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Whatchya Gonna Do (2 goats and a car)

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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!)

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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.)

No, he's not allowing you to trade your original door for the other two. That's why I said that it's essentially what he's doing. I gave Maximus and you a way of looking at a similar scenario that leaves you with the same probability of winning the car in hopes that you would see why switching doors is the wise choice.

We agree that at first there is a 1 in 3 chance of picking the door with the car behind it. We also agree that if the host lets you switch your door for the other two, your probability increases to 2 in 3, correct? Then it should be evident to you that switching is the best choice and garners you the same increased probability of winning the car whether or not he allows you to keep what's behind both doors or first reveals a goat that you can't keep.

You said this earlier:

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.

What if there were 100 doors? One door has a car behind it, the other 99 have goats. (Remember, the host doesn't reveal what's behind doors randomly before giving you a chance to switch; he never reveals the car.) You pick a door. The host reveals 98 doors all with goats behind them. He now gives you a chance to switch your door for the one that he didn't open. Do you still think "One has a goat behind it; one has a car. It's 50/50"?

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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?????

You're on the right track - you should struggle with the idea of a 1/3 chance changing to a 1/2 chance.

Why should seeing a goat that you already knew was there affect your 1/3 chance?

It shouldn't. And it doesn't.

The odds are 100% that the car is behind one of the doors:

Door 1: The door you picked has a 1/3 chance. The other two doors, combined, have the remaining 2/3 chance.

Door 2: The opened door that revealed a goat has zero. Now you know where the 2/3 chance lies - the third door.

Door 3: The third door has the remaining odds: 1 minus 1/3. Roughly 2/3.

Should you make the switch?

Hint:

The number of choices open to you is not what is important here, because the success probabilities of the choices are not equal.

Lottery tickets have only two outcomes - you win or you lose. But not with equal probability.

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No, he's not allowing you to trade your original door for the other two. That's why I said that it's essentially what he's doing. I gave Maximus and you a way of looking at a similar scenario that leaves you with the same probability of winning the car in hopes that you would see why switching doors is the wise choice.

We agree that at first there is a 1 in 3 chance of picking the door with the car behind it. We also agree that if the host lets you switch your door for the other two, your probability increases to 2 in 3, correct? Then it should be evident to you that switching is the best choice and garners you the same increased probability of winning the car whether or not he allows you to keep what's behind both doors or first reveals a goat that you can't keep.

You said this earlier:

What if there were 100 doors? One door has a car behind it, the other 99 have goats. (Remember, the host doesn't reveal what's behind doors randomly before giving you a chance to switch; he never reveals the car.) You pick a door. The host reveals 98 doors all with goats behind them. He now gives you a chance to switch your door for the one that he didn't open. Do you still think "One has a goat behind it; one has a car. It's 50/50"?

We do agree that a 2/3 chance of winning is better than a 1/3 chance of winning. However, we do not agree that switching to the other door increases your probability of winning to 2/3.

You're on the right track - you should struggle with the idea of a 1/3 chance changing to a 1/2 chance.

Why should seeing a goat that you already knew was there affect your 1/3 chance?

It shouldn't. And it doesn't.

The odds are 100% that the car is behind one of the doors:

Door 1: The door you picked has a 1/3 chance. The other two doors, combined, have the remaining 2/3 chance.

Door 2: The opened door that revealed a goat has zero. Now you know where the 2/3 chance lies - the third door.

Door 3: The third door has the remaining odds: 1 minus 1/3. Roughly 2/3.

Should you make the switch?

Hint:

The number of choices open to you is not what is important here, because the success probabilities of the choices are not equal.

Lottery tickets have only two outcomes - you win or you lose. But not with equal probability.

The three doors have a 1/3 chance of having a car before the first goat is revealed. After the goat is revealed, that 1/3 is split between the other two doors so that they each have 1/2 chance of winning.

"Pick a number between 1 and 3. By the way, it's not 1." gives you a 50/50 chance of choosing 2 or 3.

"Pick a number between 1 and 3." You pick 2. "By the way, it's not 1." Your decision here to stay with 2 or switch to 3 is the same as your original decision from the previous scenario. You are still just picking randomly between two objects, one winning and one losing. It is 50/50

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... the host, Monty Barker, showed her three doors ...

Monty who?

Ok, let me take a stab ...

If you win, you take the car. If you don't get the car, Monty wins.

You can have chose a door which has 1 in 3 chances to win. Monty then has, the

After you pick, just because Monty talks slick, opens your door, his door, Carol Merrill's door (Carol who?), or starts singing the blues, the chances of your door does not change. It's statistical; when you picked it, it had 1 in 3 and nothing can change that statistical fact.

Once that is accepted, you can quickly see that eliminating one of the remaining doors leaves the other remaining door with 2 in 3. This is because your door keeps 1 in 3 and the total always remains 3 in 3 so the remaining door must have 2 in 3.

I like

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... the host, Monty Barker, showed her three doors ...

Monty who?

Ok, let me take a stab ...

If you win, you get the car. If you don't get the car, Monty wins.

You can choose 1 door which gives you 1 in 3 chances to win. Therefore, Monty has 2 in 3 chances to win. These odds are established when the rules are set (1 door of the 3 will win). They are not established based on the order in which the doors are opened and they are not re-established after each of the 3 doors are opened.

Therefore, when Carol Merrill (Carol who?) opens one of Monty's doors (not the winning door) after you have selected yours, your chances remain 1 in 3 and Monty's chances remain 2 in 3.

It then follows that if you swapped with Monty, YOUR chances are now 2 in 3 and with only 1 door !!!

Conclusion: Everytime you play this game, make the switch and your odds will double.

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The three doors have a 1/3 chance of having a car before the first goat is revealed. After the goat is revealed, that 1/3 is split between the other two doors so that they each have 1/2 chance of winning.

When you chose a door, your chances were 1/3. There is definitely at least one goat behind one of the other two doors that the host can reveal. Why on Earth would the host revealing one increase your probability from 1/3 to 1/2 when he knows which door has the car behind it?

What about in my example with 100 doors and 99 goats? Do you also split the 1/100 probability between the remaining doors so that each has the same chance of having the car behind it?

You have been asked several questions and you haven't answered any. You have been given several proofs and you haven't even attempted to disprove one. You were even given a link to a computer simulation and you haven't answered the question asked of you twice if you've bothered to try it. Think what you wish.

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Actually, it's very difficult to explain this. There's several ways to explain but I can see how each can be challenged.

When I first saw it, I took me a couple of days to convince myself. Then, after grasping the idea, I opened the discussion with a guy who is much, much sharper than me. I could not convince him but a short while later, he convinced himself. So, even though I can understand, I have a problem explaining it.

I think Martini's example of 100 doors is a good way to explain.

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When you chose a door, your chances were 1/3. There is definitely at least one goat behind one of the other two doors that the host can reveal. Why on Earth would the host revealing one increase your probability from 1/3 to 1/2 when he knows which door has the car behind it?

What about in my example with 100 doors and 99 goats? Do you also split the 1/100 probability between the remaining doors so that each has the same chance of having the car behind it?

You have been asked several questions and you haven't answered any. You have been given several proofs and you haven't even attempted to disprove one. You were even given a link to a computer simulation and you haven't answered the question asked of you twice if you've bothered to try it. Think what you wish.

Easy, Martini, don't get too worked up. I can throw facts and figures out there and computer simulations. I think it is all irrelevant information to make a game show more entertaining. It all boils down to two doors, one with a goat, one with a car. You get to pick one. That gives you a 50% chance. I am not trying to disprove the fact that you start with a 1/3 of getting it right with your first guess. I am saying that when you eliminate a possibility, the chances of the remaining possibilities should be recalculated.

For your 100 boxes example:

The chance of choosing the box correctly on the first guess is 1:100. The host takes away 98 boxes. It's a slightly different problem because the possibilities eliminated have a different combined probability than the original choice. Now if there were 99 boxes, and you got to choose 33, then the host removed 33 and you could either keep your original or take the 33 left untouched, then the probability of the winning box being in your original 33 and the untouched 33 would be the same.

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After reading multiple articles, I must admit I'm wrong. Of course, this is no surprise to pretty much everyone that was arguing against me, but I felt obliged to admit it anyway. My stubborn brain is still wanting to use the 50/50 logic, but when enough experts agree on what you think is wrong, then you are likely wrong.

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Easy, Martini, don't get too worked up. I can throw facts and figures out there and computer simulations.

I didn't get worked up. If you can throw out computer simulations that prove your assertion to be correct, then please do it. Mine proved that my solution is the correct one. All "facts and figures" you threw out were disputed.

I think it is all irrelevant information to make a game show more entertaining. It all boils down to two doors, one with a goat, one with a car. You get to pick one. That gives you a 50% chance.

And again, according to that rationale, if it boiled down to two doors (but there were originally 100) one with a goat, one with a car. You get to pick one. That gives you a 50% chance.

I am not trying to disprove the fact that you start with a 1/3 of getting it right with your first guess. I am saying that when you eliminate a possibility, the chances of the remaining possibilities should be recalculated.

That would only be true if the host randomly eliminated a door and it happened to be a goat. I'll explain using the example with 100 doors:

You choose a door, the host randomly opens up 98 of them from the remaining 99 doors. All 98 happen to have goats behind them. Should you switch to the one he didn't open? It doesn't matter. The probability is 1/2 that you chose the door with the car behind it.

other scenario:

You choose a door, the host knows which door has the car behind it (one of the 99 or the one you chose) and purposely doesn't open it if it happens to be behind one of the 99 remaining doors. He opens up 98 doors which he knows have goats behind them. Should you switch? Yes. Switching will give you a 99/100 probability of winning a car.

For your 100 boxes example:

The chance of choosing the box correctly on the first guess is 1:100. The host takes away 98 boxes. It's a slightly different problem because the possibilities eliminated have a different combined probability than the original choice. Now if there were 99 boxes, and you got to choose 33, then the host removed 33 and you could either keep your original or take the 33 left untouched, then the probability of the winning box being in your original 33 and the untouched 33 would be the same.

But you said, "It all boils down to two doors, one with a goat, one with a car. You get to pick one. That gives you a 50% chance."

If the host reveals 33 doors which have goats behind them on purpose, then you should switch. He only revealed what's behind the doors if they had goats behind them. That makes the other 33 doors much more special than yours. Yours were picked randomly; the other 33 remaining doors were given special protection from not having the car removed from it if it were among the 66 that you didn't choose.

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I am blown away, busting a gut, I think I'm gonna have stitches from the love of finding a truly interactive web-site. I'll have another martini to go, please. Amid all the beautiful hu-hah, I did notice something that has not been commented on, that being the uncommented upon advertisement amid the "puzzle". It appears to me that it is neither the drive about Toronto in new sports car, ,or the dinner at Hoi-Polloi Hilton Restaurant, on the back of a goat that is truly the issue. Like other slick deception filled ad-games, it is neither the game nor prizes that matter, only the unseen car-wax ad slipped into the life of expectant players, that is really the governing issue here. I have not read the solution and am totally amused by the responses of all engaged. Thank you so much. Carrying on, and on..........I'll return. By the way, I have a new knock/knock joke. ok? You start it. (Knock-Knock), my response..Who's there?

Edited by Spyderzden
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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.

I think Jennifer should stick with (door #2).

I dont know why but i just feel that way.

May be loosing with door#2 wont hurt that much, than switching to door#3 n then loosing.

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The three doors have a 1/3 chance of having a car before the first goat is revealed.

After the goat is revealed, that 1/3 is split between the other two doors so that they each have 1/2 chance of winning.

"Pick a number between 1 and 3. By the way, it's not 1." gives you a 50/50 chance of choosing 2 or 3.

"Pick a number between 1 and 3." You pick 2. "By the way, it's not 1."

Your decision here to stay with 2 or switch to 3 is the same as your original decision from the previous scenario.

You are still just picking randomly between two objects, one winning and one losing. It is 50/50

Nope.

When you choose a door, you know one of the doors you didn't choose has a goat.

Your winning chances are, nevertheless, 1/3.

You are saying that somehow, magically, if you see that goat, your odds increase to 1/2.

Final try:

You pick a door.

Nothing else happens.

You are given the opportunity to pick the other two doors instead.

Do you swap? Of course. No one would argue that choosing two doors has the same winning odds as choosing one door.

Ah.... but be careful; one of those other doors has a goat! Here ... I'll even show you which one.

OK, you're right. No sense swapping now that I see that both of the other two doors aren't winners.

If you believe that two outcomes are always equally likely, you should buy more lottery tickets.

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So, in the game deal or no deal, the chance of getting a higher prize money at the end is 22 out of 23 if you switch to the last box remaining and only 1 in 23 if you stay with your own box???

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So, in the game deal or no deal, the chance of getting a higher prize money at the end is 22 out of 23 if you switch to the last box remaining and only 1 in 23 if you stay with your own box???

No, I explained this in post #36. On Deal or No Deal all of the boxes chosen are done randomly. So switching doesn't matter; the probability of winning a higher amount doesn't change if you switch. But, if Howie Mandel were the one to eliminate boxes and he didn't do it randomly, he eliminated lower amount boxes first- definitely switch.

In this riddle, the host gives special protection to the car if it is among the two doors you didn't select (which it most likely is). Since he most likely has the car behind one of those doors, and he intentionally reveals which one has the goat behind it, then of course switching to the other door is the best bet.

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Ok this is still really confusing me and i cant seem to catch on but i am happy to accept that you are right :D

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First off, let me say that I'm glad to find your site from among the Google add-ins.

With that said, this is not my first time to come across this problem. However, I just now saw the light for myself, and I thought I'd share how with those who still aren't convinced.

You initially have a 2/3 probability of picking a goat. This means that you also have a 2/3 possibility of the host revealing the only other goat on the stage. Therefore there is a 2/3 possibility that the other door on the stage holds the car. It's that simple.

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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.

Thank you martini for providing this link. Both for the simulation and for the link at the bottom of the simulation that provides articles about the puzzle's history which includes the discussion about Marilyn vos Savant. Back stories to the puzzles just makes them so much more interesting and debatable.

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No, he's not allowing you to trade your original door for the other two. That's why I said that it's essentially what he's doing.

No, that's not essentially what he is doing.

Essentially what he is doing is giving you a door, regardless of which of the other two doors you select. By your own logic, you still have 2/3 chance if you keep your original choice -- your original door AND the one that was opened to reveal one of the goats. This means that either of the remaining doors has a 2/3 chance, which makes them equally likely to contain the car.

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No, that's not essentially what he is doing.

Essentially what he is doing is giving you a door, regardless of which of the other two doors you select. By your own logic, you still have 2/3 chance if you keep your original choice -- your original door AND the one that was opened to reveal one of the goats. This means that either of the remaining doors has a 2/3 chance, which makes them equally likely to contain the car.

No you don't keep your original door because you trade it. If you switch you double your chances from 1/3 to 2/3 of getting the car.

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No you don't keep your original door because you trade it. If you switch you double your chances from 1/3 to 2/3 of getting the car.

The argument was that switching gives you a 2/3 chance because you essentially get the 3rd door AND the revealed door. You get this same chance because you get your original door AND the revealed door. Same odds.

EDIT: BTW, I played the computer simulation 5 times without switching. I won the car 4 of the 5 times. Therefore, keeping my original choice gives me an 80% chance of winning the car. :D (kidding -- about the odds -- I really did win 4/5 times.)

Edited by gmarsha11
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well it is like getting both. (aka doubling your chances).

Let's say there were 100 doors and you picked one. If i offered you the 99 other doors would you take them? What if i told you that at least 98 of the 99 other doors have goats behind them, would you still switch? Of course! because you increase your chances 99 times. When you do switch, even if i open 98 doors that have goats behind them, you still have a 99/100 chance of getting the car, instead of your original 1/100.

It's the same thing, except with 3 doors instead of two, and switching after he opens one.

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well it is like getting both. (aka doubling your chances).

Let's say there were 100 doors and you picked one. If i offered you the 99 other doors would you take them? What if i told you that at least 98 of the 99 other doors have goats behind them, would you still switch? Of course! because you increase your chances 99 times. When you do switch, even if i open 98 doors that have goats behind them, you still have a 99/100 chance of getting the car, instead of your original 1/100.

It's the same thing, except with 3 doors instead of two, and switching after he opens one.

If he opens 98 doors and I keep mine, don't I still have a 99/100 chance? 98 goats have been revealed. I now have a choice between my door and the other one for the 99th door.

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If he opens 98 doors and I keep mine, don't I still have a 99/100 chance? 98 goats have been revealed. I now have a choice between my door and the other one for the 99th door.

Surprisingly no! The other door has a 99% chance of having the car, and yours only has a 1% chance

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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.

I think I get the answer, but I'm a little unclear about this point. I didn't see anything in the question stating that the host will always start by opening a dorr that Jennifer didn't select. If she got it wrong, perhaps the host would have just said, "too bad, you picked a goat." If that's true, the answer is obviously that she has to stick with the same door. I would think that this assumption should be written into the question, but maybe I'm misunderstanding something here.

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