196 replies to this topic

[bonanova]

I will still stick to #2.
Earlier i had a blind guess of winning a car prob:1/3
Now i have prob:1/2

My hunch tell me however:: Had i selected a door with goat on first choice, the host would have happily opened it to let all know, the game would end there. They were always more interested in saving a jaguar for a goat.

Why did 1/3 become 1/2?

Remember Monty did not guess which door had a goat. He knew.
And he didn't open the door you chose, he showed a different goat.

[aaronbcj]

Sorry, I am not sure if the rules were changed in later posts, but i read it in every first post as

" .....says the host, "but before we show you what's behind Door #2, let me tell you that I am glad you did not pick Door #1, because behind Door #1 is..." The first door swings open to reveal a shabby looking goat......"

so in my "second round", i only have to choose between door #2, #3 or stick with earlier selection (door#2).

[bonanova]

Sorry, I am not sure if the rules were changed in later posts, but i read it in every first post as

" .....says the host, "but before we show you what's behind Door #2, let me tell you that I am glad you did not pick Door #1, because behind Door #1 is..." The first door swings open to reveal a shabby looking goat......"

so in my "second round", i only have to choose between door #2, #3 or stick with earlier selection (door#2).

You're fine on everything you've said. Three options have now become two options: Door 2 [stay] and Door 3 [switch].
Initially all the doors have 1/3 probability of having the car.
After we learn that Door 1 has 0 probability, where does that 1/3 go?

You assume it gets split equally between 2 and 3, making them both 1/2.
But why should Door 2 change, just because we are shown a goat behind one of the other two doors?
We know Doors 1 and 3 combined have 2/3. If we learn Door 1 has 0, doesn't that mean Door 3 now is 2/3?

Or,

If Door 2 has the car, switching to Door 3 loses.
If Door 2 has a goat, switching to Door 3 wins.

Are these cases equally likely?

[aaronbcj]

We know Doors 1 and 3 combined have 2/3. If we learn Door 1 has 0, doesn't that mean Door 3 now is 2/3?

No. Door#1 earlier had 1/3, now that it is open, it is out of equation. Only 2 doors are in race now and we have to re-assign prob. of 1/2, 1/2 for #2,#3. Both the cases you mentioned are equally likely (hence 1/2, 1/2) to happen.

Few may think that changing to #3 is better option but there is no place for finding prior prob. in this case (as #1 is removed out of equation instead of just having its prob changed).

You can read my older comment on why i still wish to stick with #2, even though it has only 1/2 of winning ..

[maurice]

you know...you can sit here and write out all the math you want, but in the end if there's 3 doors and you already know 1 to have the goat, then one door has a goat and 1 door has a car. That's a 50/50 chance. If there were 20 doors and I took away 18, you'd still be left with 2 and 1 would be the winning door. It's always going to be 50/50. It's pure logic. Mathematics may not agree with logic, but logic doesn't care. Logic is as logic does.

Funny...all this time I thoughty math was logic...hmmm

Math may not agree with intuition, or what is believed to be common sense...but Math and Logic are attached at the hip

Edited by maurice, 18 October 2010 - 05:36 PM.

[matthew laming]

This is the monty hall paradox. always switch. if you work out the probability of getting the jag in each staying or switching, switching wins. most people think its 50/50
which isn't the case.

[mmiguel1]

No. Door#1 earlier had 1/3, now that it is open, it is out of equation. Only 2 doors are in race now and we have to re-assign prob. of 1/2, 1/2 for #2,#3. Both the cases you mentioned are equally likely (hence 1/2, 1/2) to happen.

Few may think that changing to #3 is better option but there is no place for finding prior prob. in this case (as #1 is removed out of equation instead of just having its prob changed).

You can read my older comment on why i still wish to stick with #2, even though it has only 1/2 of winning ..

You cannot reassign the probabilities like that.

Let me tell you in the way that makes the most sense to me.

Let's say we have two people playing this game, Switcher and Stayer.
Switcher has firmly decided to always switch doors, while Stayer has decided to never switch doors.

Let's start with Stayer.
At the outset of the game Stayer has a 1/3 probability of picking the door with the car, and 2/3 probability of picking a door with a goat.
If he picks right, which happens about 1/3 of all the times he's taken part in this silly competition, then when Monty shows the other door, Stayer will win, since he initially picked right and stayed with it.
If Stayer picks wrong initially, which happens 2/3 of the time, he will lose after Monty opens a door, because he was initially wrong and stuck with it.
We can therefore say that Stayer will win 1/3 of the time and lose 2/3 of the time.

Now let's look at Switcher.
Just like Stayer, at the outset of the game, Switcher has a 1/3 probability of picking the door with the car, and 2/3 probability of picking a door with a goat.
If he picks right, which happens about 1/3 of the time, then after Monty opens a goat door, switcher will switch doors to the other goat door and lose, since he switched from the correct door to the particular goat door that Monty didn't open.
If Switcher picks wrong initially, which happens 2/3 of the time then he will be pointing at one of the goat doors. Monty, knowing where everything is, will open the other goat door, and Switcher will switch doors to the car door and win. (Remember this happens 2/3 of the time)
We can therefore say that Switcher will win 2/3 of the time and lose 1/3 of the time.

Let me recap, based on the strategies of the two players and a logical analysis of all possibilities,
Stayer will win with probability 1/3 and lose with probability 2/3
Switcher will win with probability 2/3 and lose with probability 1/3.

Who would you rather be?

Think of it this way, Switcher is betting his initial guess is wrong, while Stayer is betting his initial guess is right. Wrong and right have probabilities 2/3 and 1/3 each, so which would you bet happened?
You bet you were wrong! Sometimes that's the best way to be right.

I think looking at the problem like this, causes it to agree with intuition rather than disagreeing with it.

Edited by mmiguel1, 19 October 2010 - 04:12 AM.

[matthew laming]

mmiguell is right. and he just proved it.

[mmiguel1]

mmiguell is right. and he just proved it.

Thanks matt, (my name is matt too, last name is miguel).
Looking at the length of this thread, I would bet that it's been proven before somewhere in here at least 10 times over.
I don't know if they stated it the same way as me already, but I wouldn't be terribly surprised if it was already stated like this. (Bonanova likes to convey the logic in multiple perspectives)

[matthew laming]

Thanks matt, (my name is matt too, last name is miguel).
Looking at the length of this thread, I would bet that it's been proven before somewhere in here at least 10 times over.
I don't know if they stated it the same way as me already, but I wouldn't be terribly surprised if it was already stated like this. (Bonanova likes to convey the logic in multiple perspectives)

Yeah it probably has ha ha
well done on my sequence puzzle. theres another one if you fancy a challenge here. http://brainden.com/...umber-sequence/
I'm guessing you have a quite high iq?