinced, imagine there were 100 doors.
Behind 99 of them are goats and behind one is the car.
You choose a door.
Host (Monty) then opens 98 of the doors with goats and leaves only your original door and one other.
He then gives you the option of switching.
What’s more likely, that you picked the right door at the very beginning or that Host (Monty) was avoiding the winning door when he was opening up all of the losers?
Of course it’s the latter.
In 99 times out of 100 Host Monty has to avoid the correct door.
He’s practically pointing it out to you!
In this extreme example, only one time out of 99 will a switching strategy lose for you.


The same logic applies to the standard three-door example.
Statistically speaking, two thirds of the time you’ve chosen the wrong door initially and Host (Monty) is literally showing you the other wrong door. The one remaining is the correct one and you should switch to it as soon as you can.

Simple huh?

Sorry to burst your bubble, but your math is wrong. You outline 3 scenarios, there are actually 4.

In the case you selected the correct door to begin with, Monty can do one of 2 things. He can open the first incorrect door, or he can open the second incorrect door. The outcome is the same, but these are still 2 different scenarios.

When you made your original choice, your odds were 1 in 3. Once he has opened one of the incorrect doors, you have two doors remaining. He has now improved your odds to 1 in 2. No amount of gyrating changes those odds to 2 in 3.

Same applies to the 100 doors by the way. There is 1 correct door, and you have made 1 choice. By eliminating 98 wrong doors, Monty has improved your odds to 1 in 2 in the final choice.

Sorry Switt, the maths in TT's solution was right.

The person whose strategy is to switch will in the balance of probability win 2/3 of the time as opposed to 1/3 of the time if they dont switch.

Look at it this way. If you dont switch you only win the car if you pick the door with the car, which is 1/3 of time. In this case the host has two doors to pick from to reveal a goat.

If you do switch you win the car every time you initially pick a door with a goat which it 2/3 of the time. The host has the choice of only one door to open to reveal the other goat.

It takes a bit of thought to grasp it, but the odds do actually double for a player whose strategy is to switch every time.

Cheers

Doug

PS: I would still rather have the goat, they are cheaper to run and less trouble.

Don't know anything about the maths or the logic. I just know that I would always get the goat and when I opened the door it would probably spring back, hit the goat, startle it, cause it to charge at me breaking two ribs and an arm.

At least that's what happened last time.

I stand by my earlier post. Here's the proof.

To simply things I've only included the scenario where the car is behind door 1, and goats are behind doors 2 and 3. But if you work through D2 = Car and D3 = Car you will see they only extend the results with the same outcome.

D1	D2	D3	I Select	Monty Eliminates	Switch?	Result
Car	Goat	Goat	D1	D2	N	W
Car	Goat	Goat	D1	D2	Y	L
Car	Goat	Goat	D1	D3	N	W
Car	Goat	Goat	D1	D3	Y	L
Car	Goat	Goat	D2	D3	N	L
Car	Goat	Goat	D2	D3	Y	W
Car	Goat	Goat	D3	D2	N	L
Car	Goat	Goat	D3	D2	Y	W

You will see there are twice as many possible options where you have selected the correct door. That's because Monty has two different wrong doors to play with, and he can eliminate either one. So there are 4 possible paths, and two of them (the ones where you stick with your original choice) result in winning the car.

If you originally selected a wrong door (D2 or D3) Monty has only 1 option. He eliminates the remaining wrong door. So in this case you need to switch doors in order to win the car.

Bottom line, there are 8 possible scenarios.

In 2 of them, you stick with your original choice and win.
In 2 of them, you stick with your original choice and lose.
In 2 of them, you change your original choice and win.
In 2 of them, you change your original choice and lose.

I rest my case.

OK, your a former geek.

try reading the proof here Monty Hall problem - Wikipedia, the free encyclopedia.

or look up the "monty hall problem" in your favourite search engine.

Originally Posted by _fly_

OK, your a former geek.

try reading the proof here Monty Hall problem - Wikipedia, the free encyclopedia.

or look up the "monty hall problem" in your favourite search engine.

Yep, read that. The analysis is flawed, because as shown in my proof above, it ignore the fact that there are 4 possible outcomes where you have initially selected the correct door, and only two possible outcomes for each wrong door selection.

Wikipedia is wrong (not the first time).

Switt, there are only three scenarios to start out.

1, You pick the door with the car

2. You pick the door with goat 1

3. You pick the door with goat 2

If your strategy is to stick with your first pick, you win the car only when you pick the door with the car, statistically once in 3

If your strategy is to switch no matter what, you win the car every time you initially pick a goat, statistically 2 in three. If you switch and you have a goat in your chosen door, the other goat has been eliminated by Monty, so you must be switching to the car. This happens in 2 of the 3 initial selection possibilities.

If you initially chose a door with the car you would be switching to the remaining goat, but your initial chances of picking the car are only 1 in 3.

What is behind the door that Monty chooses not to open, when he has a choice, is not totally random because of information we already have.

I hope this helps.

Cheers

Doug

Originally Posted by switt775

I stand by my earlier post. Here's the proof...

You will see there are twice as many possible options where you have selected the correct door. That's because Monty has two different wrong doors to play with, and he can eliminate either one. So there are 4 possible paths, and two of them (the ones where you stick with your original choice) result in winning the car.

Right here is where you logic goes wrong Switt. You are looking at the possible options that Monty has to play with, not what the contestant is left to choose from.

Yes, if you have picked the door with the car Monty has a choice of two doors to open, but whichever one he opens, the contestant,and thats whose choices we are evaluating, still is left with only one door to switch to. If he does not have the car behind the door he initially chose then the car must be behind the only remaining door.

The puzzle is looking at the choices the contestant has left after Monty has opened a door, not how many doors Monty has a choice of opening.

Cheers

Doug

Sorry Switt, your chart lays out the possible outcomes, but does NOT lay out the chances (probability) of each one.

In simple terms:

Your first pick has a 1 in 3 chance of winning. Which means the odds are 2:3 it's behind one of the other doors.

It is a certainty that Monty will open a "losing" door... so the remaining, unopened door still has 2:3 chance of being the winner.

1:3 for your first pick, compared to 2:3 for a switch.



Hell, even if I didn't believe the maths, in the same situation I'd tell myself - at the final choice - "I have a 50/50 here, so I'll switch simply because the savants (who're s'posed to understand these things) say it's the better choice."

A bit like my consistently choosing heads in a coin toss, which everyone "knows" is 50/50 but actually is not... it depends on how well-balanced the coin in question is.

Originally Posted by doug3030

Switt, there are only three scenarios to start out.

1, You pick the door with the car

2. You pick the door with goat 1

3. You pick the door with goat 2

If your strategy is to stick with your first pick, you win the car only when you pick the door with the car, statistically once in 3

If your strategy is to switch no matter what, you win the car every time you initially pick a goat, statistically 2 in three. If you switch and you have a goat in your chosen door, the other goat has been eliminated by Monty, so you must be switching to the car. This happens in 2 of the 3 initial selection possibilities.

If you initially chose a door with the car you would be switching to the remaining goat, but your initial chances of picking the car are only 1 in 3.

What is behind the door that Monty chooses not to open, when he has a choice, is not totally random because of information we already have.

I hope this helps.

Cheers

Doug

Doug, I stand by my analysis because it is mathematically correct.

Initially there are 3 possible outcomes. But because Monty always eliminates a door with a goat, there are 4 outcomes, as shown in the matrix above. Monty is intentionally improving the odds of success to 50%. No more, no less.

There are many pitfalls in logic, but mathematics don't lie.

Here is another logic / math puzzle to ponder. You decide to flip a coin 10 times and record the results. The coin and your flipping technique have been carefully examined by Monty himself, and there is no bias in the coin, or your way of flipping it.

Q1. What are the odds that you will flip 10 consecutive Heads?

Q2. You have amazed yourself and your friends by flipping 9 consecutive Heads. What are the odds that the 10th flip will be Heads?

Before we go any further, there are a few things that must be clarified.
The first is that the car is always placed randomly amongst the three doors.
Second, the host always knows where the car is.
The third is that the host will always open a door to reveal a goat.
The last important detail is that the contestant (that’s you) knows all of this information ahead of time.

Switt 775's 'mathematical response' ignores this "precondition".

Here



Cheers!