A Probability Puzzle

A game show contestant has to choose one of three doors. Behind one is a car, and behind the others are goats. The probability of the car being behind any given door is one in three. After the contestant has chosen a door, the host - who knows where the car is - points to another door, different from the one chosen, and opens it to show that there is a goat behind it. The contestant is now givin the choice of changing from the door they originally picked to the one they did not choose to begin with and which the host has not opened.

Should the contestant switch or stay with their original choice?

Good luck!

Switching their choice will not affect their chances of winning a car.

Yes it does,

But I won't tell you the answer just yet ;)

It could be seen as:
On the contestants first choice there is 1/3 chance of winning the car, therefore 2/3 chances of winning a goat.
On the second choice there is a 50/50 chance of picking the car.
As there is a higher chance of having picked a goat on the first go, the contestant should switch their choice now to increase their chance of winning the car.
But I don’t believe this.

If the contestant goes out to win the goat in the first round - there is a higher probability of getting a goat (2/3). So in the second round the contestant has a good chance that he has picked a goat - so if he changes his mind he will win the car

I ran code to do it 1 million times and if the contestant changes his mind the probability of getting the car is about 2/3.

Rob

The car is behind door 3.
Initially there are three choices, 1,2,or 3
This choice limits the actions of the game show host:
Choose 1, he can only open no. 2
Choose 2, he can only open no. 1
Choose 3, he can open 1 or 2

Initially choose 1, shown 2, sticks with choice, wins goat
Initially choose 1, shown 2, switches, wins car
Initially choose 2, shown 1, sticks with choice, wins goat
Initially choose 2, shown 1, switches, wins car

So far switching is looking good, but

Initially choose 3, shown 1, sticks with choice, wins car
Initially choose 3, shown 1, switches, wins goat
Initially choose 3, shown 2, sticks with choice, wins car
Initially choose 3, shown 2, switches, wins goat

Switching was not such a good choice.

Out of 4 switches, 2 win the car
Out of 4 sticking with choice, 2 win the car

Chances are 50/50 whether or not the contestant changes their selection.



Did this game show ever exist?

Try this code:

Code:

---


Private Sub Command1_Click()
Winner = 0
Randomize
For X = 1 To 1000000
Dim prize(3)
'We will place the car in the middle put it can go anywhere
prize(1) = "Goat"
prize(2) = "Car"
prize(3) = "Goat"

'Choose a number between 1 and 3
choice = Int(Rnd * 3) + 1

'If the contestant changes his/her mind - (S)He will have initally chosen a Goat - then changed his/her mind to a Car. So to choose a goat (in the first round) would be a winner.
If prize(choice) = "Goat" Then Winner = Winner + 1

'If the contestant does not change his/her mind then he would have to have choosen a car to win
'If prize(choice) = "Car" Then Winner = Winner + 1

Next X

MsgBox Winner
'The result should be around 2/3 for changing his/her mind
End Sub

---

It proves you should always change your mind.

Rob

If this puzzle used electrons instead of cars and goats then it would make a difference knowing where one of the goats were (after the first choice).

Since this is not a quantum puzzle, Knowing where one of the goats are can have absolutely no effect on the outcome.

Once one goat is revealed, there will always be a 50/50 chance of finding the car (on a subsequant guess) no matter what was originally picked.

1/3 of the time the contestant will choose the car first time, so 1/3 of the time changing will cause the contestant to loose.
2/3 of the time the contestant will choose a goat first time, so 2/3 of the time changing will cause the contestant to win.

I cannot understand the people who say there is no advantage in changing. We did this problem at school in Year 10 when we were 13 years old.

This used to be a problem on an American TV show. There clearly is an advantage in changing the choice:
a)Many people have kindly written some simulation code to show there is.
b)A simple probability tree shows there is.

People who claim otherwise obviously have not investigated the problem thoroughly enough.

If the game show hosts only gives you the switch option when he knows you picked the car, perhaps you should not switch. The statement of the problem is not explicit on this point, but seems to imply that the host always gives you the switch option.

As I understand the problem, you should switch. Some posts have suggested or shown the correct analysis, others erroneously suggest that switching does not matter.

A similar probability problem occurs at the card table when playing games like bridge or whist.

At the card table it is referred to as the principle of restricted choice. It is given this name because the correct analysis always indicates that probability favors the situation in which a person has fewer choices (Id est: has his choices restricted).

The analysis by Chrisf results in correct probabilities, but this is due to the problem being a special case of a more general problem. For example, if there were 4 doors, the probabilities would be as follows.

• Stay with original choice wins 1/4 of the time, & loses 3/4 of the time.
  
• Switch wins 3/8 of the time and loses 5/8 of the time.

The above lacks the symmetry of the problem with 3 doors. For more that 3 doors, the odds are always against the contestant, but switching is the better choice.

THEROB and the rest are correct.

The answer is that you have twice as much chance of winning if you swap. If you don't beleive it just write a bit of code to check (or run the code above).

Put simply when you first pick a door you have a 2 in 3 chance of losing, when the host reveals a goat the probabilty doesn't change, you still have a 2 in 3 chance of losing. So if you swap you have a 2 in 3 chance of winning.

If your too lazy to write the code then you can see an online version, along with an explanation of the maths at:

http://www.math.toronto.edu/mathnet/games/monty.html

I’m not convinced,

If the game show host was removed from the equation, the chances would be 1:3 of picking the car with the first choice and a 2:3 chance of a goat.
If you have a goat and you switch your choice to one of the other two doors there is now a 50:50 chance of winning a car. 2/3 x 1/2 = 2/6 = 1:3.

If the game show host was re-instated, then the first choice is 2:3 of a goat, you are now left with a 1:2 choice (switch or not) 2/3 x 1/2 = 2/6 = 1:3.

There would be a 1:3 chance of winning a car whether or not you change your mind.


How is this wrong?

The bizaar thing about probability is it depends on the way you look at it. It all depends on what you know and what you don't at any given time.

You might be able to say that at the start of the show, the contestant should plan to switch because that would increase his probability of winning.

But, when it comes down to the two remaining doors, the goat is just as likely to be behind either door. OK, you were more likely to pick a goat with your first choice but that doesn't change the fact that the car could be behind either of the remaining doors with equal probability.

I have knocked a little bit of code together which I feel, reflects the game show conditions. It's a bit scary as it does seem to indicate an improvement in the chances if you switch.

I can see it, but still find it hard to believe.

Private Sub Command1_Click()
Dim col As Collection
Dim r As Integer
Dim l As Long
Dim choice As String
Dim noswitchwins As Long
Dim switchwins As Long
Dim loopsize As Long

loopsize = 10000
For l = 1 To loopsize
Randomize
Set col = New Collection
'take three goats
For r = 1 To 3
col.Add "goat"
Next
'replace a random goat with the car
r = Int(Rnd() * 3) + 1
col.Add "car", , , r
col.Remove r

'make 1st choice (from 3) and remove it from the collection
r = Int(Rnd() * 3) + 1
choice = col(r)
col.Remove r

'host removes a goat
If choice = "car" Then 'remove a random goat
col.Remove Int((Rnd() * 2) + 1)
Else 'remove the remaining goat
For r = 1 To 2
If col(r) = "goat" Then
col.Remove r
Exit For
End If
Next
End If

'decide whether to switch or not
r = Int(Rnd() * 2)
If r = 1 Then 'switch
switchwins = switchwins + (col(1) = "car")
Else 'stick with original choice
noswitchwins = noswitchwins + (choice = "car")
End If

Next

MsgBox "switchwins " & switchwins & vbCrLf & _
"noswitchwins " & noswitchwins & vbCrLf & _
"ratio switchwins:total " & switchwins / loopsize & vbCrLf _
& "ratio noswitchwins:total " & noswitchwins / loopsize & vbCrLf _
& "ratio total wins:total " & (switchwins + noswitchwins) / loopsize & vbCrLf _
& "ratio noswitchwins:switchwins " & noswitchwins / switchwins
End Sub

There is a symmetry to the 3-door game. It becomes obvious
if you make up a matrix of all the possibilities.

Code:

---

First choice:    Car          Goat
Switch:          Lose          Win
Stay:            Win            Lose

---

Note that when you pick the goat first, the host picks the other goat and the car is behind the remaining door.
The probabilities are fairly obvious.
The car column has a probability of 1/3 and the Goat column has a probability of 2/3.
Switching wins 2/3 of the time and loses 1/3 of the time,
while staying with original choice wins 1/3 of the time and loses 2/3 of the time.

If there were 4 doors, the matrix would look like the following.

Code:

---

First choice:  Car        Goat
Switch:        Lose      Win ½ & lose ½
Stay            Win        Lose

---

Note that when you pick the goat first, the car is behind one of the two remaining doors.
Now the probability calculations are slightly more complex.
The car column has a probability of 1/4 and the goat column has a probability of 3/4.
The probabilities are as follows.

• Stay wins 1/4 & loses 3/4 of the time.
  
• Switch wins (½)*(3/4) = 3/8 and loses 1/4 + (½)*(3/4) = 5/8 of the time.

Note that the action taken by the host do not change the original probabilities for the Stay Strategy.

Intuition based on the number of unknown doors suggests the following erroneous probabilites.

• In 3-Door game, host opens one door, probability is ½ for each of the unknown 2 doors.
  
• In 4-Door game, host opens one door: Probability is 1/3 for each of the unknown 3 doors.

The above intuitive analysis is incorrect.
Intuition without analysis is sometimes right on the money
and sometimes is way off base.

I think I’ve been convinced now.
Without the game show host opening the ‘wrong’ door, there is a 1/3 chance of picking the car on the first attempt, sticking with this choice will always give a 1/3 chance of winning.
On the other 2/3 chances you would have to choose between one of two doors, giving a 1/3 chance of winning if you switch.
With the game show host exposing the losing door, these 2/3 chances become a guaranteed win if you switch. Hence a 2/3 chance of winning if you change your mind.
So it is the intervention of the game show host that affects the chances.
I know this has been said by many people, but I understand it better having written it myself.

How can I apply this to the lottery and improve my chances of picking up the jackpot?

Simply put, the host has removed one wrong possibility from the equation. We're you to re-evalute, you now have a choice of 2 doors, with 50% chance each, but if you choose the door you originally chose (ie stick with the original choice), you're still in the same situation you were at the start...

yeah?

Not really, because the odds say you're more likely to be wrong on your first choice. Basically when the host removes one wrong answer from the game, you're still in a situation where the chances are 2/3 that your door holds a goat, and the other door holds a car. The host removes one wrong option, but that still leaves you with a goat most of the time.

Behemoth: You still seem to tbe missing the point of this problem.

HarryW: From your last post you may or may not be with Behemoth.

See my previous post, which shows the calculations.

In the 3-door game there are two unknown doors after the host opens one with a goat behind it.
Superficially, it seems to a matter of guessing which of two doors has the car behind it.
It looks like the probabilities are 1/2 & 1/2, when they are 1/3 & 2/3.

The odds are two to one in your favor if you switch from your original choice,
even though it seems to be a matter of choosing between two doors.

In a 4-door game played the same way, there would be three unknown doors after the host opened a goat door.
There is your original choice and the remaining two doors.

It seems as though the probabilities should be 1/3, 1/3, & 1/3,
but they are 1/4 (original choice), 3/8, & 3/8.

I wasn't sure when I first read this thread, but after just a couple of posts it became fairly clear that it's better to switch. So I wasn't 'with' Behemoth I was just trying to explain it.

Contrary to my earlier post there appears to be a 1/3 chance of winning a car if you switch every time as opposed to 1/6 chance of winning if you stick with the first choice.
Switching gives a better chance, but the odds of winning are still no better than 1/3 - the intuitive answer.


The game show must have had a good supply of goats.

The odds of winning when you switch have to be better than 1/3, in fact they must be better than 1/2 because you have more chance of having picked a goat even after the other goat is removed from the choices.

You've picked a door and there's one more door you could switch to. The chances are 2/3 you've got a goat behind your door. It doesn't matter that the extra goat was removed after you made your choice, it's still 2 in 3 that you've got a goat.

Guv is right, the chance of winning on switching is 2/3.

I expect I'm wrong again, I based that observation on results obtained from my wonky bit of code I posted earlier.

Well, i'll stick my neck out and say that i wholeheartedly disagree with the explanation that switching improves your chances of winning. Let me see if i have the problem straight to start with.

You pick from 1 of 3 doors. If you pick the car then woohoo. If not the host tells you you picked a goat and you have the option of switching ... is that right.. you always know what is in the door the host opens? If so, i would think that a new probability problem begins afresh and there is no relevancy to using the results of the initial guess. The problem now becomes that there are 2 doors and the car is behind one.. a 50/50 chance. You had 1/3rd chance on first guess and 1/2 chance on the second guess which would only arise if ur 1/3 guess was incorrect. And it matters not how many doors existed at stage one.

It may be easy to discount intuition with mathematics but sometimes maths can lead you to see only trees and not the forest. What if they took you off stage and came back with the car and remaining goat moved or in the same position. Surely your chances are still the same. The problem is not in the inituition but in the hypotheses behind the probability maths.

So, I'll politely agree to disagree and allow others their own opinions.

Regards
Stuart

http://www.gstsmartbook.com

Quote:

---

If you pick the car then woohoo

---

No this is not how the game is played. The host places a goat behind 2 doors and a car behind the other door - The player choses a door - the host reviels a goat - the player is asted if they wish to change doors - on that answer the car is shown - and the player either wins or loses.

So in the first round there is a probability of 2/3 of gussing a goat - when one goat is revieled - if the player switches there is that 2/3 chance of winning the car.

Now do you understand??

Rob

Hmmm
That sounds exactly like what i said.

I still think that u are comparing apples and oranges and that the result of the first selection being a goat has absolutely no bearing on the result of the second selection.

If you look at it from a time perspective you could ask the player what their chance of selecting a car is on the first go.... obv it is 1/3rd... They then pick a goat and have two choices left... so they have a 50/50 chance...

Now say the player is replaced after he had made his first choice and the new player had no idea what had happened previously. They are asked to select from 1 of the 2 doors. They would surely have a 50/50 chance.

My overall point is that you are using probability maths to account for a situation where the conditions change.

This situation reminds me of an old economics joke.. an economist, physics prof and chemist stranded on a desert island with only a can of baked beans and no means of opening it. After a number of failed attempts by the prof and chemist the economist lambasts them and they ask what he would do since he is so smart... the economist says "first, we assume that we have a can opener"... So basically, i don't agree with the assumptions used in the probability calcs posted above. :D

Regards
Stuart

Beachbum, it looks like you've misunderstood the game. The player picks one door, and then a different door is opened to reveal a goat, thus removing it from the game. The door the player picked is not opened until the end of the game.

The above calculations are correct, given that the game is played as explained.

hi Harry
yep i did read it wrong .. sorry. Is hard to follow visuals presented in words sometimes.
Regards
Stuart

if there wasn't a car behind the door - the host would have simply opened the chosen door and shown him the sheep.

so how do you win the pig again?

People have talked about if there were 4 doors and 1 is revealed
it gets a little harder, but here's another way to look at it that
should make it EASY to understand...

Let's say there are 100, no 1000, aw heck...1 million doors to
choose from. They all contain goats except one that has a car.

You pick one door and then the game show host opens ALL THE
OTHER DOORS except the one you picked and one other. He tells
you that one of the doors has a car behind it.

Do you switch?? OF COURSE YOU DO!! Now (at least to me) it
seems so obvious that you HAVE to switch from a probability
standpoint because there is no way you picked the car on the
first try!

Hope that helps a little!

That certainly makes it look clearer seaweed.

I looked at the problem from an alternative perspective.

With 3 doors
If you choose door1..........1/3 chance of picking right door

There is .............................2/3 chance of car behind doors 2 or 3

If host opens door2 to reveal a goat then:

This reduces to..................2/3 chance of car behind door 3

Now the problem seems trivial.


With 4 doors

If choose door1.................1/4 chance of picking right door

There is .............................3/4 chance of car behind 2,3,4

If host opens door2 to reveal goat then:

This reduces to ..................3/4 chance behind doors 3,4

So if you switch to door3 or door4 then 75% for 2 doors = 32.5% that either door3 or door4 has the car behind it vs 25% if you stick with door1.


With 10 doors
You choose door1 and the host reveals a goat behind door2

If choose and stick with door 1 : 10% chance of picking car
If switch then 90% chance of picking car over 8 doors = 11.25 % per door


Conclusion:
As the number of doors increases the benefit from switching reduces. If you had a million doors, for a one of choice, it almost wouldn't matter if you stuck with door A.

Hi Nucleus
I think the descriptive part of the solution is to say that the host opens ALL other 'goat' doors - thereby making the switch quite obvious. So, if 1000 doors he would reveal 998 goats after ur selection. Besides i think that 998 goats would be worth more than a Trabant so i may indeed pick them :p
Regards
Stuart

I don't care how old this thread is -- I do not agree!

Basically, there are two games being played.

1: Pick a door from 3

2: Pick a door from 2

In the first round, you have a 2/3 of picking a goat. I agree.

BUT. If you pick the car, you are not told so. If you pick a goat, you are not told so.

You are only told which of the doors you did not pick contained a goat. If you have the car, the other door has a goat also.

Now, for round two, you have a 50/50 chance. No more, no less.

Two doors exist. One has a goat, one a car. Period. No information from the first round tells you that you have the car, or that you have a goat. The only information you picked up was that the a goat exists behind the door that was opened.

That door has nothing to do with the present situation.

I do not need a program to know that if you pick from two items randomly, you will get one roughly have the time and the other, the other half the time.

Anyone attempting to add the first door is fooling themselves. That door proves nothing.

Quote:

---

Let's say there are 100, no 1000, aw heck...1 million doors to
choose from. They all contain goats except one that has a car.

You pick one door and then the game show host opens ALL THE
OTHER DOORS except the one you picked and one other. He tells
you that one of the doors has a car behind it.

Do you switch?? OF COURSE YOU DO!! Now (at least to me) it
seems so obvious that you HAVE to switch from a probability
standpoint because there is no way you picked the car on the
first try!

---

This is exactly the right way to approach this question, if you do have a million doors than the chances of you picking the car on the first try are a million to one, but if all the doors are opened and you stay, it makes absolutely no difference wether you can see what is behind those doors or not they are still elements. if you stay then you have a million to one odds of getting the right one, however if you switch you have a 2 in 3 chance of getting it

How do you have a two in three? You do not know that the one you chose was wrong. You don't get the car anyway if it was in the first door you chose when you switch.

In the case of a million doors, I can agree, but not in the case of three. But only IF he opens all the other doors.

Let's eval this from another angle.

If there were a million doors, you chose one, and he opened only one ... Do you switch?

You cannot evaluate a three door possibility this way.

If there were four doors, and they opened two of them, then hell yea I would switch.

But in the case of three, that's not the case.

OK, I'll write that program, but Ill write if for ME.

I need to see the events happening.

OK, I conceede.

I finally 'get' it.

I demonstrated it to someone using a shoe and 3 potplants ... it worked until they got pissed off at the logic and threw the other shoe at me :D

Chuckle, thats funny!
I've long stopped trying to explain all things mathematical to my family. Just doesn't work...

the only person who ever agreed with this logic the first time around was my computer teacher, he was a math teacher and is quite good at logic so he didn't argue with it but i had to take at least 10 or 15 minutes argueing with everyone else

No, I dissagree with the logic completly.

given that you will never be shown the door you picked
and given that whether or not you picked a goat, one goat door
will be opened.

go like this, you pick door number 1.
host reaveals dorr number 3 (goats).
you get shot in the head and your twin gets out of his seat
and takes your place, but your twin was not watching, so he
has no idea what dorr you picked the first time.

so NOW there is 1:2 chance that he will get lucky and pick some goats.
good thing your twin owns some land.

nuf said.

This would hold true for a million doors to, except yo might need a pair of
binoculars to see the last one.

Quote:

---

Originally posted by JPicasso
you get shot in the head and your twin gets out of his seat
and takes your place, but your twin was not watching, so he
has no idea what dorr you picked the first time.

---

That is the crucial part of the game that you are changing. You do know what you picked the first time. I am tempted to get out my shoe and some potplants but I think others have explained it quite well on this thread. Try it out yourself on some unsuspecting member of your family.. do this....

Put a shoe behind one potplant
Ask them to guess where it is.
Show them a potplant that they have not picked and where there is no shoe.
Ask them to pick again.
More often than not they will pick the shoe by switching.

ps Wear protective headwear :eek: :p

I'm not sure how many times this has been said (there's too many replies to read them all), but it IS BETTER TO SWAP.

think of it this way.
n>2
There are n doors.
You pick door X, with a 1/n chance of being right.
There are now two groups of doors. Chosen and Not Chosen
Chosen = door X, Not Chosen = the rest
P(Chosen) = 1/n as above.
P(Not Chosen) = (n-1)/n

By saying that what is behind all but one of the doors (let the door left be Door Y) in "Not Chosen", the host is saying:
Either stay with your "Chosen" Door, or move to the "Not Chosen" group. If you decide to change to the "Not Chosen" group, then the prize is behind door Y.

=> chosing door Y = (n-1)/n chance of winning.

Finally, try this:
Pick a door, 1 to 3
Now, either you are right (33%) or you are wrong (67%). If you are wrong, then you must change. The only way to change is to swap to the other door on offer. => if you are wrong (67% chance), and you change, then you will be right. => 67% right if you change.

And I say that working with one out of three doors
after the time where you know that the prize is behind
one of two doors is just bad math.

any time after you know that one of two doors (and an equal probability
that either door has a goat), you odds of switching to the correct door
or holding onto the correct door are exactly 50/50.....

wait a minute... let me think on this further.

the best solution is to write a program that would do it completely randomly

Deary me what a can of worms.

It's actually dead easy to work out as follows:

Suppose you pick door 1, then the possiblities are as follows:

Code:

---

Contestants stratergy: Don't Swap Doors

    Door 1  Door 2  Door 3
    Car      Goat    Goat    win
    Goat    Car      Goat    lose
    Goat    Goat    Car    lose

Contestants stratergy: Swap Doors

    Door 1  Door 2  Door 3
    Car      Goat    Goat    lose
    Goat    Car      Goat    win
    Goat    Goat    Car    win

---

So If you don't swap you have a 1 in 3 chance of winning. If you do swap you have a 2 in 3 chance of winning.

QED.

And Just for completeness the following is all possible results for any door picked.

Code:

---

pick door 1

Contestants stratergy: Don't Swap Doors

    Door 1  Door 2  Door 3
    Car      Goat    Goat    win
    Goat    Car      Goat    lose
    Goat    Goat    Car    lose

Contestants stratergy: Swap Doors

    Door 1  Door 2  Door 3
    Car      Goat    Goat    lose
    Goat    Car      Goat    win
    Goat    Goat    Car    win

pick door 2

Contestants stratergy: Don't Swap Doors

    Door 1  Door 2  Door 3
    Car      Goat    Goat    lose
    Goat    Car      Goat    win
    Goat    Goat    Car    lose

Contestants stratergy: Swap Doors

    Door 1  Door 2  Door 3
    Car      Goat    Goat    win
    Goat    Car      Goat    lose
    Goat    Goat    Car    win

pick door 3

Contestants stratergy: Don't Swap Doors

    Door 1  Door 2  Door 3
    Car      Goat    Goat    lose
    Goat    Car      Goat    lose
    Goat    Goat    Car    win

Contestants stratergy: Swap Doors

    Door 1  Door 2  Door 3
    Car      Goat    Goat    win
    Goat    Car      Goat    win
    Goat    Goat    Car    lose

---

Possible outcomes with no swapping 3 wins, 6 lose.
Possible outcomes with swapping 6 wins, 3 lose.

There for you have twice as much chance of winning if you swap.

Yes, i agree.

Just wanted to say:

Quote:

---

the best solution is to write a program that would do it completely randomly

---

Actually, the BEST way would be to write a program that checks all combinations, like what Slaine did.
Randomly is still good, but mainly for real-life testing, where not all events can be checked.

Okay, I'm convinced.

The "trick" is knowing that your first choice is probably (67%) wrong.
and that switching your choice when you know what other door NOT to pick is a big help.
The other "trick" is not getting shot in the head so that your
twin can take all your fabulous prizes.

Fun.