Went to see 21 last night, and none of the us got the gameshow host door explanation. Is it really better to switch? Yes, it is. Here is my explanation. This is not a math problem, but rather a logic problem

Game setup

> Three doors. One with prize and two with goats.

> Prize and goats randomly hid behind the doors.

> The host knows which door has the prize

Game play

> First, contestant picks a door, but not open it

> Next, the host MUST open a door with a goat, BUT, cannot open the contestant's door

> After the host opens the door showing the goat, the contestant can stay with his original choice, or, switch to the other door.

> Should the contestant stay or switch?

What the movie said

> The move said that it's better to switch. That switching will win 2/3 of the time. And I agree.

Why is it better to switch?

> The critical point here, is that the host CANNOT open the contestant's door, which limits what he can do, when the contestant's first round choice is a goat door.

> So let's say, that the contestant always chooses door 1, which is as good a random choice as any.

> Now, consider the case when the prize is NOT in door 1.

> In that case, the host only has one choice of door to open.

> For example, if Prize is in door 2, the host HAS to open door 3.

> Let's think about what happens if we switched (this is the critical point!) -- in this case, the contestant is GUARANTEED to win, if he SWITCHED

> (What causes that "guaranteed win" from the switch? It's because the game show host cannot open the contestant's door. So he is always eliminating the other goat door.)

> And now, how often is the prize NOT in door 1?

> 2/3 of the time!

> And that is why, 2/3 of the time, the "switch" choice will win

But wait! What about when the contestant picked the right door to begin with? switching will lose!

> Yes, of course. But, how often is the Prize is behind door 1?

> Only 1/3 of the time.

Come again. Why is the second round pick NOT 50/50?

> It's because the host could NOT open the contestant's door. If he could, then the "guaranteed win" scenario above would not be true.

Other fun explanation and debates on this problem. I also created a simulation just to make sure I'm not out of my mind.

## Monday, April 14, 2008

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Your explanation isn't entirely correct. The fact that the game show host knows the right answer and has avoided a door to open increases its odds.

ReplyDeleteWell, yes of course - The gameshow host must know the right answer *as well*.

ReplyDeleteBut the host knowing the right answer is the "setup" part of the game. Without it, he can't run the second round choice at all.

So the critical point that turns the second round choice from 50/50 to 33/66, is that the gameshow host could not open the contestant's door.

in my country, goats are big prizes... thus your logic goes down the drain ;)

ReplyDeleteHaha. I agree with the fallacy of my logic there, 100%!

ReplyDeleteWhy is the prize not usually in door 1?? It is a baseless assumption... could you explain that one please?

ReplyDeleteSo lets say we start by selecting door 2... the host opens door 3. Now would you change your choice to 1 - the prize is usually not in door 1!

Hi Adeep, the answer is "Yes". The contestant should always switch.

ReplyDeleteI was using "door 1" to explain the fact, that the contestant is guaranteed to win if he switched, when door 1 has a goat.

The exact same thing is true if the contestant picked door 2 or 3 (every first round door has 66% of having a goat)

Hope that helps?

thank you for this, i saw 21 too and was confused, cause i didn't get the proof - i'm very much a maths person... lol, but this really helped, except i don't get the maths behind it... purely hypothetically: say there was 4 doors, would the odds change from 25%each to 25% and 2x 37.5%? or is it more skewered?? i have no idea... i hate stats!!!

ReplyDeleteYeah, but still, you only play once, and this explanation with the odds says that if you do it x times (x being sufficiently large) then you win about 2/3 if you always switch. But for the one time that you play, there is no sense in considering statistical approaches. Just like the exponential decay of the radioactive nuclei. It is exponential only when there are lots of them. Otherwise there is no way to predict anything or speak of chances (of course there still is the chance for a nuclei to decay in the next second but that's another story)

ReplyDeleteI haven't given it much thought to be 100% percent sure, but that's what i think..!

a famous problem that maths phd's got so very wrong in the 90's. Important info is that the host will never open the door with the prize.

ReplyDeleteYour mistake is that you calculate the chances by picking cases where the car is. This is wrong. There are only two possibilities that you should review: picking the car and picking the goat. Chances don't increase by switching doors - they reamin 50:50 :)

ReplyDeleteI know it seems like it should be 50/50 Vicho, but it simply isn't :)

ReplyDeleteIf you don't believe my explanation, check out my simulation spreadsheet (which is in my post)

INCORRECT,

ReplyDeletehere are your 4 possibilities:

3 doors, Car(C) Goat1(G1) and Goat2 (G2)

You possible host choice:

C G1

C G2

G1 G2

G2 G1

THose are the 4 possible scenarios, of which 2 you have the car and 2 you have a goat. Switching makes no difference. It is as if a stranger walked up at that point and had a choice between the 2 remaining doors. 50-50.

By the way, I actually graduated from MIT 1973, went to las vegas in 1971 with 2 frat brothers to count cards, and got taken to the back room of the Circus Circus. No kidding.

my formatting was off:

ReplyDeleteyou pick Car ... host picks G1

you pick Car ... host picks G2

you pick G1 ... host picks G2

you pick G2 ... hoat picks G1

Those are the 4 scenarios, switching gets you the car 2 of 4 times. If one incorrectly "groups" cases 1 and 2, it will appear to be 2/3.

p.s. we used a 10 count system which involved division to 2 decimal places, which was a bit more challenging with 4 decks than simple point count.

hmnnnnn ... I am rethinking this a bit ... I wanted to use 100 doors, 99 goats to prove 50/50 but it seems to go the other way... 1/100 that you picked the Car, and then the host filters out 98 other wrong choices. 99/100 that you should switch ... ??? stay tuned !

ReplyDeleteok ... I was right ... I was wrong.

ReplyDeleteits 2/3.

Try 100 doors .. makes it VERY clear who wins ... lol ...

p.s. I and the 2 las vegas frat brothers are now playing "cars and goats" to prove it out !

All that needs to be said is...

ReplyDeleteWhen u made ur choice u only had a 33% chance of being right - 66% of being wrong, more than likely u will choose the WRONG door which means more than likely the host will have to eliminate the LAST wrong door

I have read several posts on this. Having strugled with this subject in college I have a good way to explain. Both 50% and 66% can be correct but 66% is a better answer. Lets look at it as follows:

ReplyDeleteTake a quater. 50% chance heads, 50% tails. Flip it 10 billion times. Your results should roughly be 5 billion heads, 5 billion tails. This can be proven with a radom varaible program. Now lets look at it on a smaller scale. Flip the coin 10 times. The first 9 are heads. Now do you think the odds are 50/50 on the 10th flip? Chances are much higher it will be tails.

Thank you Wendy for clearing this.

ReplyDeleteI did not get it when I watched the movie.

Your logic works for me, but your game set up is not quite what the movie mentioned:

You say:

"Next, the host MUST open a door with a goat, BUT, cannot open the contestant's door"

The movie said:

"Then the host DECIDES to open another door."

I read this as - the host does not have to open the other door.

Does this make a difference? The movie says no, but I am not too sure.

Let's say the host's objective is to not let you win (something that's kind of implied in the movie) then why would he open the door 3 unless he knew you had the right door and he wanted you to change your mind?

I think if we know that the host does not have to open another door and that his objective is to beat you at the game, staying with the original choice gives you better chances than switching.

Agree?

Now, what happens if we don't know those 2 things or only know one of them?

Your thoughts?

I think it's important to remember with this problem that the reason it doesn't become 50/50 upon the second chance is because the first and second events (choices) are linked.

ReplyDeleteSInce it is more likely you picked the wrong door on the first try (regardless of the number of doors) if you were able to pick all the remaining doors instead you would have a better chance of getting the prize. In the three door example since the host eliminates 1 of the 2 remaining doors that 66% is thus concentrated into one door, making it a better choice than the original 33% you had to start with. It's almost like he's allowing you to pick all the remaining doors!

On the other hand, the flipping a coin example doesn't work in this case because each flip is an independent occurrence. Even if you flipped heads 9 times in a row there is still statistically a 50% chance you will flip heads on the 10th try because it is a fresh round. It would be pretty darn weird if you pulled it off though.

On flipping heads 10 times in a row -- I actually once asked a Vegas dealer once how often he sees 15 straight reds or blacks on a roulette table.

ReplyDeleteAll the time, he said.

So don't go betting all your money on black just because you see a big pile of red on the board -- Every new toss has nothing to do with the last toss.

great logical math problem...it seems to me that if the lone variable is that the host does not either open the contestant's choice or the prize door, the contestant originally has a 33% chance of success and a 66% chance of failure. But when the choice is given to switch, and the lone variable stays intact, the odds change to 66% chance of success or 33% chance of failure if the contestant makes the right choice : )

ReplyDeleteThis comment has been removed by the author.

ReplyDeleteAnonymous at 1:51 gets it, the answer is much more simple than all these other explanations....*simply put*

ReplyDeleteYou have a *better chance of picking a wrong door first*

which means

the host has a *better chance of having to eliminate the remaining wrong door*

leaving the prize door untouched

U might be thinking - well wut if u pick the prize door first?... and u may - but there is still a BETTER chance u wont

Tell me if I'm wrong

anonymous at 9:38 you are entirely wrong.If you are taking permutations into account you missed two. It would really be:

ReplyDeleteGoat1 Car

Goat2 Car

Goat1 Goat2

Goat2 Goat1

Car Goat1

Car Goat2

in your idiocy you've actually stumbled across the key to this problem. on your first choice you have a 2/3 chance of picking a a door with a goat behind it. If this is the case the host has to eliminate the other goat therefor if you switch you get the car. meaning that every time that you pick a door with a goat behind it (2/3 of the time) you get the car.

i thought about this and finally got it when i thought:

ReplyDeleteYou choose door 1 and the host would seemingly get 2and3. He has a 67% chance of the prize while you only have a 33% chance. Now when he throws out, say, door 3 know its not in that one, door 2 still owns its same 67% chance. It represents doors 2 and 3 but is still only one door. So when you take door 2 you get the 67% chance while your initial door 1 only holds the 33% chance.

Hope it helps those who were confused

what if you have four doors and the host opens two wrong doors after your initial choice and gives you a chance to switch after each?

ReplyDeleteIf you were simply given a choice between 2 doors, 1 of which contained a car, it's quite clear that your odds are 50/50.

ReplyDeleteIt doesn't matter what set of circumstances led to the 2 doors, 1 car situation. It is 50/50.

Done.

Aaaack! Wait a second...

ReplyDeleteIf there were one million doors and you picked one and the host eliminated all the other doors except yours and one other door, the prize would almost certainly be behind the door the host left.

Okay, I see the folly of my previous post.

Done.

Wow, this had the guys at my work debating for hours, productive in the office haha,

ReplyDeletei thought about this and finally got it when i thought:

You choose door 1 and the host would seemingly get 2and3. He has a 67% chance of the prize while you only have a 33% chance. Now when he throws out, say, door 3 know its not in that one, door 2 still owns its same 67% chance. It represents doors 2 and 3 but is still only one door. So when you take door 2 you get the 67% chance while your initial door 1 only holds the 33% chance.

Hope it helps those who were confused

thats the explanation which finally ended our discussions.

Consider a graphical treatment of this problem.

ReplyDeleteDoor 1 Door 2 Door 3

Goat Prize Goat ...just one case

1stchoice Host_choses NoSwitch Switch

1 3 1 2

2 1 or 3 2 1 or 3

3 1 3 2

We see that without switch a contestant chose {1,2,3} finally and since prize is behind door 2..out of 3 chances he won only once hence without switching a contestant has only 1 out of 3 chances.

Now consider the switching contestant: choices would be {2, 1, 2} or {2 , 3, 2} and both set of final choices being equally likely so the chance in this case is 2 out of 3 or again 2 out of 3 ....therefore 2 out of 3

To anonymous at aug23 10:36:

ReplyDelete"You choose door 1 and the host would seemingly get 2and3. He has a 67% chance of the prize while you only have a 33% chance. Now when he throws out, say, door 3 know its not in that one, door 2 still owns its same 67% chance. It represents doors 2 and 3 but is still only one door. So when you take door 2 you get the 67% chance while your initial door 1 only holds the 33% chance."

You made the assumption that by eliminating door 3, the probability that door 3 had were projected onto door 2. Now why would that be? A more "logical" explanation would be that the probablity were equally distributed onto door 1 and 2, so they are still equally likely to win.

A counter-scenario would be, what if the constant chooses door 2 in the first place?(which has an equal chance as if he/she chooses door 1). According to the quoted theory, door 2 has 2/3 chance of winning. Now it doesn't matter which door host eliminated, and again according to the theory switching has 2/3 chance of winning, which contradicts each other.

Also, the spreadsheet in the blog post has a lot of duplicated data if you pay attention. I guess the spreadsheet were made to "proof" the theory in the blog post instead of to come up with it.

demetris said:

ReplyDelete'Yeah, but still, you only play once, and this explanation with the odds says that if you do it x times (x being sufficiently large) then you win about 2/3 if you always switch. But for the one time that you play, there is no sense in considering statistical approaches. "....." I haven't given it much thought to be 100% percent sure, but that's what i think..!'

The statement that statistics do not apply in a one-time-event is rubbish. That is exactly what statistics mean: if you try it x (e.g. a million) times, it resembles the odds that you have at the first attempt. So if you would apply the 'tric' in the 3-door-question, you would win 666666 times, and lose 333333 times by switching in a 999999 time try. This therefore means that one try also gives you 66% chance to win (and 33% chance to lose) if you are going to switch. These odds just don't show after one attempt, that's why mathematicians multiply the experiment to make the odds visible. Doensnt mean that they're not there at the first go though.

Hope this is understandable.

Your odds of winning are 50/50.

ReplyDeleteIn the beginning, your odds were 50/50, even though you had 3 doors and here's why:

The 3rd door is just an illusion. The host knows that when you make a choice, that at least one of the remaining doors will be a goat, and he reveals that.

This did nothing to change your odds. It wasted time, but never changed your 50/50 odds.

In other words, offering 3 doors was just an attempt to confuse you.

Think of what your odds would have been if you were offered only 2 doors in the beginning, then after you make your choice, the host says "By the way - here's an extra door with a goat behind it. Now let's get back to the game."

You still have one car, and one goat.

Your odds have remained 50/50.

In another scenerio, imagine someone flipping a coin. What are the odds for heads? What are the odds for tails?

The answer is 50/50.

Why? Because the coin only has 2 sides.

Now imagine I tell you that I've flipped the coin 10 times, and 8 out of 10 times, it landed on tails.

Do your odds change?

NO.

My story did not affect the physical properties of the coin.

The coin has no memory. There are still only 2 sides. Heads or tails. 50/50.

If you still believe that "Odds" would make a difference, then try it in a game of Sudoku.

You will be proven wrong.

Here's a follow up to my last post.

ReplyDeleteImagine you are initially told there are 2 cars and only 1 goat.

You choose one door, and the host reveals a hidden car.

This again leaves you with one car and one goat.

Would changing your answer reduce your odds?

Or have you once again been fooled by an illusion?

If your strategy is to always switch, you will pick a goat 2/3 of the time initially, then switch to the car. (67% chance of winning the car)

ReplyDeleteIf your strategy is to always stick, you will pick the goat 2/3 of the time initially, and stick with the goat. (33% chance of winning the goat)

***33% chance of winning the CAR***

ReplyDeleteimagine 3 doors

ReplyDelete1 car, 1 goat, and 1 unknown (could be a goat, could be a car).

Once you make your choice, the host (who knows there were 2 cars) reveals the 2nd car.

You're down to 2 doors. 1 goat and 1 car.

Or,

once you make your choice, the host (who knows there were 2 goats) reveals the 2nd goat.

You're down to 1 goat and 1 car.

Are you starting to see that your chances are (and have always been) 50/50?

The only way there would ever be different odds (33% or 66%) would be if you had 3 prizes, and nothing was eliminated.

2 prizes only

2 prizes only = 50%

2 sides of a coin

Heads or tails = 50/50

Car or goat = 50/50

Most people look at the whole scenario, and believe that 2 goats vs 1 car means 2 to 1 loss.

Switching your choice would somehow mean your loss becomes a 2 to 1 win.

This will never be true. Unless you are allowed to pick 2 doors from the 3.

But picking 1 door from 2 = 50% chance.

ok, if you think it's 50:50 when given the choice between 2 doors, u dont understand stats, sorry.

ReplyDeletelets try another way.......

it doesnt matter WHERE the prize is because you have to look at the LIKELIHOOD (probability) - there's no way of guessing

CHOICE 1

chances of being right - 1/3 (door A)

chances of being wrong - 2/3 (doors B and C)

CHOICE 2

One door is eliminated, NOTHING ELSE CHANGES - its not a "new" 50:50 choice because the door was eliminated based on your original choice, so its still the same "game" - same odds

chances of being right - 1/3 (door A)

chances of being wrong - 2/3 (doors B and C)

the third door is still a part of the "game" because you can't choose it.

except now you can't pick door C (or B)

ODDS NOW........

chances of being right - 1/3 (DOOR A)

chances of being wrong - 2/3 (DOOR B OR C)

Thats why the remaining door has 2/3 chance of being right, and your original door has 1/3 chance of being right. thats fact, no guesses, pure 100% fact and mathematically proven.

and to end this debate, here is conclusive solid proof of this the fact, ALL the situations and outcomes are explained and calculated below, just look at the results for yourself.

ReplyDeleteScenario 1

Prize – behind A

Chosen Door - door A

(B or C eliminated)

Switch – lose

Stick – win

Chosen Door - door B

(C eliminated)

Switch – win

Stick – lose

Chosen Door - door C

(B eliminated)

Switch – lose

Stick – win

SWITCH SUCCESS - 2/3

Scenario 2

Prize – behind B

Chosen Door - door A

(C eliminated)

Switch – win

Stick – lose

Chosen Door - door B

(A or C eliminated)

Switch – lose

Stick – win

Chosen Door - door C

(A eliminated)

Switch – win

Stick – lose

SWITCH SUCCESS - 2/3

Scenario 3

Prize – behind C

Chosen Door - door A

(B eliminated)

Switch – win

Stick – lose

Chosen Door - door B

(A eliminated)

Switch – win

Stick – lose

Chosen Door - door C

(A or B eliminated)

Switch – lose

Stick – win

SWITCH SUCCESS - 2/3

minor mistake in the last post, heres is the corrected version

ReplyDeleteScenario 1

Prize – behind A

Chosen Door - door A

(B or C eliminated)

Switch – lose

Stick – win

Chosen Door - door B

(C eliminated)

Switch – win

Stick – lose

Chosen Door - door C

(B eliminated)

Switch – win

Stick – lose

SWITCH SUCCESS - 2/3

Scenario 2

Prize – behind B

Chosen Door - door A

(C eliminated)

Switch – win

Stick – lose

Chosen Door - door B

(A or C eliminated)

Switch – lose

Stick – win

Chosen Door - door C

(A eliminated)

Switch – win

Stick – lose

SWITCH SUCCESS - 2/3

Scenario 3

Prize – behind C

Chosen Door - door A

(B eliminated)

Switch – win

Stick – lose

Chosen Door - door B

(A eliminated)

Switch – win

Stick – lose

Chosen Door - door C

(A or B eliminated)

Switch – lose

Stick – win

SWITCH SUCCESS - 2/3

This reminds me of of the old math riddle known as the missing dollar.

ReplyDeleteThe story is about 3 salesmen who are looking for a place to sleep.

They each have only $10 with them.

The desk clerk tells them they can share a room which is $25.

She takes their $30 (total) and gives them back $5 in change.

They generously give the $5 to the bellboy as a tip.

Regulations prevent the bellboy from keeping more than $2, so he gives back $3 (or $1 to each salesman).

Now if every salesman paid $10, and got $1 back, they paid $9 each.

3 x 9 = $27

The bellboy has $2

$27 + $2 = $29

What happened to the missing dollar?

Eventually, you will see that the logic in this "joke" doesn't add up, because the storyteller is getting you to look at his breakdown of the situation.

Ha ha - good laugh.

Someday, you will also look at the 3 door game show in the same way.

Until then, I feel sorry for anyone who buys into odds of 33% or 67% when it isn't.

But the odds really are 33/67! Maybe I can make it clearer...

ReplyDeleteFirst off, at the beginning of the game, there are 3 doors, 2 with goats and 1 with the car. Your odds of getting a goat are 2 out of 3; that's just basic probability.

Second (and more importantly), whenever you guess a goat door first, the other closed door is always the car! There are 2 goats and 1 car; you selected one of the goats and the host opened the door with the other, which only leaves the door with the car. Think about that, it's important!

This brings us to the conclusion that whenever you guess a door with a goat behind it, the other door has the car. And remember that you have a 2 in 3 chance of guessing a goat.

And this is where the switch comes in. If you don't switch, you have a 2 in 3 chance of guessing a goat, which leaves you with a 1 in 3 chance of guessing the car.

But if you always switch, the odds are reversed. You have a 1 in 3 chance of guessing the car door right off; if you do, you lose, so you have a 1 in 3 chance of losing. You have a 2 in 3 chance of guessing a goat door, in which case you switch to the car door. So you have a 2 in 3 chance of winning.

If you break it down, you'll find that

If you never switch,

-you have a 2 in 3 chance of guessing the goat and losing

-you have a 1 in 3 chance of guessing the car and winning

If you always switch,

-you have a 2 in 3 chance of guessing the goat and WINNING because you switched to the car door

-you have a 1 in 3 chance of guessing the car and losing because you switch to a goat door.

It takes a bit of thinking but it becomes clear after a while. Hope this helps convert some of the 50/50 disciples :)

-Dave

Here is my analysis.

ReplyDeleteSuppose you play this game many times. Which door the car is behind is chosen randomly each time, but the host always gets to look behind the doors prior to your choice. But you will always choose door 1 initially. Why not? If the game is fair it's as good as any.

The three possibilities are that the car is behind door 1, 2, or 3. We'll call these possibilities D1, D2, and D3. Each has a 1/3 chance, so you are right 1/3 of the time.

Now the host is going to open one of the unchosen doors THAT HAS A GOAT BEHIND IT.

If D2, the host MUST choose door 3. We will call this D2H3. It happens 1/3 of the time.

Similarly if D3, the host MUST choose door 2. We will call this D3H2. It happens 1/3 of the time.

Now, suppose D1. That is, suppose your initial choice is correct. This is the tricky one. The host now HAS A CHOICE of which door to open. It is assumed this choice is MADE AT RANDOM. That is, based on a coin toss. So we have two possibilities: C1H2 and C1H3. Each will happen 1/6 of the time (1/2 X 1/3).

So on this particular play, the host opens door 3 and reveals a goat. We can now elimimate C3H2 because the host opened door 3.

But also we can eliminate C1H2 again since the host opened door 3.

So the only possibilities left are C1H3 which happens 1/6 of the time and C2H3 which happens 1/3 of time.

That is, C2H3 is now TWICE AS LIKELY as C1H3. And so, after the host has opened door 3, there is a 2/3 chance that the car is behind door 2.

That's it.

Please stop trying to claim that the odds are 50:50. They are mathematically proven as 66:33 and you can even test the game out yourself. Read the above comments carefully and think about it before making yourself look stupid.

ReplyDeleteI wanted this problem to be true in the movie but I believed in the 50/50 statistic. Thank you for helping us understand how it really works. I feel a lot better about it now.

ReplyDeleteThis is not a coin toss type situation. In a coin toss every toss is independent of the last. So if I just tossed 9 heads the chances of the next toss coming up heads is still 50/50. This is often mistaken with the initial odds of tossing 10 heads in a row before starting which are 1024:1.

ReplyDeleteIn the goat/car game you are presented with two choices but they are not independent choices. The host gives a big clue if we assume he has to open a goat door.

I think you can agree that when you initially pick from 3 doors your chances of the car are 33% of winning and 67% of losing (roughly).

The odds of your initial door being a winner cannot change after this decision. You chose 1 door from 3, 33% chance. If nothing else changes now and the host chooses NOT to open a losing door your odds are still 33%, right.

The chance of it being in one of the other 2 doors is 67%. When the host opens one of those doors the chances of the car being behind one of those two doors is still 67% except now you know that its not behind one of them. In other words the chances of it being behind the door he opened is now 0%.

Another way to look at it. If he opened a losing door but didn't give you a chance to swap would your odds of winning suddenly become 50%? No because you choice was one from three.

This is different from the coin toss because this is all one probability event, not two as we seem to hold up if your chances became 50/50 all of a sudden. Three doors one choice, that's it.

Yet another way to look at this.

ReplyDeleteThe host is essentially giving you the choice to trade your one door for the two doors you didn't choose.

His opening one of the other two is a smokescreen, he knows it doesn't have the car. In essence he gave you that door and the choice to take the other you didn't choose in the beginning.

Yes...

ReplyDeleteInstead of actually opening one of the unchosen doors imagine the host said.

"Now you didn't choose doors 2 or 3, and I can tell you that at least one of those two doors has a goat behind it (you already know this BTW) Would you like to trade your door (1) for both doors 2 and 3 knowing that at least one of them has a goat."

This is the real choice you are getting. Showing you a goat is classic misdirection.

okay, seriously, I read all these post and I have a final explanation newly thought out so that you won't even have to read any of these previous post(as informative as they are).

ReplyDeleteThe contestant is YOU.

The scenario:You're on a gamshow and the host tells you to choose one door out of three doors, behind which randomly contains 1 car, 1 goat and another goat.

When you select a door, the chance of you choosing correctly(by choosing correctly, I mean choosing the door with the car inside), is a 33.3% chance. That door that you choose specifically belongs to you and not the host. Consider it yours and only yours.

Now, the host has the other two doors and those two doors belong to him and only him. The host has a 66.6 percent change of having the car since he has two doors that belong to him.

Then, the host opens one of his own doors to show that it contains a goat. The next part is the main part.

Critically, The host chances of choosing correctly(66.6%) has not changed even after he opened one of his doors and saw there was a goat in it. So now that the host only has his one door left, that door contains his 66.6 percent chance of choosing correctly(because as i said, HIS CHANCES OF CHOOSING CORRECTLY HAVEN'T CHANGED).

Therefore, when the host offers you to switch doors it is in your best interest to switch doors because his only door has a 66.6% percent chance of being the correct choice.

End of discussion.

if you think i made a mistake, please do comment.

You have 100 suitcases, one contains money the rest are empty. You choose one suitcase. You then have the choice of opening the one you chose or the 99 others. You obviously choose the 99.

ReplyDeleteIn the goat/car example, by switching you are choosing to open 2 doors instead of 1. The order or time in which they are opened does not matter. There is no transferring of odds or percentages.

I'm pretty sure I am convinced that switching is the correct decision 66% of the time. And that if presented with 100 doors and 98 are opened for me, switching would be correct 98 percent of the time. So, what if I'm on that horrible show with Howie Mandel, I have a suitcase I chose at the beginning in front of me.... I then get lucky enough to open the remaining 18 suitcases without losing the million (leaving 1 girl's case and my initially picked case). Should I chose to switch suitcases if Howie gives me the opportunity?

ReplyDeleteOk, so after much thought I figured out that it is right to switch if you make one basic assumption: the gameshow host is always going to reveal the door with a goat.

ReplyDeleteIn that case it is very simple. There are only three scenarios. Either the car is behind door 1,2, or 3.

Scenario 1: Car behind door 1

you pick door 1

host reveals goat behind door 3

you switch to door 2 and lose

Scenario 2: Car behind door 2

you pick door 1

host reveals goat behind door 3

you switch to door 2 and win

Scenario 3: Car behind door 3

you pick door 1

host reveals goat behind door 2

you pick door 3 and win

You win 2 of 3 under with the assumption that the host will only reveal the goat. Your odds are 67% to switch.

Now if that assumption is incorrect and the host always reveals door 3 - regardless of what is behind it, then the odds are only 50:50. This is essentially because you only get the chance to switch 2 of 3 times.

Take the previously run scenarios. But this time on scenario 3 the host again picks door 3 (the car) and you lose. Now you only win one of three. You only had a chance to switch twice and you won once and lost once - 50:50.

So as long as the host is planning to show you a door that has a goat, then your odds are best to switch. (This seems like a safe assumption if he reveals a door that you didn't choose, right?) But if the host just picks one of the remaining doors at random, and then gives you a chance to rechoose if the car is not revealed - then your odds are only 50:50.

I think the previous explaination will explain why on that stupid show with Howie Mandel, it doesn't matter if you switch your briefcase with the last remaining case at the end.

ReplyDeleteIf you picked one of twenty cases at the start of the show, odds would be 1:20 that it contained a million dollars. Then, if you managed to randomly eliminate all of the cases without revealing the million dollars, you will still have a 50:50 chance between the last two. This is because each time you chose a case, you would have had progressively better odds to reveal the million dollars (thereby losing it). If you run out the scenarios, you will reveal the million dollar case 18 of 20 times before getting to the last two cases.

However, if at the beginning you picked one case out of 20, again the odds would be only 1 of 20 that it had one million dollars. But if this time Howie purposefuly eliminated 18 that did not contain a million dollars, then the chances would be 19 in 20 that the remaining case contained the million dollars. So switching would be right 19 of 20 times in that case. Since that is not how that dumb show works, there is no advantage to switching cases at the end.

Let me simplify this for you. The 3 prizes are Car, Goat or Donkey. Now lets go through all possible scenarios.

ReplyDeleteScenario 1:

Host gives you the choice to pick. You end up picking Car. Host opens door with Donkey. Gives you the switch, you switch and pick Goat. YOU LOSE!!

Scenario 2:

Host gives you the choice to pick. You end up picking Car. Host opens door with Goat. Gives you the switch, you switch and pick Donkey. YOU LOSE!!

Scenario 3:

Host gives you the choice to pick. You end up picking Goat. Host opens door with Donkey. Gives you the switch, you switch and pick Car. YOU WIN!!

Scenario 4:

Host gives you the choice to pick. You end up picking Donkey. Host opens door with Goat. Gives you the switch, you switch and pick Donkey. YOU WIN!

2 wins out of 4 with the switch, i.e. 50%

What is causing you to arrive at the wrong conclusion is that since there are two goats you are incorrectly grouping Scenario 1 and Scenario 2 together as 1 single scenario. But its not, because if you picked the car, the host can either remove goat 1 or goat 2(donkey or goat).

This is what you are doing

1)You pick car , host opens goat 1, give you switch , you pick goat 2 , you lose

2)You pick goat1, host opens goat 2, gives you switch, you pick car, you win

3)You pick goat2, host opens goat 1, gives you switch, you pick car, you win.

Yay 2 out of 3 with switch. But you are incorrectly forgetting scenario 4.

4) You pick car, host opens goat 2, gives you switch, you pick goat 1, you lose.

Sorry its 2 out of 4.

Im hoping all of the above will somehow register in some heads and they will become smarter human beings. Some will however be stuck in their ways.

Oh wait, I wasnt thinking about the fact that there is equal probability for each goat to be picked 2, so infact I was forgetting 2 more scenarios where the switch would make you win. It would indeed be 4 out of 6 and 66%. It took a nights sleep after watching the movie to register that I was wrong.

ReplyDeleteWow, thanks to all for explaining this and making it clear. I was watching '21' tonight, and it didn't make sense. How enlightening. I feel like I should have always known this.

ReplyDeleteI'd say that 2/3rds of you have been right about 50% of the time. And about 99% of us are confused out of our goards. I've always considered myself a "math guy", but now I think I'll just take up, oh, I don't know, professional beach volleyball? I'm just as qualified for that as I am for trying to figure out this stuff here. If you get this stuff, I'm not sure whether I feel sorry for you, or admire your intellect. I'd say the odds on that last part are about 50/50.

ReplyDeleteOk, I programmed up a simulator in C++ and ran a few MILLION runs of this situation, using pure randomness.

ReplyDeleteYou have a 66% chance of winning if you always switch on the second door pick. It does NOT MATTER if you think your math is right or the other person's math is wrong, I wrote the simulator to just follow the rules of the game, and I ran this simulator multiple times, each time running for for MILLIONS OF RUNS.

Could you all please stop insulting each other, hell I'll explain it in a way that'll make everyone happy.

Yes, the odds of picking the right door on the second choice are 50%, but that is NOT what is going on. By approaching this situation with a 'always switch' plan, you alter the game so that it's not 2 independant choices, but rather falls to 1 independant choice.

Understand that? By using the 'always switch doors on second pick' rule, you eliminate the choices down to 1, the first one, that's it. So we understand that now, yes? The game that was 2 choices is now 1, because we've decided to use this 'always switch' rule.

Ok, now that there's only 1 choice, the first choice, that's the choice we'll look at. Your chances of getting the wrong door on the first choice are 66%, right? Right. Now, if you use the 'always switch' rule, whenever you pick the wrong door, the host will open the other wrong door, leaving you with only the right door left to open, ASSUMING YOU ALWAYS SWITCH.

See the connection? Half of you have considered the game as 2 choices, when by using this rule you got the choices to 1, the first one, the only one that matters. Using the 'switch doors' rule you ALWAYS WIN if you pick a wrong door, and since you have a 66% chance of picking the wrong door, you thusly have a 66% chance of winning if you always switch doors.

Hell maybe later I'll upload the little simulator for people to run and see for themselves.

I agree with winning 2/3s of the time. But one thing that was bothering me was the guy who posted the riddle about the 3 guys getting a hotel room. I'll repost it.

ReplyDeleteThe story is about 3 salesmen who are looking for a place to sleep.

They each have only $10 with them.

The desk clerk tells them they can share a room which is $25.

She takes their $30 (total) and gives them back $5 in change.

They generously give the $5 to the bellboy as a tip.

Regulations prevent the bellboy from keeping more than $2, so he gives back $3 (or $1 to each salesman).

Now if every salesman paid $10, and got $1 back, they paid $9 each.

3 x 9 = $27

The bellboy has $2

$27 + $2 = $29

What happened to the missing dollar?

The salesmen pay $27 and keep $3. They pay $25 to the hotel and $2 to the bellboy = $27. They keep $3. = $30 total. The $2 the bellboy got was part of the $27 they paid.

But to the original question. You're wrong in your first choice 66% of the time. Then the host opens one door and gives you the other. That's the same as if he let you have both of the doors without opening them. Because even though you know one door is wrong, it would be wrong even if the host didn't open it. And one would be right 66% of the time.

Posted by Anonymous Feb 14, 2009 1:03:00 PM

ReplyDeleteCritically, The host chances of choosing correctly(66.6%) has not changed even after he opened one of his doors and saw there was a goat in it. So now that the host only has his one door left, that door contains his 66.6 percent chance of choosing correctly(because as i said, HIS CHANCES OF CHOOSING CORRECTLY HAVEN'T CHANGED).

-------------------------------------------------

That's where you're wrong. The whole clue to this is that the host KNOWS what doors contain goats.

If his picks were random aswell, it would be a 33% chance of him opening the car-door the first time around. If that doesent happen, he has lost his edge and it is now 50/50.

This is just one of a few things in this movie that I thought was a big stretch. But of course, even in reality, truth is stranger than fiction. BUT... You can't tell me that given 3 choices, 2 of which would be bad, you would ever get better than 33-1/3 percent chance of winning, even after the second chance given. If the game show host always reveals a goat and probably will on the 1st try, you will at that point have a 50/50 shot at winning. However, changing at that point is not an advantage, it only builds drama. This scenario "DOES NOT" have a memory that can be manipulated like counting cards in Black Jack. Because you are given a 50/50 shot by the host does NOT increase your chance of obtaining a win. Someone earlier said they setup a spreadsheet and simulation on this, but I'd like to see how they set it up. Cheers!

ReplyDeleteAn earlier post had it right.

ReplyDeleteThe way one wins is to pick a door with a goat behind it. The host then has no choice but to open the other door with a goat, leaving the player to switch to the door with the prize.

So what is the chance the player picks a goat? 2/3.

For those that think it is 1/2, you should go to New York and play 3-card monte with a street vendor...you'll lose your shirt.

Be all End all...

ReplyDeleteIt's a reasoning of mathematics and statistics.

-You have 3 doors.

-33.3% chance of picking the RIGHT door.

-Those are bad odds!

-The host takes away a door (not yours).

-You now have a 50% chance of picking the RIGHT door.

-MATHEMATICALLY You have a 50% chance.

-STATISTICALLY you HAD a 33.3% chance when you choose the first time, so statistically, you chose the wrong door.

-Changing your door, Statistically, makes it 66.6% chance you have the right door.

-Mathematically, it's still 50% (2 doors)

So changing your door does matter. Anyone need a simpler explanation?

listen everyone, it is simply NOT 50% at all. for those of you who are attempting to explain that...you are missing key parts to the problem. There are only 3 possible outcomes of this came. Here they are:

ReplyDelete1: contestant picks door 1 and it contains prize

host removes a goat door,

contestant switches and LOSES.

2: contestant picks door 1 and its a goat

host removes another a goat door,

contestant switches and WINS

3: contestant picks door 1 and its goat (2nd goat)

host removes another goat door,

contestant switches and WINS.

that makes a 2/3 chance of winning - theres NO WAY AROUND IT.

Now, this is what happens if the contestant NEVER switches:

1: contestant picks door 1 and its the prize

WIN

2: contestant picks door 1 and its a goat

LOSE

3: contestant picks door 1 and its the other goat

LOSE.

because i'll get attacked for a typo, I meant to say game*

ReplyDeleteThere is no reason to switch doors (or NOT to switch doors) after the host opens a door. With reference to the example above that references coin tosses, the first nine of which come up heads, the chance of the tenth trial coming up tails is NOT greater than 50%. Each flip of the coin is an fully independant event. The past has NO effect on what will happen next.

ReplyDeleteIf a new "contestant" wonders in off the street after the host opens a door and is given a chance to win the car, that contestant can choose either of the closed doors with an equal probability of winning, namely 50%. Likewise, the original contestant has a 50% chance of winning the car with his/her original choice. Do NOT make the erronious assumption that the past influences the future in this way!!!

I can understand the confusion, so I'll explain this in the shortest, simplest way possible.

ReplyDeleteYour chances of picking the right door at first are 33%, and the chances of picking the wrong door are ~67%. So if you pick door 1, for example, there is a 67% chance the prize is in either door 2 or 3.

Let's say the host opens door 3 to reveal a goat.

Since the host MUST open either door 2 or 3 and the door they open MUST have a goat behind it, and since there is a 66.7% chance that either door 2 or 3 had the prize, door 3 has now been eliminated and there is a 66.7% chance that door 2 has the prize. Therefore, odds double that you will win if you switch your choice. Remember, your initial pick only had a 33% chance of being correct.

This is all assuming the host is unbiased and has no motive in asking if the contestant would like to switch.

Pretend Monty never opens a door. What he's offering is that he'll give you doors 2 & 3, and if the car is behind either, you win. Would Monty's opening a door *after* you decide to take 2 & 3 make your chances of having won at the [previous] moment of your choosing go down? Of course not.

ReplyDeleteNow pretend that you kept your eyes closed and earplugs in and don't realize what door has the goat. Does your chance of winning by switching go down? You're still choosing to take 2 & 3. You still know there's a goat behind one. So what if you see which door of the two was the loser?

I'm not sure I buy it, but wikipedia right now says even pigeons do, fwiw. Go figure. The above is the best I can do convincing myself your chances go up by switching.

Mathematically speaking, there's no doubt it's better to switch. Imagine there are one million doors instead of three. Sorry, but you did not pick the door with the car; you picked the goat. And now the game show host will reveal all the other goats. The remaining door is the car. However, in reality, it may not be better to switch; it all depends on the game show and the host. Let's assume the host doesn't have to offer you a choice; sometimes he does and sometimes he doesn't. In that case, it may be that the host only offers you the choice when you happened to pick correctly. Relying on aforementioned advice, you'll lose every time. So the final answer depends whether or not the game show host must offer you a choice every time. Yes: switch. No: research all previous games to see if the host follows any pattern.

ReplyDeleteI just watched this movie and I didn't get this at first. However, I think I figured it out.

ReplyDeleteNow lets say everyone picks door 1.

Door 1 Door 2 Door 3

Goat 1 Goat 2 Car

Goat 2 Car Goat 1

Car Goat 1 Goat 2

Now cross off one goat from each row. You'll see that switching lands you a car 2/3 of the time.

that didn't space well. sorry.

ReplyDeleteDoor 1 Door 2 Door 3

Goat 1 Goat 2 Car

Goat 2 Car Goat 1

Car Goat 1 Goat 2

Three things can happen:

ReplyDelete1) you originally choose car, you are shown goat 1, you switch to goat 2 and lose.

2) you originally choose goat 2, you are shown goat 1, you switch to car and win.

3) you originally choose goat 1, you are shown goat 2, you switch to car and win.

What happened to you originally choose car, you are shown goat 2, you switch to goat 1 and you lose.

ReplyDelete**** ALL YOUR QUESTIONS ANSWERED --29/12/2011-- ****

ReplyDelete[SECTION a1]

--------------

D1 D2 D3

Car G1 G2

--------------

Pick D1 - Opens D2 - Stick WIN

Pick D1 - Opens D3 - Stick WIN

Pick D1 - Opens D2 - Switch LOSE

Pick D1 - Opens D3 - Switch LOSE

Pick D2 - Opens D3 - Stick LOSE

Pick D2 - Opens D3 - Switch WIN

Pick D3 - Opens D2 - Stick LOSE

Pick D3 - Opens D2 - Switch WIN

--------------

[SECTION a2]

--------------

D1 D2 D3

Car G2 G1

--------------

Pick D1 - Opens D2 - Stick WIN

Pick D1 - Opens D3 - Stick WIN

Pick D1 - Opens D2 - Switch LOSE

Pick D1 - Opens D3 - Switch LOSE

Pick D2 - Opens D3 - Stick LOSE

Pick D2 - Opens D3 - Switch WIN

Pick D3 - Opens D2 - Stick LOSE

Pick D3 - Opens D2 - Switch WIN

--------------

[SECTION b1]

--------------

D1 D2 D3

G1 Car G2

--------------

Pick D2 - Open D1 - Stick WIN

Pick D2 - Open D1 - Switch LOSE

Pick D2 - Open D3 - Stick WIN

Pick D2 - Open D3 - Switch LOSE

Pick D1 - Open D3 - Stick LOSE

Pick D1 - Open D3 - Switch WIN

Pick D3 - Open D1 - Stick LOSE

Pick D3 - Open D1 - Switch WIN

--------------

[SECTION b2]

--------------

D1 D2 D3

G2 Car G1

--------------

Pick D2 - Open D1 - Stick WIN

Pick D2 - Open D1 - Switch LOSE

Pick D2 - Open D3 - Stick WIN

Pick D2 - Open D3 - Switch LOSE

Pick D1 - Open D3 - Stick LOSE

Pick D1 - Open D3 - Switch WIN

Pick D3 - Open D1 - Stick LOSE

Pick D3 - Open D1 - Switch WIN

--------------

[SECTION c1]

--------------

D1 D2 D3

G1 G2 Car

--------------

Pick D3 - Open D1 - Stick WIN

Pick D3 - Open D1 - Switch LOSE

Pick D3 - Open D2 - Stick WIN

Pick D3 - Open D2 - Switch LOSE

Pick D1 - Open D2 - Stick LOSE

Pick D1 - Open D2 - Switch WIN

Pick D2 - Open D1 - Stick LOSE

Pick D2 - Open D1 - Switch WIN

--------------

[SECTION c2]

--------------

D1 D2 D3

G2 G1 Car

--------------

Pick D3 - Open D1 - Stick WIN

Pick D3 - Open D1 - Switch LOSE

Pick D3 - Open D2 - Stick WIN

Pick D3 - Open D2 - Switch LOSE

Pick D1 - Open D2 - Stick LOSE

Pick D1 - Open D2 - Switch WIN

Pick D2 - Open D1 - Stick LOSE

Pick D2 - Open D1 - Switch WIN

--------------

I cannot believe the stupid*ty of some of you. Even after countless correct explanations, programs that have been created and even spreadsheets, some of you fools still think its "50/50" or not even that! The others amongst you cant even get your stats right even though your explanations make sense. You leave more unanswered questions than you do answered! I bet most of you dumb*sses are americans too. Bunch of fat retarded potatoes. I feel like fuc*ing mashing your faces in. Dumb Fuck*ng Twats.

Whats more laughable is that some of you claim to be "Maths People"...L-O-L - Bunch of fuck*ng doucheb*gs.

The reason im so piss*d off is because I am not a maths person. I hate math. Always have. My mental arithmatic isnt very good. My timestables are weak and im a grown adult.

So what did it take to work this out?

PEN and PAPER.

How many of you claim to fame idiots actually stepped away from your shitty computers for a second and decided to use the age old pen and paper method of working things out?

It took me all of 3 minutes to write down the above figures and conclude.

I am now running upto 30 minutes to format and type up. See how simple a pen and paper formulation can be?

**HOW TO READ THE ABOVE FIGURES**

ReplyDeleteI have created SIX SECTIONS -

Section a1

Section a2

Section b1

Section b2

Section c1

Section c2

I will let you figure out what D1, G1 et al mean. Time to use that "genius" logic of yours!

Upon first glance, you may find that the results are in fact 50/50. You win four times and you lose the same. This is what confused most of you people.

You forget one key factor...YOU CHOOSE A DOOR.

By choosing a door, the win loss ration changes too.

All you have to do is the following -

1) PICK [Door 1] and STICK WITH IT through EACH SECTION.

2) PICK [Door 1] and SWITCH in EACH SECTION.

Do the same with each door. For your safety, for each Section i've already made it easy by putting the car behind each door and even working out the all the outcomes for you.

**THIS IS WHAT HAPPENS**

DO NOT count the goats. I have labelled them as G1 and G2 but do not count the goats. There could be 50 goats behind Door 2 and 50 goats behind Door 3. That does not mean I run results for each goat. My figures may come across at first that they are always equal in each section. But I have counted the goats for your reasoning. You need to keep in mind, there are only 3 doors.

The IMPORTANCE lies with the DOORS. ONLY THREE DOORS. REMEMBER THAT.

I pick Door 1 and STICK with it in each section.

I WIN in [Section a1+a2]

I LOSE in [Section b1+b2+c1+c2]

I pick Door 1 and SWITCH DOORS in each section.

I WIN in [Section b1+b2+c1+c2]

I LOSE in [Section a1+a2]

THEREFORE

If i STICK - 2W:4L = 1W:2L

If i SWITCH - 4W:2L = 2W:1L

Movie name? 21. Yes. 21 / 2-1 / 2:1. The answer is within the name. If i SWITCH after picking Door 1 I have set myself a 2 in 1 chance of winning. This works for each foor. Try it out!

IE.

33.3% + 33.3% = 66.6% (2W)

Versus

= 33.3% (1L)

Use your Dull Brain(33.3%) + Pen(33.3%)

instead of just

Dull Brain(33.3%)

***Extra 1***

The chances of you picking the wrong door in round one is 66.6% (2 Goats = 33.3% + 33.3%)

The chances of the host picking the wrong door in round one is 66.6%

Because the host shows you one of the doors as a goat and the fact that youve already chosen a door which had the percentage of 66.6% being wrong, after the host shows you the wrong door (33.3% now), your door then becomes 33.3% wrong and the winning door becomes 66.6% win. Chances are that if you switch you will win. The host HAS to show you a door with a goat as per rules.

***Extra 2***

This puzzle was answered correctly by writer of a column for a newspaper magazine back in the 90's. The indicidual was then put under duress by the so called PhD's and Professors and so aclled smart people. Remember, even the smartest can be wrong at times.

I think the smart *ss above has not even thought that he is considered not good at Maths for a reason....

ReplyDeleteI really cannot understand that ppl thinking switching would give them a 66% chance of winning give the same probability regardless of the player choosing Door 1..2 or 3..

Suppose i Chose Door 2 to start with...So i had 33% of being right..then Door 1 was opened and now we are left with two doors out of which one has Goat..the other Car..i.e. 50:50.

Now assuming Door 2 had a CAR...if i switch to Door 3 i will loose despite some ppl thinking i had 66% chance of winning if i switch...

but if i dont switch i win...

So two options in the second leg...switch or dont switch 50:50 chance.

Wow. Lots of time gone in here....

ReplyDeleteI think the most revealing way to look at this is through the eyes of the Stranger who walks into the room just after the Host has opened a goat door. Now, our Stranger only knows that behind one door is a car and behind the only other unopened door is a goat. Ask the Stranger to choose and she will get the car 50% of the time. That is, the probability of the car being behind either door is 50%.

That is exactly the choice that the contestant now has. What has gone before is irrelevant. A decision previously made in the contestants head doesnt affect the location of the goat.

All the argument about one door (previously) being 33% is misleading because by revealing the goat door, that first unopened door now becomes a 50% door. Just as when you play a game of Russian Roulette without spinning the revolver, probability that you will die on the first trigger pull is 17%. Then it's 20% for the next pull and then 25% and then 33% etc. What matters is the current state.

Gosh, this is much more than I wanted to write, but it seems so basic....... Annoyingly so.

(For this to work you must assume that there is only one universe.)

It's clearly 33/66. How some people can still think it's 50/50 after more than 20 years is beyond me. It's been mathematically proved, appears in text books, there are countless millions of computer simulations that empirically demonstrate it's 66% chance of winning if you switch, And yet you still get people making ridiculous statements like "What has gone before is irrelevant". Jeez, it's not rocket science.

ReplyDeleteWhen you selected the door, the door was 1/3 to have the car. The other 2 doors have a 2/3 chance of winning. When the game show host removes on of the other doors (0/3 chance), your door's probability is still 1/3. However, the other closed door is now 2/3 to win, so you always switch. This is known as the Monty Hall problem.

ReplyDeleteThis comment has been removed by the author.

ReplyDeleteThe past does affect this mathematics problem. After the first choice the location of the car ceases to be random. Effectively the host had two chances of picking the car while the player only has one. So the dealer is twice as likely to have the car as the player. Even after he reveals a door, he is still twice as likely to have the car. Because it was only random at the start. In Russian roulette the probability resets each time you choose. The stranger off the street is being cheated because the game was rigged in favour of the host. For a hundred doors there is a much smaller improved chance of winning if you switch after the first choice (1 door open), thereafter changing does not matter so long as you do not pick your original choice. If you switch (for the first time) at the last opportunity (98 doors open) you are 99 times more likely to get the car. Because the host had 99 opportunities to pick the car and you only had one.

ReplyDeleteIn the suitcase game the host could turn over the "car" so the game is different.

I't not the odds of where the car is that matter. It's the choice of switching that has a 2/3 chance of being correct. If you write down all the possibilities of where the car is and which door you choose, the ACT OF SWITCHING will be correct 2/3 of the time.

ReplyDelete