Game Show Probability: Newsletter Challenge (05/01/07)

Exactly English Rose,

It's no different to the countless examples of casinos making a killing on the roulette wheel when say red has up up 3 times and everybody places large bets on black on the belief that black "has" to come up next.

Red and Black have both always got a 50/50 chance of coming up next.

Hi Guest,

The difference is that subsequent rolls of the roulette wheel are independent (hopefully), but in this case the host has filtered out one choice (a deliberate act, not random).

Say you always select door 1 and stick with it. You will always have a 1/3 chance of winning.

Say I always select door 1, then change to the one NOT selected by the host. If the car was behind door 1, then I lose (1/3). Say the the car was behind door 2. The host opens door 3 to reveal a goat. I chose door 2 and win (1/3). If the car is behind door 3 then the host opens door 3 to reveal a goat. I chose door 2 and I win (1/3). This strategy has a 1/3 + 1/3 = 2/3 probability of finding the car.

Also, it's not 50/50 red and black. The payout is 1:1, but the odds are 18/37 or 18/38 (American/European). No I'm not a gambler, but I had a friend who WAS. I don't recommend it.

Davo

If the car is behind door 3 then the host opens door 3 to reveal a goat

? I'd think about that.

> If the car is behind door 3 then the host opens door 3 to reveal a goat.

say what?

With my statement that red or black is 50/50, I acknowledge that I wrote it incorrectly and what I ment was 50% chance of red being next, 50% chance of black being next. No different to picking one door with a 50% chance or picking the other door with a 50% chance

The flaw in the logic of balls in a bag is that after one ball is pulled out & revealed the player's odds are NOT 50-50 because s/he has already selected & commited to one of the remaining balls--at that point the choice is not random but determined by the selection made when three balls were present.

With a 1/3 chance of pulling the white ball a player will actually get the white ball one out of three times. The fact that a black ball is revealed after the choice is made has no effect on the odds when the choice was made.

This is the correct answer, but maybe I can clarify it. Your original choice will be correct 1/3 of the time, incorrect 2/3 of the time. If you choose same door strategy, you will be correct 1/3 of the time (the time your original choice was correct). If you choose other door strategy, you will be correct 2/3 of the time (the times your original choice was wrong).

> If you choose other door strategy, you will be correct 2/3 of the time (the times your original choice was wrong).

actually, one of those two thirds was if the host had opened the door with the car, and it is void in this scenario

so, > If you choose other door strategy, you will be correct 1/3 of the time (the times your original choice was wrong). [and the host picked the door with a goat]

And if the host would have opened the door with the car then you would have been wrong the 2^nd 1/3 of the time.

If the car is a Gremlin I would rather have a goat.

Ahh so thats the trick... and we were all thinking about propabilities.

At least the goat will run no matter how you park it... bahhhhhhh

At least the gremlin will stay where you park it (who's going to steal it?). A goat is almost impossible to keep penned, they are escape artists. When they get out, they head for the most succulent, expensive young tree to strip its bark off. Goats are the reason the land around the Mediterranean has no forests (prior to the raising of goats, it was forested).

Goats are contrary, they say nahhhhh. It's sheep that are skeptical, they say bahhhhh.

Oh, so that's why the Sahara has no trees. It's those Darned Bedouin goats!

ROFL

The Sahara was once a much smaller desert, however being more Euro-centric, I was thinking of from Spain to the Middle East.

In ancient times, these areas were as heavily forested as North America was at the time it was discovered. Goats are almost universally blamed for deforestation throughout the world, although there is some thought that it has more to do with overcrowding of domesticated farm animals. However, having had goats, I know that they have a predilection for stripping the bark off of saplings, even when plenty of alternate and more nutritious food is available. No saplings and eventually you have no forest.

If you switch, you have a higher probability. The door you first picked has a 1/3 probability of having a car, and there is a 2/3 probability the car is behind one of the other two doors. If the host shows you another door without the car, there is a 2/3 probability that it is behind the remaining door.

correct. the trick is the host is not opening a random door. he knows where the car is, and is intentionally NOT opening that door. the answer to the puzzle is to switch, however I would like to see some real data.

btw, if the goat is in fact a 1970 Pontiac GTO I'd go with the option of decking the host.

-A-

Assume that you pick door d1 since everything is symmetrical. There are only

three cases to consider:

d1 d2 d3

1. g g C

2. g C g

3. C g g

Case 1: host opens door d2, you switch, you win

Case 2: host opens door d3, you switch, you win

Case 3: host picks either door d2 or d3, you switch, you lose

You win in 2 out of 3 cases. If you stick:

Case 1: you lose

Case 2: you lose

Case 3: you win

You only win 1 out of 3 cases if you don't switch.

The explanation above does a great job of explaining the puzzle from the viewpoint of the three original doors, but I think that once the host opens a door and offers the contestant the opportunity to choose a new door, that the 3-door game has ended and a new 2-door game has begun. Once the host opens one of the doors to expose a goat, each of the two remaining doors has a 2/3's probability of being correct. The host has removed one of the failures, so each of the remaining doors has an equal chance of containing the car. By staying in the game, the contestant's probability of success has improved from 1/3 to 2/3.

It is as if there are two games being played sequentially. If the first contestant was removed after the one door was opened and a new contestant asked to choose between the remaining doors, he would have a 50-50 chance of being right. The previous knowledge of the original doors no longer applies to the new round, except when looking from the viewpoint of the 3 original doors.

Now, I don't know if it is legal to change one's perspective in the middle of a mathematical puzzle, but it helps me justify my gut feeling on the puzzle;)

If the host did not know which door the car was behind, then it's 50/50 that the car could be behind either door. Sticking with the d1 d2 d3, your door is d1 and the host reveals d2:

1. g g C - switching wins, sticking loses

2. g C g - host reveals the car, you lose

3. C g g - switching loses, sticking wins

Because the host revealed a goat, Scenario 2 is eliminated, and you're left with a 50/50 chance of winning the car either way. But I think you have to make the assumption that the host does know the location of the car, therefore switching improves your odds of winning.

Same with the 100 ball scenario - if your friend randomly pulled out 98 black balls, then there's a 50/50 chance that you have the white ball. But if your friend looks into the bag as he pulls out the balls, then there's a 99% chance that the last ball in the bag is white.

You are correct and a very nice explanation. This is indeed the "Monty's Dilemma" problem and this discussion just goes to show how deceptive probabilistic problems can be. The game does change when a known "goat door" is revealed.

Interested parties can see a detailed discussion here:

http://illuminations.nctm.org/LessonDetail.aspx?id=L377

and a java-base simulation here:

http://www.shodor.org/interactivate/activities/GeneralizedMontyHall/

Just so you all know I'm not slacking....the next episode in the story can be found here - the mad fools have given me a space all of my own!!

Note to self: need to add chimps...and doughnuts

I think this is called "Monty's Dilemma," a technique derived from the show "Let's Make a Deal." The answer, though hard to contemplate, is to switch. Even Monty Hall couldn't believe it when it was explained to him, but it's true.

Golly, if a game show host [presumably not an intellectual giant of any significance] can understand it you'd think the engineers would too!

You'd have a 50-50 chance either way. There are only two doors left. It' s not like the doors have some memory for probability! 50-50. It can't matter -- its pure chance. If you flip a coin 10 times and it comes up heads 10 times, what is the probability that it comes up tails on the next flip. 50%.

HC

If objects behind the doors are not moved and the contestant would know that their first choice was wrong, and choosing the same door would guarantee their odds of winning was zero. Choosing a different door would provide a 50-50 chance of winning the car.

If the objects behind the doors can be moved, listen for the sounds of movement and hope that those provide a strong hint about which door to choose.

If the doors can be moved, choose a door that has a different location from the first choice.

If everything can be moved, try listening closely.

Is the contestant blindfolded and spun around while all this is happening?

Why are these things worded so ambiguously?

Stay with your original answer. It shows character.

Really.... it doesn't matter which one; you have a 50% chance of getting it right.

Try this:

Pretend that you are going into some sort of trance or something and have the audience get very quiet for you.....

Then suddenly yell out what ever goat herders say to get their goat to make a noise ..... like "here goaty, goaty, goaty....."

Then when you detect which box the goat makes a noise from, quickly chose the other.....

Dwight

Or use your nose - I have a thought that there is a subtle difference between a goat and a car.

definately take the other remaining door, the odds are 2/3 that the car resides here. this is because the odds of your first guess being right are 1/3. 2/3 is the odds that the car is behind one of the two remaining doors, and its obviously not in the one he just showed you.

Hi Guest,

See if you agree with this logic: Two viewers are sitting at home watching the game show. Before any doors are opened viewer 1 picks door 1, while viewer 2 picks door 2. Host opens door 3 to reveal a goat. By your logic viewer 1 can improve their odds by changing to door 2, while viewer 2 can improve their odds by changing to door 1.

This is illogical.

The fact that either viewer (or the guest on the show) picked a door doesn't change the odds of finding anything in particular behind the door.

Nice and mischievous, Davo

Yes, using the valid logic of this post as a counterargument is illogical, because it omits the dependency of the host's actions on your choice. If the two of you are sitting at home, two times out of three the host will open a door that one of you selected. At this stage, the other choice is only improved to 50%, however, because the host could as well have opened that door (which he couldn't if it was the person choosing in the studio).

For what it is worth, the numbers here stack as follows:

Overall probability of success for pair = 2/3

Host opens one of chosen doors on 2/3 of occasions -> 50% probability of success for pair (1/3 total).
Host opens third door on 1/3 of occasions -> 100% probability of success for pair (1/3 total)
Checks out at 2/3 total.

Fyz

That is a different game, having two players. In your case, with the players both picking different doors while watching the game on TV, 2/3rds of the time the host would pick one of the doors that they picked. If he didn't, then if the players pick different doors, one of them will win the car 100% of the 1/3 of the time this situation occurs.

I guess what this overall post shows is how hard it is to prove a truth, when the other side doesn't want to believe it. No wonder O.J. got off.

Hello from Canada!

Bob Barker of the Price is Right, might suggest you have your Cats and Dogs spayed or neutered, with four hundred pound Tigers exempted and protected.

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

There are only two doors left and a goat is behind one of them. The answer is simple. A car won't cry like a goat! Simply approach the doors and bleat like a goat. The door that answers back is not the one you open.

Now after you win the car, I suggest you convert it to Hydrogen or Electric power or better yet, refuse the car and ask for the cash equivalent.

Signed: Joseph Raglione

Executive director: The World Humanitarian Peace and Ecology Movement.

Baaa Baaaa Baaaa Baaaaa ;-)

In the problem the number of doors is three (N=3). Lets for illustration of the principle increase it to 100 doors but, 99 doors have goats and one door has a car. You make a choice of one door, your probability of that door having is a car is 1 in 100 and 99 in 100 a goat. What is the probability that you have picked a car 1 percent. Got a goat 99 percent. The host opens doors for you, goat, goat, goat………. two doors are left. In all the doors shown the host has not shown the car. Your door and the other remain (N=2).

When you first choose your door you had 1 in 100 chance of getting the right door with the information that you had at the time (N=100). Now you have NEW information (N=2) one goat, one car. Your pick was made on OLD information 1 in 100 chance. The new information is 1 in 2 or 50 percent. By changing to the other door you have increased your chance from 1 percent to 50 percent.

The point is the information has changed between the choices.

For the case of thee doors:

Step 1. Pick a door 1 in 3 chance or 33 percent.

Step 2. Open a door - NEW information.

Step 3. New choice 1 in 2 or 50 percent. Old choice 1 in 3 or 33 percent.

Step 4. Change to the other door increase your chance by 17 percent.

How much does this increase your odds over your fist choice? A 17 percent improvement out of a 33 percent baseline is (17/33) or 51.5 percent better choice than before.

Will changing doors win every time no. But it is a better choice.

So, switching "increases" chances to 50%, eh?

Well, there are two (remaining) doors. And if we switch to the other door, our odds are 50%. Therefore, the odds of the door we are not switching to are also 50% (because 100%-50% =50% and there sure is a car behind one of the doors).

So if we select the door that isn't the door you say we should switch to, the odds are still 50%.

Hold on a minute though, thats the originally chosen door!

Yes we have NEW information - but you cannot selectively apply that to some probabilities and not others. The probability for the NEW situation depends solely on the latest information - even for a door we chose earlier.

.....you are correct......1 chance out of 2 is still the same as 1 chance out of 2 .... either door's a winner and either door is a loser.... the only NEW information is that you get to chose 1 of 2 instead of 1 of 3....

However, I guess you could pick the first door and get the announcer's goat.

You should switch. It's easiest to see why if you change the question a bit and imagine there are 100 doors, say. You pick one, then 98 others, all with goats behind them, are opened. Wouldn't you change your choice?

you started with 3 choices

one incorrect choice was taken away after you made your choice

you now have 2 choices and a chance to choose again

the odds are better now so you have a better chance of choosing correctly

i would just leave it though..... the goat told me to say that

Very simple answer to this question. Goats are social animals. Provided that the open door (with the goat) is on middle of the stage, open the door to which the goat is NOT facing. (logic... exposed goat is facing the hidden goat)

If the goat is on either of the far sides of the stage and the goat is turned away from the closest adjacent door, choose that door. (logic... the hidden goat is not immediately near so the exposed goat could care less about what's next door)

Other solution, yell very loudly and the goats should faint (assuming that they are common domesticated fainting goats). Listen for a thump behind one of the doors, and choose the other door.

The answer is at http://www.cut-the-knot.org/hall.shtml

It doesn't mater. You have a 50 50 chance for each door.

How bad would that be to win a goat? They eat grass, so that cuts down on your lawn maintenance; they also give milk, which is great for babies and lactose intolerant people; and if all else fails, they taste a little like deer.

Goat or car, you're a winner either way!

Clearly, you are a person for whom the glass (of goat's milk) is half full, rather than half empty!

you've probably heard this one before, but .....

The skeptic says the glass is half empty,

The dreamer says the glass is half full,

The engineer says the glass is twice the size it needs to be.....

.

Not quite!

The engineer says it is the right size, to allow against spillage.

It is the accountant who says it is too big.

My wife would agree with the second engineer

Pick the largest door - presumably the goat is smaller than the car?

In the "Monty Hall Dilemma" the host knowingly opens a door with a goat behind it, AND the guest knows that this is the case. The guest therefore obtains new information and in fact gains an advantage by switching his choice.

In the problem we are dealing with we are not told if the host knows what is behined the door. So the problem is ambiguous and can not be answered with certainty.

jshloram

THE HOST'S MOTIVES/KNOWLEDGE IN OPENING A DOOR AFFECT THE PROBABILITY. While a player might not know the host's motives and cannot answer the specific situation with certainty, the three possible options can be modeled to address all possibilities

1) The host will reveal a goat every time all the time. Player's chance of winning car by switching rise to 2/3. This is the one possible scenario everybody seems to be addressing.

2) The host will only open a door to reveal a goat IF the host knows the player picked the car. Player's chance of winning car is zero if they switch.

3) The host guesses which door to open. Presumably if a car is revealed the game is over & the player doesn't get the car. If a goat is revealed switching has a 50-50 chance of winning the car.

This is explained in response #83's links...or...in response #46 directly above. The following link addresses the host's random selection in simple detail (better than #83): http://en.wikipedia.org/wiki/Monty_Hall_problem

It makes no difference.... because the host revealed one of the goats before offering the

contestant the choice to switch doors.... the probability doesn't change whether the

contestant switchs doors or not.

What's all this going on about changing the number of doors ?

I'm confused enough already.

The invitation to change is an old game show ruse to get the audience laughing hysterically as you fall flat on your face . Producers love it when people have to shout inane comments from the crowd.

Sounds like 'the money or the box'

Worse still is 'deal or no deal' - it's all about naked greed and stupid superstition trying to get people to believe willpower can improve their chance. Watch Noel Edmund's for a really good laugh at how crowd manipulation and statistical nonsense can work,

What you do today is not going to change what happens tomorrow (in a game show context)

Hi Kris, So you have equal chance whether you change your mind or not? Davo

I always hedge my bets Davo . My declared position will appear later. Kris

I'm confused enough already.

Me too! All this talk about three doors has me puzzled. Although the host may know whether the car has three doors or four, how would the contestant? This determination of door number has to be central to the question of probability, doesn't it? If you already have a three door car and really want a four door car, then you are probably going to try to win the goat -- but if and only if you know the number of doors on the car. Is there something about the way in which the host reacts that is expected to tip off the contestant to the number of doors?

Does the uncertainty principal enter into this discussion?

...it's all about naked greed and stupid superstition

I hope I am reading too much into your statement, but it seems that you are trying to imply that this is a bad thing??! I hope the rift between the UK and the colonies has not become so broad that we can't both share our core values.

it's all about naked greed and stupid superstition - This was aimed at a particular (and very British ) TV show. I was not enticing you to come skinny-dipping on Friday 13th (or any other day , and certainly not for money ).

I hope the rift between the UK and the colonies has not become so broad that we can't both share our core values. The only rift I know of is in the mid-Atlantic. Fyz and STL have demonstrated that we still (just about ) speak the same language.

What if, for ratings purposes, the game show host wanted to help you win the car.

If you picked correctly the first time, he would have just opened your door of choice.

If however you picked incorrectly, he'd eliminate the other wrong choice and offer you the chance to switch.

I've seen host Eddie McQuire do it on Australian Who wants to be a Millionaire, and interestingly enough, on the Million Dollar question. Obviously he couldn't eliminate choices for the contestant but he certainly used his very well developed powers of persuation to push the contestant in the right direction. A million dollars in exchange for national ratings would have been well worth the effort, however the contestant declined and walked with half a mil.

Eddie was shattered.

Alrighty,

I've also thought more about my answer above, where we question the host's motives. ... Question Eddy McQuire ... !?!

3 scenarios are

He doesn't want you to win the car

He does want you to win the car

He really doesn't care either way.

In scenario one, if you'd picked incorrectly the first time, he would have just opened your door and you've lost, therefore if he doesn't open your door, it's because the car is behind it.

In scenario two, if you've picked correctly the first time he would open your door and you've won. Likewise, if this doesn't happen it's because the car is not behind your door

In scenario three, it could go either way.

So your challenge as a contestant is to decide if the show via the host want you to win a car or not. I would guess that they do, therefore I'd switch my door when given the opportunity, but I'd also acknowledge that two doors and one last choice give me even odds of getting it right, easy come easy go if I got it wrong.

Do you get to keep the goat ?

"So your challenge as a contestant is to decide if the show via the host want you to win a car or not. I would guess that they do,"

It would not help the ratings if everyone won - there still needs to be some luck involved, but if no-one has won for a few weeks.........

Sometimes I think these engineering discussions degrade into management discussions. I've had one line of my discussion quoted with the reasoning behind the quote totally ignored. That's not engineering, that's management whitewash.

Forget motives, that's for management - this would be an ongoing show, so you have history to work with. The question suggests that the host's behaviour is consistent - he opens a door (other than yours) that has a goat behind it on each occasion. Unfortunately, it doesn't actually say that this happens every time, so it is good practice to consider other possibilities that you could base on the history; that would include the version that is most probably what the questioner intended.

I believe the answer lies in game show dynamics and the question "why did the host open a door different from the door chosen by the contestant?"

A question arises - do the rules of the game say that the contestant can chose twice or until he hits the door with the car (whichever is earlier)?

If yes, then there is a higher probability of the car being behind the door chosen by the contestant. Because, had that door been opened (and a car found behind it) the show would have come to an instant end - something game shows do not want.

If no, I cannot think of any reason to say which door has a higher probability of having a car.

Addendum to my previous comment:

I am led to conclude that the car HAS to be behind the other door (third door).

Because, had it been behind the door originally chosen by the contestant and if the contestant now chooses the third door; it can cause a major argument between the contestant and the host that the host purposefully did not give him the prize. Since in the original question, it does not say that the game show rules allow the host to choose any door other than the contestant's choice - the show and the host have clearly made a mistake. They would not only end up giving the prize but also earn public wrath - very very undesirable.

But, if the car is behind the other door - the two possible scenarios are both 'safe' for the show.

Scenario 1 : The contestant choses the first door (his original choice) and loses. He cannot argue because had the host opened the same door last time, he would have still lost. Infact, the host gave him a second chance to win by breaking the rules.

Scenario 2: The contestant choses the other door and wins a car. Though the host did break the rules on the first choice, nobody is complaining and the show gets better ratings.

Having thought about it again, I am very sure that the car HAS to be behind the third door.

- Gurdas

"To maximize the probability to win the car"

this is the question. how would switching the selection affect the probability in winning the car.

say the three doors are named Goat1, Goat2 and Car. If he picked originally Goat1, then the host reveals Goat 2 and the contestant is down to a Goat and a Car.

Same goes for each case, he'll be down to a goat and car wether he picked the car in the first place or not by hitting the 1/3 chance. he's naturally by the game rule pushed into 1/2

i think its just luck from here on, he cant do anything to maximize his probability. he would have reached this point whatever he had chosen in the first place.

Schroedinger's cat is dead!

Good Point !!

Refuse to choose, that way you've got both a car AND a goat.

Just up and leave your forwarding address for them both.

PS If they protest, just tell them that if they hadn't interfered you'd have a car and TWO goats, and you're thinking about sueing them because they changed the rules half way through [Double Slit experiment]

This is a case of combination theory.

Look at the probabilities below :

Door 1 Door 2 Door 3

Comb 1 Goat Goat Car

Comb 2 Goat Car Goat

Comb 3 Car Goat Goat

Initially, the probability of picking the right door is obviously 33.33%.

Let us assume that we picked Door 1. Now, look at the situation where the host opens one of the doors (say Door 2) to show a Goat. Since Door 2 has a Goat, it means Door 2 and Comb 2 are no more part of the probability set. Therefore, in the new probability set, only Comb 1 and Comb 3 are relevant. Both Door 1 and Door 2 could have either a Car or a Goat behind them.

Therefore, the probability of finding a Car behind either of the doors is 50%. It does not really matter whether you stay with the first guess or switch to the other door.

Mathematically, it is quite straightforward. Initially, the number of possible outcomes is 3C2 ie. 3 and so the probability is 1/3. When the host reveals one Goat, the number of possible outcomes is 2C1 ie. 2 and so the probability is 1/2.

What this exercise means is that the host actually increases the chances of the participant by opening one of the doors. I am not sure whether TV game shows do this deliberately.

C D Sripathi

Hi C D Sripathi,

"I am not sure whether TV game shows do this deliberately."

Whether he does it deliberately or is the real question. If the host chooses randomly then he has a 1/3 chance of finding the car himself, leaving 2/3 chance that it is left for you to find. You have a 1/2 chance of finding it from the two remaining doors so your total odds are 1/2 x 2/3 = 1/3.

If the host deliberately chooses the door with the goat then the door he ignores has a higher probability of having the car.

Try this analogy. You have 3 people and you want to find the most intelligent of them. You put one in a room by himself and give the other 2 people an IQ test. You reject the one with the poorer results from the IQ test. The person who had the higher IQ test result is more likely to be more intelligent that the untested person because he has been through a selection process.

Davo

much different...

=-==

by definition, the odds are that the untested person has an IQ of 100.

you know the IQ of the higher test score, so...

If we get to keep the two goats and they are of an amourous disposition, we could be quids in. And maybe if we get to keep the doors too, we can start the goat pen (triangular of course). The host would be condemned to be a hired hand for his tricky psychological dealings, the producer would keep house and make the cheese, the hosts and hostesses are actually from the countryside and have great goat rearing knowledge and we would all live happily ever after. Especially as fuel rationing in the future means that the game theory idiot who won the car won't be able to use it.

Its what's not written in the question which is the most important.

1] u pick door - 1/3 right; 2/3 wrong

2] monty picks door - 1/3 monty can pick either of two goat doors to open leaving you with 1/2 odds; 2/3 monty is forced to open goat door instead of car door [constrained by the question] and leave you with a car if you switch to 'monty's bain'

=-==

If I remember the show right, monty did not always pick the goat, some times he would show you that you lost the car...sadistic old coot.

You must consider what reason the host had for opening the door not requested by the the gamer.

If he wants the show to fill an alotted time with the same gamer , & he knows where the goats are, then he is stalling for time to pass. Under this scenario , since he cannot stretch the show beyond one more guess anyway, I would stay with the door first chosen. Why - The host will want to give the car away because to have steered the gamer away from the right door will destroy the audience confidence in the shows management.

If the host truly does not know where the goats are, then you are reduced to a 50-50 pick of the doors with pure chance being all you can hope for.

Since the host made a decision to open an alternative door, he presumably had a reason & therefore knowledge of where the cars were. Stay with the door chosen

STAY WITH THE FIRST GUESS...THEY ONLY OPENED THE OTHER DOOR TO GET THE CONTESTANT TO CHANGE HIS MIND BECAUSE HE/SHE WAS RIGHT...

Having already switched, I have decided to switch back to my original answer, which was to switch when given the opportunity.

The first thing that we picture is "Let's Make a Deal" with Monty Hall, so we assume that there are three separate doors on a stage sitting next to each other. The statement said that there is "a goat behind two of them" implies that there is only one goat and that there are two doors in alignment with each other so that the single goat can is behind two simultaneously.

Also,

There is nothing that says that the car can not be behind the same door as the goat. The statement says: "A goat is behind two of them, a new car behind the third". This means that it was never determined which doors the goat(s) were behind in relation to first, second, third. We only know that the car is behind the third.

My conclusion that this is an exercise in grammar that leaves a vague conclusion.

JN

JN

I believe you make points that are at least 2/3 valid (or 1/3 valid, depending upon your point of view). When I sent the question in, it said "an old goat is behind each of two doors." My question rambled on for two entire paragraphs, describing the scenario in painful detail. The CR4 folks did a great job editing it down to just one paragraph, and any ambiguity introduced can be easily addressed by the responder: "Assuming there are two goats and one car, and further assuming that these objects do not move around between guesses in either a random or programmed way, and further assuming..."

The CR4 editing improved the question by leaving out some detail re possible motives of the host and routines he might follow. Without this stuff spelled out, we might ask, "Do his motives change the probabilities?" "If his actions are random, how would that affect the probabilities?"

You write: "... the single goat can is behind two simultaneously." By "goat can", were you referring to the sort of can used to collect milk?

You write: "My conclusion that this is an exercise in grammar that leaves a vague conclusion." In attempting to parse that, I wondered if you had intended to create a sentence? I suppose you might have meant, "My conclusion is that..." You might also have been creating a sort of self-referential sentence re vagueness. I enjoy such stuff too. This sentence is missing it's

To be honest, I suspect there is no car at all. I think it's all in our heads.

Ken

Ken

No matter what door you pick tha host has a goat behind at least one of his doors which he can open to mess up your mind which is all part of the show. No matter what you have picked he is always going to open a door with a goat behind it because "THE SHOW MUST GO ON".

You had a one in three chance originally, but now it is a one in two. Your chances don't change by switching doors now because what the host did had nothing to do with what you had originally picked and either side of a one in two is still one in two.

See 46 and 136 amongst others. If in every single round of the game the host opens a door with a dummy-prize behind it, the action contains information. Use it. If unconvinced, try it. You don't even need a friend to help really, because the behaviour is entirely deterministic. Then think again.

Writing so that your initial selection is on the left. The initial possibilities are:
G1 G2 C (1/6)
G2 G1 C (1/6)
C G1 G2 (1/6)
C G2 G1 (1/6)
G1 C G2 (1/6)
G2 C G1 (1/6)

After the host removes one goat, the possibilities are:
G1 __ C (1/6)
G2 __ C (1/6)
C __ G2 or C G1 __ (the sum of these remains 1/6)
C __ G1 or C G2 __ (the sum of these remains 1/6)
G1 C __ (1/6)
G2 C __ (1/6)

If you stay as you were, the probability of a car is 1/6+1/6 = 1/3 (exclusive options)
If you change your mind, the probability of a car is 1/6 + 1/6 + 1/6 +1/6 = 2/3

Fyz

ur argument is quite to the point, however the mathematical model u did have one mis-assumption.

see when you reached the second phase, your choices were down to:

G1 C

G2 C

C G2/G1

C G1/G2

G1 C

G2 C

It is obvious that the first and last Two are the same, if you change you hit the jackpot, thus logically you're down to just 4 possible scenarios, and thus the outcome would be:

G1 C ---> 1/4
G2 C ---> 1/4 thus if you change you win 1/2

C G1 ---> 1/4
C G2 ---> 1/4 thus if you change you loose 1/2

so it doesnt really matter, it is only a mind game the host is playing

cheers

Try all three choices with marked counters. You'll find that if you choose correctly first time, changing your mind will lose the car. But if you choose wrong, changing your mind will always get you the car. As the first happens 1/3 of the time, and the second 2/3 of the time, the advantage in changing your mind is a factor of 2.

You mistake is assuming that the action of the host changes the probabilities - i.e. that the 1/3 initial probability that you chose correctly expands to 1/2 just because the host shows you that there is a goat behind one of the other doors.

You mistake is assuming that the action of the host changes the probabilities - i.e. that the 1/3 initial probability that you chose correctly expands to 1/2 just because the host shows you that there is a goat behind one of the other doors.

what the ___?! just a couple posts back you said the hosts showing where a goat is changed the probs up to 66%. Now you say it has no effect?? The action of the host does change the prob -- up to 50% instead of 33!

goat... car. one or the other. 50%. can I make it any easier?

Yes, you can make it easier for yourself. Forget what you were taught as formal statistics (it was probably geared to specific cases where it may or may not have been right), and think about the possibilities on the basis that, although the initial situation is 1/3 for each window, the host is giving you information that refines your choice. It really is horribly simple - you have a probability of 1/3 of being correct in the first instance. That does not change. In the absence of additional information, each of the other two positions also has a probability of 1/3 (2/3 in all). If at this point, host gave you the option of opening both the other doors and taking the car if it is behind either one, you would naturally switch. If host then knowingly opens the door with a goat first, that doesn't change the total probability - it simply tells you that the car definitely wasn't behind that particular door. If he chooses between the doors at random, and happens to display a goat (i.e. he shows the car 1/3 of the time as you would expect from a random choice), that would leave you with the 50% probability, and there is no purpose in switching. The reason the residual door now has double its original probability is that the host's informed choice has systematically eliminated a number of the original possibilities - the majority of which were that the car was not behind the door he failed to choose. (If the car had been behind the door he eventually opened, he would not have done so).

Of course, this only works on the assumption that the host will knowingly open a door with a goat, and that it is not the door you chose, irrespective of the choice you make. If host's behaviour depends on your choice, you need to know the algorithm before you can draw any conclusions.

Fyz

Then you'd have a probability of getting the goat 75% of the time, right?

You make the assumption that each result that you describe as 'different' in the second phase has the same probability. I can find no basis for this.

If either the initial choice or the initial car position is random (or, indeed, both), your initial choice had a 1/3 probability of being correct. This will not change unless the game-show company move the car around - in which case the probabilities and best decision are whatever the company decides they should be. The other condition under which there can be a change is if the decision to show the goat depends on what is behind the window you chose, or if the host randomly opens one of the other two windows - in which case this would reveal the car one time in three

let me go back to your original statement and show you what is not right:

G1,G2,C ----eliminating one goat---- G1,-,C ==> 1/8
G2,G1,C ------- // ----------------- G2,-,C ==> 1/8
C,G1,G2 ------- // ----------------- C,G1,- ==> 1/8
------- // ----------------- C,-,G2 ==> 1/8
C,G2,G1 ------- // ----------------- C,-,G1 ==> 1/8
------- // ----------------- C,G2,- ==> 1/8
G1,C,G2 ------- // ----------------- G1,C,- ==> 1/8
G2,C,G1 ------- // ----------------- G2,C,- ==> 1/8

If as you are saying that different results can not have same probability or in other words do not merge into one condition, then these eight cases that I put up there are the possible cases when the host moves to the next phase.

thus you can clearly see that switching doors is 4/8 and sticking to the original choice is also 4/8. which means it is as obvious just a 50-50 probability. in anycase its not the point as while progressing to the next phase, the original problem do not exist, and you're down to just 4 possible cases:

G1 C
G2 C
C G1
C G2

You got to chose one of these two doors, you can't just stick 1/3 in there.

cheers.

Where do you get 1/8 from? The sum of probabilities of all events based on an initial event is equal to the probability of that event in the first instance. That is where the 1/3 factor comes from, and it will not go away. If you don't believe this, try it experimentally - and be careful that you count the results from each initial selection just the once.

Fyz

On a game show, a contestant is offered the choice of three doors

I would have said jim , as he's the only one I can name , and demanded my choice of prize.

.

I think that now the gameshow 'host' has opened one of the three doors and revealed a goat that now gives me a 50/50 chance of my guessing which door has the car or perversly the remaining goat. If I stick with my original door choice the odds remain the same its still 50/50 I don't know if the show host knows which door has the car/goat behind it so in being offered another choice of door selection is irrelevant. Short of listening at the doors before picking one to for the animals sound,which in a niosy studio may be difficult, just pick a damn door.

It depends on what the host is doing. If he always follows the same pattern - i.e. opens a door, and there is never a car behind it, then his choice of which door to open contains information. So you are now looking at dependent probabilities, not independent. I think simplest way of looking at this is as follows: you choose a door, the chances are 1 in 3. The host now says - you can either stay as you are, or have the option of my opening whichever of the others does not have a car behind it, and you opening the other of them. If (and effectively only if) he always does this, then the chances of the car being behind your original door have not changed. Nor have the chances of it being behind one of the other two. So the offer is just the same as giving you the option of opening both the other doors, and taking the car if it is behind either.

If preconceptions prevent you accepting this (which would not be surprising, given the way statistics is generally taught by rote), I suggest you and a friend try an equivalent experiment. The results should allow the fresh air in.

so I can clearly not choose the wine in front of you. But, you've also bested my Spaniard, which means you must have studied, and in studying you must have learned that man is mortal, so you would have put the poison as far from yourself as possible, so I can clearly not choose the wine in front of me.

The pellet with poison is in the chalice from the palace, the flagon with the wagon had the brew that is true. So you had a 1/4 chance of getting the combination right and the same as last time.

Guest:

Inconceivable! He built up an immunity to the poison, and so spoke the truth when he said he did not even truly know which had the poison and which did not. Your best chance at that point was NOT to drink either, but to RUN AWAY!

You need to understand that what he was supposed to remember was,

"The pellet with the poison is in the vessel with the pestle. The chalice with the palace has the brew that is true."

But then they dropped the chalice from the palace, and the plan had to be changed....and the pellet with the poison was in the flagon with the dragon.

There are TWO goats. Whether you pick the car or the goat that still leaves MONTY the ability to open the OTHER goat. In the final phase you still had the same odds. Your just being allowed to second guess your first decision.

...to open the OTHER goat

Ewe! Sounds gruesome.

No, niet, nein non niente. If the host ALWAYS opens a door, that gives you information. It doesn't change the position of the car.

If you selected the door with the car initially, it is a mistake to change. That happens one time in three.

If you do not select the door with the car initially, there is a goat behind one door and a car behind the other. The host tells you which door has a goat - so the other must have a car (assuming no cheating). So you change your mind, you get a car. This happens two times out of three.

It's not really complicated - except that preconceived ideas (due to bad teaching?) can make it hard to accept.

Fyz

Bayes' theory or the the theory of restricted choice says that the odds slightly favor switching to the third door.

Joe

You meant to say second door right? Bayesians would have to pick the second door. 50% of the time the car would be there-- which is better than 33%.

Depends on assumptions. If the host's behaviour is consistent, and he always shows a goat, the benefit in switching will be 50%. If the host has no knowledge, and just happens to show a goat on this occasion, the probabilities in each case will be equal, so there's no benefit in switching (but would reduce to zero 1/3 of the time when he shows the car...)

It doesn't matter, you have a 50/50 chance. For all of those who disagree, I would love to play a few of card games with you, for money of course.

After the first goat is revealed, ALL YOU KNOW is that its behind one of the two remaining doors but you have no clue one way or the other. Multiply fractions till the cows come home, it won't make a difference.

Think of it like this: We have doors X,Y,Z. You pick door X and some guy in the audience picks door Y. The host opens up door Z and reveals a goat. This talk of switching doors to improve your chances of winning would guide you to switch from X to Y and the guy in the audience to switch from Y to X... hmmm... how can door Y be more probable for you but X more probable for the guy in the audience?