Brain Teasers

Vassago,

The choices have changed. These are not INDEPENDENT events.

In an extreme example, consider 1000 doors; one is the winner.

Your choice has 1/1000 of being right.

If the host removes 998 doors, you are faced with two options:

Keep your door ... still a 1/1000 chance.

Switch to the other door ... 1/2 chance.

Even if they remove only one choice your original 1/1000 is still worse
than switching to 1/999 ... It is a "brand-new" scenario.

You should always switch.

It seems counter-intuitive, but it is not like the consecutive coin-flip
scenario. Those are independent events.

Wayne
 
I understand the 1000 door example - well put.
The 1 other door has the perception of a better chance of being the winning door because the 998 removed doors were false.

To switch to this door is a higher probability 1/2 then your original choice which was 1/1000 and the chance of your original choice Was correct remains at 1/1000.

In effect, for the first choice to be correct it has to perform two criteria.
Be the 1/1000 and now the 1/2 - long shot - against the 2nd door which only has to beat the 1/2 odds.

Still think it will be a hard choice to make at the time.
 
What entertained me the most was reading from the beginning of this thread... pondering one question that stumped me for a few moments.... and then thought... "A DEAL!" only to scroll down a few posts to find TessB suggest "A deal?"

I looked at the date.... 2006???? WOW!
I scrolled down a bit more, to look for confirmation of my correct answer.
It said "Cards" I frowned..... lol. I thought... But..... a deal fits the logic as well!

Scrolled down more.... watched myself fight for recognition.
Will I EVER change?
 
Vassago,

The choices have changed. These are not INDEPENDENT events.

In an extreme example, consider 1000 doors; one is the winner.

Your choice has 1/1000 of being right.

If the host removes 998 doors, you are faced with two options:

Keep your door ... still a 1/1000 chance.

Switch to the other door ... 1/2 chance.

Even if they remove only one choice your original 1/1000 is still worse
than switching to 1/999 ... It is a "brand-new" scenario.

You should always switch.

It seems counter-intuitive, but it is not like the consecutive coin-flip
scenario. Those are independent events.

Wayne

Not quite -

Imagine 1000 doors, 1 of which has a car behind it 999 have a goat.

Your chance of choosing the door with the car behind it are 1/1000.

And 999/1000 you were wrong and choose a goat.

Inevitably 998 of the remaining 999 doors at least have a goat behind them, we knew this before the original choice, after the original choice and we still know it after the host reveals the inevitable 998 goats.

So we know nothing new about the original choice. So the odds of it being correct haven't changed.

- with just 2 doors left unopen - the chance we were originally correct is still 1/1000 and still 999/1000 we were wrong.

All the host in effect asks you - is do you want to stick with the 1/1000 chance that you were originally correct or switch to the 999/1000 chance, that you were originally wrong.

Although theres 2 choices, the odds aren't 50/50.
 
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I understand the brain teaser logically, but it is using disingenious means to make the math look a particular way. Assume someone says "select one of these three options", and one of the three options is labeled "Don't pick me". If the person making the pick is guaranteed that the "Don't pick me" label is not a trick, then it really isn't an option anymore.

Or to put it another way, it would be like having three women stand on stage, a blonde, a brunette, and a red head. Each of the women is holding a bag that contains an amount of money between $0 and $1 million. The blonde's bag is labeled "$0". Then, after watching several different people make selections, drawing the conclusion that no one likes blondes. That would be anagalous (and just as faulty) as the brain teaser.
 
I understand the brain teaser logically, but it is using disingenious means to make the math look a particular way. Assume someone says "select one of these three options", and one of the three options is labeled "Don't pick me". If the person making the pick is guaranteed that the "Don't pick me" label is not a trick, then it really isn't an option anymore.

Or to put it another way, it would be like having three women stand on stage, a blonde, a brunette, and a red head. Each of the women is holding a bag that contains an amount of money between $0 and $1 million. The blonde's bag is labeled "$0". Then, after watching several different people make selections, drawing the conclusion that no one likes blondes. That would be anagalous (and just as faulty) as the brain teaser.

Uh? Theres nothing disingenuous at all about the question is there?
 
An old guy is playing the game he selects door3, door1 is opened to reveal a goat, the show cuts to a commercial break.
After the break he is asked if he wants to switch , he says" I can't remember my original choice so I'll have door3". He has chosen 1 of 2 , why is it not 50 - 50.

Brian
 
From a choice of two it is 50-50, however.... in a choice 1 of 1000 it is 1 of 1000 and stays 1 of 1000 that your pick is right.

So if -before the commercial break- he picked one is has a 999 of 1000 chance to be wrong. So you could argue that if they take away 998 that he still has a 999/1000 chance to be wrong about his current choice. Hence statisticaly, your first choice would most likely be wrong (999 out of 1000 odds on that your wrong).

However... If out of the 998 this also includes your current choice or if you are just to start your pick at the 2 options, then yes 50-50
 
As PngBill and I have both said, you guys are using history to make your point, history can play no part in the next odds.

With 3 doors there is a 1 in 3 chance for each door.
With 2 doors there is a 1 in 2 chance for each door.

The odds for every door change as doors are taken out of the choice.

Brian
 
An old guy is playing the game he selects door3, door1 is opened to reveal a goat, the show cuts to a commercial break.
After the break he is asked if he wants to switch , he says" I can't remember my original choice so I'll have door3". He has chosen 1 of 2 , why is it not 50 - 50.

Brian

To him now it is 50/50 - I think.

Hes forgotten which door , door 1 was "associated with"

the important bit is which remaining door, door 1 was associated with has twice the chance of having the car

Door 3 - had a 1/3 chance.
And together Door 1 and 2 had a 2/3 chance

but he cant remember if it was

Door 2 - had a 1/3 chance
And together Door 1 and 3 had a 2/3 chance

So hes 50/50 on which door, door 1 was associated with. Or hes 50/50 on which door he originally picked.

The audience remembers that it was door 2 - that had twice the chance.

--
Think of it in the 1000 box example - he cant remember which door was the 1/1000 chance or which was the 999/1000 chance.

So he guesses - and its now 50/50 to him.
 
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A new one -




Its the year 2008. Last month was still 2008 and next month is 2008.

If the date of the last Monday of last month is added to the date of the first Thursday of next month, the result is 38.



What is the current month?
 
A new one -




Its the year 2008. Last month was still 2008 and next month is 2008.

If the date of the last Monday of last month is added to the date of the first Thursday of next month, the result is 38.



What is the current month?

August.

The only way to achieve this is for the monday to be the 31st and the Thursday to be the 7th.

So the current month starts on a Tuesday and must end on a thursday so that the first thursday can be the 7th. This means it must have 31 days also. Only July and August are consecutive months, in the same year, which both have 31 days therefore it must be August.
 
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Here's another.

At the Infinity Hotel there are an infinite number of rooms. A man walks upto reception and asks for a room. He is told there are already an infinite number of guests staying at the hotel so all the rooms are occupied.

How do they get him a room to stay in?
 
Uh? Theres nothing disingenuous at all about the question is there?

I'm not sure if it is the way you phrased your question, or if I am just slow, but I'm not sure what you're asking. If you didn't understand what I was trying to say, Brian hit on the key point with:

Brianwarnock said:
As PngBill and I have both said, you guys are using history to make your point, history can play no part in the next odds.
 
If the probability was 50/50 then there will be an equal number of winning outcomes based on switching or not switching. That means I have missed 3 possible scenarios from my list. If you can provide the 3 scenarios I will agree that the probability is 50%.
 
An old guy is playing the game he selects door3, door1 is opened to reveal a goat, the show cuts to a commercial break.
After the break he is asked if he wants to switch , he says" I can't remember my original choice so I'll have door3". He has chosen 1 of 2 , why is it not 50 - 50.

Brian

In this situation the probability is 50/50. It's only if the person knows which was originally chosen and which door was not eliminated by the host that the probability is not 50/50.
 
If the probability was 50/50 then there will be an equal number of winning outcomes based on switching or not switching.

There is. Here:

Code:
Winning Door                  Door Chosen    If you switch do you win?
1                                   1                          No
1                                   2                          Yes
2                                   1                          Yes
2                                   2                          No

The moment door x is revealed to no longer be a viable option, it ceases to exist.
 
If the probability was 50/50 then there will be an equal number of winning outcomes based on switching or not switching. That means I have missed 3 possible scenarios from my list. If you can provide the 3 scenarios I will agree that the probability is 50%.

I think the difference is, is that it becomes a brand new choice, a brand new game, so to speak. So once the question is asked, "do you want to switch?" It presents a new problem, your scenarios are reduced to two (Edit: the prize is behind 1 or behind 2), therefore 50/50 chance that you are correct.
 
There is. Here:

Code:
Winning Door                  Door Chosen    If you switch do you win?
1                                   1                          No
1                                   2                          Yes
2                                   1                          Yes
2                                   2                          No

The moment door x is revealed to no longer be a viable option, it ceases to exist.

yes but you don't know what door will be eliminated therefore you must list all possible combinations from the starting position. Above you have excluded door 3 but you don't know that door will be excluded therefore it must be included in possible outcomes of the game.
 
yes but you don't know what door will be eliminated therefore you must list all possible combinations from the starting position. Above you have excluded door 3 but you don't know that door will be excluded therefore it must be included in possible outcomes of the game.

But that's just it, at the point that this question becomes viable, you do know that door three is excluded, so it doesn't have to be included in the possible outcomes.
 

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