Cognitive Illusions :

[Kramer]

A cognitive illusion arises when some parts of our minds cannot access information in other parts of our minds, and our perception of reality is led astray.

Cognitive illusions are remarkably persistent, and very common. And their existence implies that our intuitive judgements about some things may often be wrong - indeed, are likely to be wrong - because of the way our minds are hard-wired.

Let me give you a great example that you can use on yourself and others to get an argument going. It's most often called the Three Box Problem.

Imagine that I set up three boxes in front of you, and call them 'A', 'B', and 'C'. I've placed a twenty dollar bill inside one of the boxes.

I know which one has the $20, and you do not. I ask you to pick a box, any box. Let's agree for ease of discussion that you pick Box 'A'. Now, what are the odds that you selected the box with the money in it?

You, of course, answered one in three. So far, so good. Now, imagine that I do my best Monty Hall imitation and tell you that I'm going to give you a chance to change your guess, but first I'm going to open one of the boxes. Then, with much ceremony, I proceed to show you that - AHA! -

there's nothing inside of Box 'C'.

There are now two closed boxes in front of you: Box 'A' and Box 'B'. The choice before you now is to either stay with Box 'A', or change your guess to Box 'B'. What are the odds associated with your choices? Should you stay, should you change, or does it many any difference at all?

Before I tell you the answer, let me assure you that cognitive scientists report that very smart people mostly get this wrong. And they don't like the answer even after it's explained to them.

That's the very nature of cognitive illusions. They're persistent because of a hard-wiring issue, and not because of intellect or education.

Here's the answer: There is a one-third probability that the $20 is under Box 'A', and a two-thirds probability that it's under Box 'B'.

When Monty opens one of the boxes and then lets you change your original guess, you should always change your guess. In fact, doing so would cause you to win two-thirds of the time.

I know! You thought the odds of the $20 being under either Box 'A' or 'B' were 50-50, right? I mean, after all, there are two boxes in front of you. One of them has $20 in it, and the other one doesn't. It's a no-brainer that if it ain't in one, it's in the other - and your guess is a 50-50 shot, right?

Except that it's not. You've been pulled into a little mental tunnel where you can't see what's really going on, and the obvious thing that your intuitive judgement is insisting on is just flat wrong.

Let's look at how things really are.

Remember when you first picked Box 'A'?
We agreed then - and can agree again - that there was a one-third chance that the $20 was in Box 'A'. You'll notice that there's also a one-third chance that the money is in Box 'B', and a one-third chance that it's in Box 'C'.

Now, with that last part in mind, what was the probability that the $20 was in "Not Box 'A'"? Never mind whether it's in Box 'B' or 'C'.

Just focus on the idea of "Not 'A'".

The probability that the money is in "Not 'A'" is two-thirds.

Do you see it now? After an empty box is removed, you are left with your original choice of Box "A", which had - and has - a one-third probability of containing $20.

And now, thanks to the removal of the empty box, the two-thirds probability of the "Not 'A'" choice is contained entirely in one box: Box 'B'.

You can tell Monty that you happen to be a student of both cognitive science and probability, and you would be most happy to change your original guess and thereby double your probability of winning.

No game show is going to find you stuck in any mental tunnels. (If you're still kind of stuck, you can amuse yourself with trials. Do a bunch sticking to your original guess, and then a bunch where you always switch. Enlightenment will come.)

Okay, that's a bit of fun, and you're now well-equipped for the inevitable revival of Let's Make A Deal. And recounting this effect to others will create lively conversation about how you're wrong, and how the odds on the second guess cannot be anything other than 50-50.

But the implications of the presence of cognitive illusions that are common and persistent aren't amusing. They are, in fact, expensive.

[mrwho]

I dont get this one Kramer, why wouldnt the ball not in Box B 2/3 chances also? I must be dumb.

[Bobtheicon]

I think it sucks that I'm not smart enough to agree with this.

[mrwho]

Originally Posted by Bobtheicon

I think it sucks that I'm not smart enough to agree with this.

you're not the only one bob.

[Bobtheicon]

Wait a minute...
What if when you are offered the chance to change your pick, some other guy walkes into the scene. This guy never saw the original choice of the 3 boxes; when he gets on the scene there are only 2 boxes. Wouldn't he have a 50/50 shot of picking it right? So, regardless of the past, there are currently 2 boxes in front of me. I think I should have the same 50/50 chance as the guy.

[mrwho]

Originally Posted by Bobtheicon

Wait a minute...
What if when you are offered the chance to change your pick, some other guy walkes into the scene. This guy never saw the original choice of the 3 boxes; when he gets on the scene there are only 2 boxes. Wouldn't he have a 50/50 shot of picking it right? So, regardless of the past, there are currently 2 boxes in front of me. I think I should have the same 50/50 chance as the guy.

yes i was thinking this also...or even if the guy is there to pick in beginning..you pick a he picks b. They eliminate c and this is saying the guy has a better chance of having the money bc he's on box b?

[skippy]

When there are 3 boxes and the money is in ONE of them.
The chance of it being in A is 1/3.....in B is 1/3.....in C is 1/3.

If you pick box A...the chance of the money being there is 1/3.
Thus the chance of the money NOT being there is 2/3 because you have B & C (1/3 + 1/3 = 2/3) or 1 - P(A).

One out of 3 means that in the worse case scenario..you would need to open up 3 boxes to find the money.

....................

Now if one empty box is opened and removed....there are only 2 boxes left hence the chance of the money being in the first box is 1/2 and the second box is 1/2.

One out of 2 means that in the worse case scenario..you would need to open up 2 boxes to find the money.

....................

Now if another empty box is opened and removed....there is only 1 box left hence the chance of the money being in that box is 1 or 100%....not 1/3

With the elimination of each box....your chances improve.

[Kramer]

Its because the way our minds have been taught

You pick box A

The probability that the money is in "Not 'A'" is two-thirds.

There’s nothing in box C

[mrwho]

yea but probability not in box b is 2/3 also...

[skippy]

Originally Posted by Kramer

Its because the way our minds have been taught

You pick box A

The probability that the money is in "Not 'A'" is two-thirds.

There’s nothing in box C

P(A) = 1/3
P(~A) = 2/3

But once a box has been removed....the probabilities have to be recalculated based on how many boxes are left.

[Yankee23]

I'll take what's behind the curtain

[skippy]

Originally Posted by skippy

P(A) = 1/3
P(~A) = 2/3

But once a box has been removed....the probabilities have to be recalculated based on how many boxes are left.

Put it this way....

Contestant #1 picks C and its empty.

Now its contestant #2's turn.

There are only 2 boxes left.

What are his chances now???

1/2 or 2/3???

[uva3021]

[Kramer]

90% of the bets are on the Cowboys -10
1 of 4 things will happen :
1: Cowboys win and cover
2 : Cowboys win don't cover
3 : Bills win + cover
4 : Push (at -10)
Therefore you have 2/3 chance of winning your bet at Bills + 10
Who do you bet yet 90% Bet the Cowboys ?

[mrwho]

Nm..