Fun with Probability

[lesliedellow]

The probability of the host opening door three is only 1 if the car is behind door 2.

It is not about the host opening door 2, it is about the event of a goat being behind door 2. When he opens the door, and a goat is revealed, that probability becomes 1, whereas it was previously 2/3. Probabilities are not immutable, and they are not inherent properties of anything.

 

[crjmurray]

It is not about the host opening door 2, it is about the event of a goat being behind door 2. When he opens the door, and a goat is revealed, that probability becomes 1, whereas it was previously 2/3. Probabilities are not immutable, and they are not inherent properties of anything.

Whatever door Monty opens will always be a goat. If you pick A you have a 1/3 chance of being correct. If the car is behind A then the odds of Monty opening door B are 1/2. If the car is behind B the probability of Monty opening B is 0. If the car is behind C the probability of Monty opening B is 1. Good so far?

 

[lesliedellow]

Whatever door Monty opens will always be a goat. If you pick A you have a 1/3 chance of being correct. If the car is behind A then the odds of Monty opening door B are 1/2. If the car is behind B the probability of Monty opening B is 0. If the car is behind C the probability of Monty opening C is 1. Good so far?

Sure you wouldn't like to revise that?

 

[crjmurray]

Sure you wouldn't like to revise that?

Revised.

 

[lesliedellow]

Whatever door Monty opens will always be a goat. If you pick A you have a 1/3 chance of being correct. If the car is behind A then the odds of Monty opening door B are 1/2. If the car is behind B the probability of Monty opening B is 0. If the car is behind C the probability of Monty opening B is 1. Good so far?

So?

 

[crjmurray]

So?

So what is the probability that Monty will end up opening door B?

 

[crjmurray]

I'll make it easier.

The probability that Monty opens B is:
p(Monty opens B) = p(A) * p(Monty opens B|A) + p(B) * p(Monty opens B|B) + p(C) * p(Monty opens B|C) = 1/6 + 0 + 1/3 = 1/2

Good so far?

 

[lesliedellow]

So what is the probability that Monty will end up opening door B?

A - Car behind A
B - Car behind B
C - Car behind C

B* - Monty opens door B

Pr(B*) = Pr(B*&A) + Pr(B*&B) + Pr(B*&C)

= Pr(B*|A) * PrA) + Pr(B*|B) * Pr(B) + Pr(B*|C) * Pr(C)

= 1/2 * 1/3 + 0 + 1 * 1/3 = 1/2

 

[crjmurray]

A - Car behind A
B - Car behind B
C - Car behind C

B* - Monty opens door B

Pr(B*) = Pr(B*&A) + Pr(B*&B) + Pr(B*&C)

= Pr(B*|A) * PrA) + Pr(B*|B) * Pr(B) + Pr(B*|C) * Pr(C)

= 1/2 * 1/3 + 0 + 1 * 1/3 = 1/2

Oh good. Now that we have some actual values we can use Bayes' Theorem to work through this.

So what is the probability of A being the correct door if B is opened?

P(A|Monty opens B)= p(A) * p(Monty opens B|A)/p(Monty opens B)

That's a fairly simple one.

Hint: this is where that whole fun part starts.

 

[lesliedellow]

That's a fairly simple one.

Sorry, I failed my eleven plus.

 

[crjmurray]

.

 

[crjmurray]

Sorry, I failed my eleven plus.

Is that a concession or would you like me to finish the math?

 

[Loudmouth]

Since we have so many experts on probability here, I thought we could play around with some puzzles. We'll start with an oldie.

"Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice?"

Well? Do you stay with your original choice or do you switch? By staying or switching does one give you any better/worse odds?

Without looking at the other answers . . .

Your chances of picking the right door on the initial guess is 1 in 3. By eliminating one of the doors, your odds are reduced to 1 in 2. Therefore, switching to the other door reduces your odds from 1 in 3 to 1 in 2.

 

[crjmurray]

Without looking at the other answers . . .

Your chances of picking the right door on the initial guess is 1 in 3. By eliminating one of the doors, your odds are reduced to 1 in 2. Therefore, switching to the other door reduces your odds from 1 in 3 to 1 in 2.

Hey! Good jo- nah just kidding that's not right.

 

[Loudmouth]

Hey! Good jo- nah just kidding that's not right.

You are right. After finding the answer elsewhere, I can say that I was close.

 

[crjmurray]

You are right. After finding the answer elsewhere, I can say that I was close.

You did realize that the initial choice is 1/3 and that switching increases your odds. You get points for that.

 

[crjmurray]

I'll go ahead and work through the last problem I posted. This is the probability of A being correct after B has been opened.

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

1/3 not the seemingly correct 1/2. Let's prove this further. The probability of B being correct is 0 if Monty is opening B since Monty will always reveal a goat. This means we can move on to C. If switching is the appropriate move, we should be able to show the increase in probability. This is the probability of switching to C after Monty has opened B.

P(C|Monty opens B) = p(C) * p(Monty opens B|C)/p(Monty opens B)= (1/3*1)/(1/2)= 2/3

As I said earlier, you are twice as likely to be correct if you switch from your initial choice after a door has been opened.

 

[Albion]

But in practice, you are twice as likely to win if you switch.

But it's not because of statistics, right? It has something to do with the psychology of game shows or what makes the most interesting scenario for the viewers, etc.

 

[crjmurray]

But it's not because of statistics, right? It has something to do with the psychology of game shows or what makes the most interesting scenario for the viewers, etc.

If it's not because of statistics then what's all that math doing up there?

 

[Subduction Zone]

Oh boy, I love the Monty Hall Paradox. Even mathematicians have been known to get this simple exercise wrong.