Whatchya Gonna Do (2 goats and a car)

[Guest]

Jennifer was selected to be on the popular TV game show "Whatchya Gonna Do?". As she jumped up and down in excitement, the host, Monty Barker, showed her three doors.

"Now, Jennifer, behind one of these doors I have personally placed a brand new Jaguar XJS"

"Eeeeeeee!", Jennifer squealed in delight.

"But", continued MB, "behind the other two doors, there are goats! Select a door and the prize behind it is yours. Which shall it be: Door #1?".

"Oh-oh-oh-oh", Jenny jumped.

"Door #2?"

"Ah-ah-ah-ah"

or "Door #3?"

"Um-um-um-um....well...ah...okay - TWO! I choose Door #2". The audience cheers.

"Door #2. Okay Jennifer", says the host, "but before we show you what's behind Door #2, let me tell you that I am glad you did not pick Door #1, because behind Door #1 is..." The first door swings open to reveal a shabby looking goat. Jenny bounces around the stage in joy because she was thinking of picking that door.

"Ah, but the name of the game is 'Watchya Gonna Do?'. You picked Door #2, but there is still Door #3. I'm going to offer you a choice: either you can keep the door you picked, Door #2, and if the car is behind it, I'll throw in a dinner for two at the Hoi Poloi Restaurant..." The audience gives a collective "oooooo" (who wouldn't want to eat at the exclusive Hoi Poloi?)

"... or you can trade it for what's behind Door #3 - AND if it is the car, I'll throw in a year's supply of Platypus Wax! Tell us about the Platypus Wax, Joe"

Joe's voiceover proceeds to tell us all about the wonders of using Platypus Wax on a Jaguar. As he drones on, Jennifer is nervously trying to decide. Until, MB spins around, points to her, and says (with the audience chorusing behind him)

"What - Chya - Gon - Na - Do?"

Should Jennifer stay with her original choice (Door #2), switch to Door #3, or does it make any difference?

Assume the dinner at the High Poloi and the Platypus Wax are worth about the same dollar amount.

Aaaaaaand, before any of you wise-acres start, Jennifer lives in Toronto - the Jaguar is of far more use to her than the goat.

Jennifer should switch. Contrary to what may seem intuitive, switching actually doubles her chances of winning the car.

This problem is just a re-wording of what is known as the Monty Hall Problem. The key to understanding it is that the host knows the locations of the car and goats. His knowledge changes his actions and thus affects the odds.

Here is a breakdown of all the possible scenarios that Jennifer faces and why Jennifer should switch:

• Door #2 has goat B (probability 1:3) - MB shows goat A behind Door #1 (1:1) - the car is behind Door #3 (1:1) - switching wins the car - total chances (1:3 x 1:1 x 1:1 = 1:3)[/*:m:6a353]
  
• Door #2 has the car (probability 1:3) - MB shows goat A behind Door #1 (1:2) - goat B is behind Door #3 (1:1) - not switching wins the car - total chances (1:3 x 1:2 x 1:1 = 1:6)[/*:m:6a353]
  
• Door #2 has the car (probability 1:3) - MB shows goat B behind Door #1 (1:2) - goat A is behind Door #3 (1:1) - not switching wins the car - total chances (1:3 x 1:2 x 1:1 = 1:6)[/*:m:6a353]
  

There are (1:3 + 1:3 = 2:3) chances that switching will get Jennifer the car, and only (1:6 + 1:6 = 1:3) chances she would get the car by not switching. She should switch.

A more general presentation of the reasoning is this:

At the start of the game, there is a 2:3 chance that Jennifer will pick a door with a goat behind it. If she does, the host will reveal the other other goat, and switching doors will get Jennifer the car.

There is a 1:3 chance she will pick the car. The host will then reveal a goat. Switching would win Jennifer a good supply of Ch?vre (and the disdain of her neighbours).

So, 2 out of 3 times switching gets the car. Simple - unintuitive, but simple.

Why does the host's knowledge change the odds. Because he does not randomly select a door to open - he always opens a door with a goat. By doing this he reduced the possible scenarios for Jennifer to the four listed above. If he randomly picked, then Jenny's chances, if the show progressed as presented, would be 50/50. However, there would also be a 1:3 chance that MB would open the wrong door and reveal the car's location (followed by a 1:1 chance that MB would be sacked and re-runs of McGyver would fill out the remainder of the season!)

[Guest]

I will refrain from answering this riddle as I encountered recently and after several hours of debating, I finally saw the light and understood the justification behind the correct answer.

I did want to give cpotting a round of applause for the great setup.

[Guest]

Its a very interesting puzzle. However I have a question in order to solve this, otherwise my answers dont make any sense.

Question: Since the question states

Should Jennifer stay with her original choice (Door #2), switch to Door #3, or does it make any difference?

Will he show her what is behind that remaining door, after she picks one?

If Yes, then I think she should Stay with Door #2, Because it seems like the host is just playing with her mind.

If No, then yeah it doesnt make any diffrence hoping that both other doors have cars, and since this is just a show, which ever sponsor gets lucky will pay the host

did I make any sense at all?

peace...

[Guest]

Will he show her what is behind that remaining door, after she picks one?

If Yes, then I think she should Stay with Door #2, Because it seems like the host is just playing with her mind.

There's no need for him to reveal what's behind the remaining door if she decides to switch (other than to prove the show was honest about what was behind the doors). After she's seen what's behind two of the doors, what's behind the remaining door is apparent.

All that matters is that the host's policy is always to reveal a goat behind one of two doors that weren't picked and give the contestant the opportunity to switch. If the host always behaves in this manner, should you switch, not switch, or does it not make a difference?

[unreality]

this is an old casino problem originally made with three turtle shells and a pea. its a fun one

[Guest]

No takers (except for those already familiar with the problem)?

C'mon! I would have thought someone out there would take a stab at it.

[Guest]

Honestly, I am very interested in seeing your answer for this, cpotting.

The only possible hints that might lead to the answer is using the name "Hoi Poloi" for a restraunt.

After looking into it the term "hoi polloi" is used to refference the general public in a derogatory sense. As if to catogorize the group as "commoners."

In that sense, commoners would not own Jaguars, so if this was some sort of clue then I would say that door #3 would be the right pick.

But other than that, I would say it doesnt make a difference. If it is not stated that the contestants are given any valid hints, then no matter what anybody says it wouldn't matter which door she picked because there is a 50% chance that it is behind either of the doors and without any hints she has no way of knowing which one it is.

Thats my go for an answer. But I do want to find out what you have to say about it. Im dying to know your answer.

[Guest]

Honestly, I am very interested in seeing your answer for this, cpotting.

Okay then. I put the answer in the spoilers for the original post. I hope my explanation makes sense.

[Guest]

hmmm... So if she switches, there are more chances of her winning the car?

I'm sorry but i dont think this answer makes any sense....

Door #2 has goat A (probability 1:3) - MB shows goat B behind Door #1 (1:1) - the car is behind Door #3 (1:1) - switching wins the car - total chances for this scenario (1:3 x 1:1 x 1:1 = 1:3)

Door #2 has goat B (probability 1:3) - MB shows goat A behind Door #1 (1:1) - the car is behind Door #3 (1:1) - switching wins the car - total chances (1:3 x 1:1 x 1:1 = 1:3)

Door #3 has the car (probablility 1:3) - MB shows goat A behind Door #1 (1:2) - goat B is behind Door #3 (1:1) - not switching wins the car - total chances (1:3 x 1:2 x 1:1 = 1:6)

Door #3 has the car (probablility 1:3) - MB shows goat B behind Door #1 (1:2) - goat A is behind Door #3 (1:1) - not switching wins the car - total chances (1:3 x 1:2 x 1:1 = 1:6)

you never show possibility of Door #2 having the car.

and overall I dont think its a math question, in my opinion its more of a Logic or WordGame puzzle, meaning you should make a decision depending on Hosts reactions and your own intuition. What if the car is behind Door #2, then your whole math will be wrong, and she is left out with out a car.....

i know that i'm not 100% right. however i think thats a legit solution to that puzzle.

please feel free to debate that

peace....

[Guest]

you never show possibility of Door #2 having the car.

If there is one thing that I have learned over the last few days, it is that posing problems over an internet forum is easy - editing them is harder than getting a cat to take a bath!.

You are right - the text should have read:

Door #2 has goat A (probability 1:3) - MB shows goat B behind Door #1 (1:1) - the car is behind Door #3 (1:1) - switching wins the car - total chances for this scenario (1:3 x 1:1 x 1:1 = 1:3)

Door #2 has goat B (probability 1:3) - MB shows goat A behind Door #1 (1:1) - the car is behind Door #3 (1:1) - switching wins the car - total chances (1:3 x 1:1 x 1:1 = 1:3)

Door #2 has the car (probablility 1:3) - MB shows goat A behind Door #1 (1:2) - goat B is behind Door #3 (1:1) - not switching wins the car - total chances (1:3 x 1:2 x 1:1 = 1:6)

Door #2 has the car (probablility 1:3) - MB shows goat B behind Door #1 (1:2) - goat A is behind Door #3 (1:1) - not switching wins the car - total chances (1:3 x 1:2 x 1:1 = 1:6)

Sorry for the confusion - I will edit the original and hopefully it will make more sense. All these errors... it's embarrassing

[Guest]

My explanation of the logic is a bit different and possibly easier to understand.

Let's say the scenario were a little different.

After Jennifer picks Door #2, Monty Barker does not open Door #1. Instead, he decides to be generous and offer Jennifer to trade what is behind Door #2 for what is behind both Door #1 and Door #3.

Assuming that Monty is not intentionally trying to entice Jennifer away from Door #2, it obviously makes sense to trade.

The odds of the Jaguar XJS being behind Door #2 are one in three.

The odds of the Jaguar XJS being behind either Door #1 or Door #3 are two in three.

Jennifer knows she will get at least one goat if she trades, but she also doubles her chances of getting the car.

Now, back in the real scenario, it does not matter that Monty shows Jennifer the Goat behind Door #1. We already knew there was a goat behind at least one of the remaining doors. Switching still doubles her chances of getting the car. The goat behind Door #1 has just been discarded before she wins it, instead of afterwards.

If the above still does not convince you, then maybe this one will. This is the method my friend used to convince me after I argued with him for quite a while about it.

Change the scenario from three doors to a hundred boxes.

Now, there are one hundred boxes to pick from. One has a key to the fabulous Jaguar XJS, the other ninety-nine are actually boxes of Kleenex.

After Jennifer picks box #54, Monty opens all of the other boxes, except #43, to show that they are all tissues.

Finally, he offers Jennifer a chance to trade her box for #43.

Should she make the trade or keep the box she has.

It should be obvious at this extreme that there is still only a one percent chance of box #54 having the key, but a ninety-nine percent chance that #43 has the key.

[Guest]

Riddari,

That second spoiler is amazing. makes it pretty clear for some people... like roommates

peace...

[bonanova]

edit: ha! now I've read Riddari's first spoiler. This post adds nothing to what he said.

The way I finally understood the answer is this:

if you stick with your first selection your chances of winning are 1 out of 3. If you switch, you cover the other two possibilities and your chances have to be 2 out of 3.

That one of the two doors you get by switching has a goat is a red herring. you knew that already. Doesn't change your chances.

Riddari's explanation #2 is compelling, also - it's the same situation, magnified, and makes intuition favor the correct answer. It took me an hour to believe the 1 of 3 case.

[Guest]

If the above still does not convince you, then maybe this one will. This is the method my friend used to convince me after I argued with him for quite a while about it.

Change the scenario from three doors to a hundred boxes.

Now, there are one hundred boxes to pick from. One has a key to the fabulous Jaguar XJS, the other ninety-nine are actually boxes of Kleenex.

After Jennifer picks box #54, Monty opens all of the other boxes, except #43, to show that they are all tissues.

Finally, he offers Jennifer a chance to trade her box for #43.

Should she make the trade or keep the box she has.

It should be obvious at this extreme that there is still only a one percent chance of box #54 having the key, but a ninety-nine percent chance that #43 has the key.

The concept is correct, but I think the reasoning is stated imprecisely. I would word it as follows: there is a 1:100 chance she picked the right box (#54) and, conversely, a 99:100 chance that she didn't. Obviously, she should switch - there is a 99% chance that some box other than the one she picked has the car - but which one? Once 98 of the 99 boxes are shown to be the wrong ones and only box #43 remains, it becomes obvious which she one she should switch to.

I'm not disagreeing with you Riddan, I just changing the wording a bit - it seems clearer to me, but, as they say, to each their own.

The same logic applies to the original problem - just replace 100 with 3, 99 with 2 and 98 with 1.

[Guest]

I agree that your wording makes more sense, cpotting. That is what I was trying to get at.

[Guest]

I disagree with carrying the odds from the first choice to the second. I think the choices represent two separate independent events. In the first you have a 1:3 chance of picking the car. The second event is a choice between two doors with one prize. I say 50 / 50.

[Guest]

I disagree with carrying the odds from the first choice to the second. I think the choices represent two separate independent events.

You pick a door. If you pick a door with a goat (most likely), the host will show you which of the other door's also have the goat behind it. That makes switching to the other door the best bet.

For instance, if you pick the door with goat #1 behind it, the host will show you the door that has goat #2 behind it and by switching you will win a car.

If you pick the door with goat #2 behind it, the host will show you the door that has goat #1 behind it and by switching you will win a car.

If you pick the door with the car behind it, the host will show you a door that has one of the goats behind it and by switching you will win the other goat.

By switching doors when offered, you have a 2 out of 3 chance of winning the car.

Play with this a little and you'll see how it works.

[Guest]

The two events are completely independent of each other. You are all looking at this the wrong way (except skini). It is already decided that no matter what your first choice is, you will be shown a goat. That means that there is no probability. It will happen. It is just a little trick to add to the suspense. It will always happen. So since we are dealing with probability and possibility you can rule it out. It's 1. It will happen. So you are essentially looking at a really snazzy toss up.

Someone tells you that they are thinking of a letter, A, B, or C. You guess B. He or she says, "Well, it's definitely not A." So that doesn't mean anything. It means it is either B or C. You're odds are now 50/50. It is a probability between two choices.

[Guest]

The two events are completely independent of each other. You are all looking at this the wrong way (except skini). It is already decided that no matter what your first choice is, you will be shown a goat. That means that there is no probability. It will happen. It is just a little trick to add to the suspense. It will always happen. So since we are dealing with probability and possibility you can rule it out. It's 1. It will happen. So you are essentially looking at a really snazzy toss up.

Someone tells you that they are thinking of a letter, A, B, or C. You guess B. He or she says, "Well, it's definitely not A." So that doesn't mean anything. It means it is either B or C. You're odds are now 50/50. It is a probability between two choices.

The second choice is definitely not independent, because the options presented to you in the second choice are dependent upon your first choice. And having two choices does not mean that there is equal probability between the two.

The following is a probability tree that shows all possible outcomes if you initially choose door B. C represents a car. G represents a goat. X represents a door eliminated by the host.

 1/3    1/3    1/3

 CGG    GCG    GGC

  |     / \     |

 1/3  1/6 1/6  1/3

 CGX  XCG GCX  XGC[/codebox]

The total probability that the car is behind door B in the second choice is 1/3. The total probability that the car is behind the "other door" (which is A or C) is 2/3.

[Guest]

Look at the final scenario not the initial ones. You have two doors. One has a goat behind it; one has a car. It's 50/50. The rest is just smoke and mirrors.

[Guest]

Look at the final scenario not the initial ones. You have two doors. One has a goat behind it; one has a car. It's 50/50. The rest is just smoke and mirrors.

Did you try to follow my scenario? Did you try the computer simulation I linked to? There's no smoke and mirrors. Try to follow this:

Behind door #1 there is a goat named Betsy. Behind door #2 there is a goat named Fred. Behind door #3 there is a car.

Three possible scenarios if you choose to switch doors when asked:

1. You pick door #1 (Betsy). The host will show you Fred behind door #2 and then asks if you'd like to switch your door for door #3 (which has a car behind it). You choose to switch. You win a car.

2. You pick door #2 (Fred). The host will show you Betsy behind door #1 and then asks if you'd like to switch your door for door #3 (which has a car behind it). You choose to switch. You win a car.

3. You pick door #3 (the car). The host will show you a goat behind one of the remaining doors and then asks if you'd like to switch your door for the other one (which also has a goat behind it). You choose to switch. You win a goat.

Out of the three possible scenarios in which you choose to switch doors, you will win the car in two of them.

[Guest]

Although the simulation does seem to help with your side of the arguement (everyone but a few)

I am still struggling to understand it surely at first there is a one in three chance of the car being behind the door you pick, but when the host reveals the goat the chances then chance from one in three to one in TWO as there are now only 2 possible places the car could be?????

[Guest]

Door numbers do not matter. There are four possible outcomes:

You switch and get a car.

You switch and get a goat.

You do not switch and get a car.

You do not switch and get a goat.

50/50. It's even.

[Guest]

Door numbers do not matter. There are four possible outcomes:

You switch and get a car.

You switch and get a goat.

You do not switch and get a car.

You do not switch and get a goat.

50/50. It's even.

That's not how you go about figuring out probability. Using your rationale:

There are two possible outcomes when buying a lottery ticket:

You buy one and win.

You buy one and lose

50/50. It's even.

Did you or did you not try the simulation I linked to? I went over the three possible scenarios for you and outlined how two out of three of those scenarios the car is won. You're not even bothering trying to refute it.

[Guest]

Although the simulation does seem to help with your side of the arguement (everyone but a few)

I am still struggling to understand it surely at first there is a one in three chance of the car being behind the door you pick, but when the host reveals the goat the chances then chance from one in three to one in TWO as there are now only 2 possible places the car could be?????

Yes, after one door is eliminated, there are two possible doors which the car could be behind. However, the host does not eliminate a door randomly as he never eliminates one with the car behind it.

Think about it this way: When you pick one of the doors, you have a 1 in 3 chance that you picked the door with the car behind it. If the host gave you a choice to switch your door for both of the others, wouldn't you do it so you can have a 2 in 3 chance of winning? Well, that's essentially what the host is doing, only he's revealing what's behind one of the losing doors first.

[Guest]

Yes, after one door is eliminated, there are two possible doors which the car could be behind. However, the host does not eliminate a door randomly as he never eliminates one with the car behind it.

Think about it this way: When you pick one of the doors, you have a 1 in 3 chance that you picked the door with the car behind it. If the host gave you a choice to switch your door for both of the others, wouldn't you do it so you can have a 2 in 3 chance of winning? Well, that's essentially what the host is doing, only he's revealing what's behind one of the losing doors first.

But he's not telling you to trade your original door for the other two. He's giving you the revealed door free no matter what based on your hypothetical. If it there no matter what, then it should no longer be figured into the probability.

(This is by far my favorite post, btw.)