Fun with Probability

crjmurray · May 19, 2015

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?

amanuensis63 · May 19, 2015

crjmurray said:
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?

Always switch after the reveal. THis is called the MONTY HALL PROBLEM.

It relates to a change in the odds with new information. Something to do with Bayes Theorem.

lesliedellow · May 19, 2015

The odds shorten from one in three to one in two. So why switch?

Michael · May 19, 2015

crjmurray said:
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?

That was a good brain teaser. I wouldn't have thought it really made any difference, but as suggested I read up on the Monty Hall Problem, and indeed there is a statistical advantage to switching. How interesting. The image they put on WIKI that seemed to make it click for me was this one:

crjmurray · May 19, 2015

lesliedellow said:
The odds shorten from one in three to one in two. So why switch?

Do you think the chances of picking the right door are equal between staying and switching?

MerlinJ · May 19, 2015

lesliedellow said:
The odds shorten from one in three to one in two. So why switch?

It's much more obvious if you increase the number of doors. Try it out yourself.

http://onlinestatbook.com/2/probability/monty_hall_demo.html

lesliedellow · May 19, 2015

crjmurray said:
Do you think the chances of picking the right door are equal between staying and switching?

Before the door was opened, the possibilities were:

Code:

Door 1.       Door 2.       Door 3
Car.         Goat 1.       Goat 2
Car.         Goat 2.       Goat 1
Goat 1.       Car.         Goat 2
Goat 1.       Goat 2.        Car
Goat 2.       Car.          Goat 1
Goat 2.       Goat 1.        Car

After the door is opened, possibilities 4 and 6 are ruled out. So the remaining possibilities are:

Code:

Door 1.        Door 2.        Door 3
Car.          Goat 1.        Goat 2
Car.          Goat. 2.       Goat 1
Goat 1.        Car.              Goat 2
Goat 2.        Car.              Goat 1

Therefore the conditional probability of a car being behind door 1 is 1/2

crjmurray · May 19, 2015

lesliedellow said:
Before the door was opened, the possibilities were:

Code:

Door 1. Door 2. Door 3 Car. Goat 1. Goat 2 Car. Goat 2. Goat 1 Goat 1. Car. Goat 2 Goat 1. Goat 2. Car Goat 2. Car. Goat 1 Goat 2. Goat 1. Car

After the door is opened, possibilities 4 and 6 are ruled out. So the remaining possibilities are:

Code:

Door 1. Door 2. Door 3 Car. Goat 1. Goat 2 Car. Goat. 2. Goat 1 Goat 1. Car. Goat 2 Goat 2. Car. Goat 1

Therefore the conditional probability of a car being behind door 1 is 1/2

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

lesliedellow · May 19, 2015

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

Prove it. The diagram which seems to have convinced Michael presupposes that conditional probability is unaffected by the availability of now knowledge, and that, of course, is nonsense.

crjmurray · May 19, 2015

lesliedellow said:
Prove it. The diagram which seems to have convinced Michael presupposes that conditional probability is unaffected by the availability of now knowledge, and that, of course, is nonsense.

The odds for picking the initial door are 1/3.
After door number three is opened, revealing a goat, the odds for that door are now 0.
The odds for door number 2 are now 2/3.

The player is effectively making the choice between their initial 1/3 chance or the 2/3 chance of picking both doors 2 & 3.

lesliedellow · May 19, 2015

crjmurray said:
The odds for picking the initial door are 1/3.
After door number three is opened, revealing a goat, the odds for that door are now 0.
The odds for door number 2 are now 2/3.

The player is effectively making the choice between their initial 1/3 chance or the 2/3 chance of picking both doors 2 & 3.

Rubbish. The player is making a choice between a door which has a car behind it, and a door which has a goat behind it, and there is nothing weighting the probability in favour of one door rather than the other. Probability isn't a physical something which sticks to the two doors like glue, and then gets transferred to the one door.

Tinker Grey · May 19, 2015

Look at it this way. If you have a 1/3 chance that the door you pick is correct, there is a 1/3 chance that the host will have a choice as to which door he can show you. I.e., if you have a 1/3 chance of being right, you have a 2/3 chance of being wrong. The host then, since he cannot show you the prize, has a 2/3 chance that he has no choice at all. There is therefore a 2/3 chance that the remaining door is correct. You should switch.

Try it with cards and a friend. (Say 2 twos and an ace.)

lesliedellow · May 19, 2015

Tinker Grey said:
Look at it this way. If you have a 1/3 chance that the door you pick is correct, there is a 1/3 chance that the host will have a choice as to which door he can show you. I.e., if you have a 1/3 chance of being right, you have a 2/3 chance of being wrong. The host then, since he cannot show you the prize, has a 2/3 chance that he has no choice at all. There is therefore a 2/3 chance that the remaining door is correct. You should switch.

Try it with cards and a friend. (Say 2 twos and an ace.)

The idea seems to be abroad that the 2/3 probability attaching to the doors the contestant does not pick becomes somehow their inalienable property, and so if one door "dies" the other door "inherits" its share of the 2/3 probability.

crjmurray · May 19, 2015

Tinker Grey said:
Look at it this way. If you have a 1/3 chance that the door you pick is correct, there is a 1/3 chance that the host will have a choice as to which door he can show you. I.e., if you have a 1/3 chance of being right, you have a 2/3 chance of being wrong. The host then, since he cannot show you the prize, has a 2/3 chance that he has no choice at all. There is therefore a 2/3 chance that the remaining door is correct. You should switch.

Try it with cards and a friend. (Say 2 twos and an ace.)

Good explanation

crjmurray · May 19, 2015

lesliedellow said:
Rubbish. The player is making a choice between a door which has a car behind it, and a door which has a goat behind it, and there is nothing weighting the probability in favour of one door rather than the other. Probability isn't a physical something which sticks to the two doors like glue, and then gets transferred to the one door.

The initial choice is and always will be 1/3. Narrowing the choices from 3 to 2 does not change that you have a 1/3 chance of being correct. If there are a thousand doors the chance of being correct on the initial decision is 1/1000. Narrowing down to door 1&2 while revealing doors 3-1000 does not create a 50/50 decision. It creates a 1/1000 versus 999/1000 decision.

lesliedellow · May 19, 2015

crjmurray said:
The initial choice is and always will be 1/3. Narrowing the choices from 3 to 2 does not change that you have a 1/3 chance of being correct.

Oh yes it does. What do you think Bayes' Theorem is all about?

crjmurray · May 19, 2015

lesliedellow said:
Oh yes it does. What do you think Bayes' Theorem is all about?

Apply the theorem. Show your work .

keith99 · May 19, 2015

Answering this question is like determining how one will play a poker hand only based on the probabilities of what they see.

For an extreme example let's say we know:

The show is short of cash. Very short. As they can't afford a win. Host will not offer the choice if you already have a goat.

Or B. Show has not had a big winner in too long, they need a winner and this car is a Jag. Host will not offer the choice if you have the Car. Switching is a sure thing.

Both those of course knowing what drives the host.

EDIT: Note that the simulation covers the case where the host MUST give you the option every time. That is totally different from the host may give you the option.

Side note, if you can figure out what is usually the real reason why they sometimes offer this and sometimes do not you should take the change. (Hint the show is a half hour).

lesliedellow · May 19, 2015

crjmurray said:
Apply the theorem. Show your work .

A = car behind door 1

B = goat behind door 2

C = goat behind door 3

Pr(A) = 1 – Pr(B&C) = 1 - Pr(B|C) * Pr(C)

Host opens door 3, Pr(C) = 1

Pr(A) = 1 – Pr(B|C) * 1 = 1 - Pr(B|C)

Pr(B|C) = ½, since there are only two possibilities – a car or a goat, and they are available in equal numbers.

Pr(A) = 1 – ½ = ½

crjmurray · May 19, 2015

lesliedellow said:
A = car behind door 1

B = goat behind door 2

C = goat behind door 3

Pr(A) = 1 – Pr(B&C) = 1 - Pr(B|C) * Pr(C)

Host opens door 3, Pr(C) = 1

Pr(A) = 1 – Pr(B|C) * 1 = 1 - Pr(B|C)

Pr(B|C) = ½, since there are only two possibilities – a car or a goat, and they are available in equal numbers.

Pr(A) = 1 – ½ = ½

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

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