OMG.

yes! take the bet.

To Click and Clack:

I disagree with your published answer

> TOM: So you’re saying the chances of winning are two to one in favor of

> red?

about dependent probabilities:

Multiplication Rule 2:

When two events, A and B, are dependent, the probability of both

occurring is:

P(A and B) = P(A) * P(B | A)

A = picking a card with red on either side

B = probability of a card with red on one side is also red on the other

P(A) = 2/3

P(B) = 1/2

P(A and B) = 2/3 * 1/2 = 1/3

(at the bottom I’ll directly respond to Mechaniker’s reponse).

I imagine you’ll get a lot of other people also writing to

tell you.

The odds are that the con man is wrong. it’s only 1/3 likely that

the other side of the card is red.

run a simulation. you’ll see.

RAY: Believe it or not. You can believe it because what

you’re dealing with is not cards–

you’re dealing with the sides. When you see the red side

up,

you could be seeing one or the other face of the red card.

That’s what most people don’t grasp

It’s true that you don’t know if a card showing red has red on the other side.

That’s the point of the question.

Your conclusion is wrong.

You seem to conclude that because there are more red sides, it’s more likely

to see red on the other side.

To help you visualize, consider a situation with 1000 cards.

Let’s say 500 are green on both sides; 499 have both red and green;

and 1 has red on both sides.

As in your puzzler, there are more red sides than green sides.

given that it’s not a green/green card

green sides = 499

red sides = 499 + 2

Here, P(A) = 1/2

P(B) = 1/500

P(A and B) = 1/2 * 1/500 = 1/1000

Alternatively, consider only 1 card out of 1000 that have red on both sides

What do you think your chances are of picking that card?

Doesn’t it seem to you that it would be unlikely. say 1/1000.

Given that it’s not a green/green card, there are more red sides possible,

but do you really think you’d have been more likely to pick that 1 card in 1000

instead of one of those cards with red and green ?

This is similar to the “Monty Hall” situation. shown 3 doors, all closed.

the contestant picks one.

Monty opens a door, showing a booby prize, and asks if you should switch.

The answer is yes. Your chances are 2/3 that the door you didn’t pick is the

winner.

To visualize, again consider 1000 doors.

only one good door.

you pick a door.

Monty opens 998 doors with bad prizes.

Should you switch ?

yes.

The chances are 1/1000 that you picked the good door.

The chances of the other door you didn’t pick is 999/1000 of having

the good prize behind it.

The fact that doors were opened doesn’t increase your original odds.

But, you now have more information, and you know that 998 doors were wrong.

Chances were 999/1000 that you didn’t pick the right door, and now that

you know 998 of the other doors were wrong, the one remaining door is

likely to hide the good prize with a probability of 999/1000

Switch your choice.

For the puzzler, take the bet.

You’ll be right 2/3 of the time.

Mechaniker says:

use Bayes’ theorem: P(A|B) = P(B|A) * P(A) / P(B)

Let A be the event: red on both sides before seeing face; P(A) = 1/3

Let B be the event: a red face shows up; P(B) = 1/2

Let B|A be the event:

red face shows up given that red is on both sides; P(B|A) = 1

Then A|B is the event: red on both sides given that red face up;

P(A|B) = 1*(1/3)/(1/2) = 2/3

With all due respect, I think this is a mis-application of Bayes’ theorem.

You don’t get to pick the side.

You pick a card.

you say P(B|A) = 1

but that assumes you’ve picked the red/red card.

You define A to be the event of picking the red/red card.

Of course, if you pick the card with red/red, the probability of the other

side being red is 1.

But you don’t know that A has occurred.

All you know is that red is visible.

It should be:

The initial event, A, is picking a card. some have red, some don’t.

It should be that:

A = picking a card with red on some side.

P(A) = 2/3

now that you’ve picked a card with red showing, what’s the probability that

the other side is red? there are two possibilities, 1/2 are red.

B = probability that other side is red

P(B) = 1/2

using Multiplication Rule 2:

When two events, A and B, are dependent, the probability of both

occurring is:

P(A and B) = P(A) * P(B | A)

The application of Bayes’ theorem should be:

P(B | A) = 1/2

P(A) = 1/3

P(B) = 1/2

P(A|B) = 1/2 * (1/3) / (1/2) = 1/3

so the chances of having picked the red/red card are 1/3

you’d be right 2/3 of the times to take the bet with the con man.

This also is an answer to Mechaniker February 22

The question: What now are the cowboys odds of winning if he stays with his

original selection of card #1? What are the cowboys odds of winning if he

switches to card #2? Explain your reasoning.

the cowboy should switch his choice.

just like the Monty Hall situation.

I agree with your post on Mechaniker February 23