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# Whatchya Gonna Do (2 goats and a car)

## 197 posts in this topic

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Ooooh Grasshopper gets cranky.

Anyway, let me now really confuse you.

Let's look at it from the OTHER SIDE.

You have 3 choices. You don't know what you picked BUT what about MB?

Here are ALL of his choices for which one he KEEPS :

You pick CAR he picks GOAT A

You pick CAR he picks GOAT B

You pick GOAT A he picks CAR

You pick GOAT A he picks GOAT B

You pick GOAT B he picks CAR

You pick GOAT B he picks GOAT A

Hold on hold on. Don't go off the deep end again.

What we have here is called a CONDITIONAL. In other words the odds are not what they seem because Action B is conditional on Action A. There is a condition involved stating the CAR must be in the final 2. Thus we eliminate any result that does not have the CAR in the final 2.

So that leaves us with

You pick CAR he picks GOAT A

You pick CAR he picks GOAT B

You pick GOAT A he picks CAR

You pick GOAT B he picks CAR

Now what YOU have done is combine the first 2 into 1. Can't do it. They are seperate probabilities. What you CAN do is say this.

IF you pick the car he picks A or B.

IF you pick A or B he picks the Car

Can you do math? You end up with CAR = A+B and A+B = CAR

What you are trying to do is say that it makes sense that based on the initial odds, he will have the car to choose from and thus MUST choose the car. But the math doesn't work that way. Only knee jerk reaction works that way. The math says that he has a 2/4 chance of picking EITHER goat. And a 2/4 chance of picking the car.

Gotta love conditionals.

Anyway, this is just going to back and forth until the goats go home and the new model year rolls around. Take some classes in probability and see what you come up with.

Have fun!

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Posted · Report post

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Posted · Report post

Ooooh Grasshopper gets cranky.

Knock it off!

Anyway, let me now really confuse you.

You're in no position to confuse anyone as you are the one that hasn't yet explained why all reliable internet sources disagree with you and a simulator that deals with this very problem consistently shows that 2/3 of the time switching wins one the car.

If you want to discuss this, the way to do it is to by rebutting ones arguments. I went through the trouble of showing you your mistakes in my last post, it would be nice and maybe get you somewhere if you responded to them.

I brought up the simulator several times and you have still not mentioned trying it, even when I ask you directly. As a matter of fact, you are not answering any of my direct questions.

Did you try the simulator? Did you look at the Dr. Math website I linked to?

Let's look at it from the OTHER SIDE.

You have 3 choices. You don't know what you picked BUT what about MB?

Here are ALL of his choices for which one he KEEPS :

You pick CAR he picks GOAT A

You pick CAR he picks GOAT B

You pick GOAT A he picks CAR

You pick GOAT A he picks GOAT B

You pick GOAT B he picks CAR

You pick GOAT B he picks GOAT A

Hold on hold on. Don't go off the deep end again.

What we have here is called a CONDITIONAL. In other words the odds are not what they seem because Action B is conditional on Action A. There is a condition involved stating the CAR must be in the final 2. Thus we eliminate any result that does not have the CAR in the final 2.

So that leaves us with

You pick CAR he picks GOAT A

You pick CAR he picks GOAT B

You pick GOAT A he picks CAR

You pick GOAT B he picks CAR

Go off the deep end again? No one has gone off the deep end. How about listening a little instead of continuously trying to teach?

Monty can't pick the car, remember? He can only pick a goat. You are mistakenly looking at two situations in which you pick a car. There are only three scenarios in which you choose to switch:

Pick Door A and switch

Pick Door B and switch

Pick Door C and switch

You're also changing your tune. Earlier you only gave three choices for switching and three for sticking. Now you're giving four for switching? Who is it that's going off the deep end?

It doesn't matter that If you pick Door A, Monty can reveal what's behind either Door B or C- either way you lose. You don't include choosing Door A twice as you can only pick it once. You pick Door A and switch and you lose. Pick Door B or C and switch and win. Switching guarantees a 2/3 probability of winning. Again, if you don't believe it, try the simulator multiple times, record the results and get back to us. But stop coming back and repeating incorrect explanations of how probability works until you've done so.

Or try the simulator once and look at the results recorded of the last 300 or so times it was played. Care to tell us why those who switched won roughly 2/3 of the time? This is not a rhetorical question, btw. These results of this experiment being played out requires an explanation from someone that claims the results should be different, no?

Now what YOU have done is combine the first 2 into 1. Can't do it. They are seperate probabilities. What you CAN do is say this.

IF you pick the car he picks A or B.

IF you pick A or B he picks the Car

Can you do math? You end up with CAR = A+B and A+B = CAR

I can do the math but apparently you can't.

Pick the car and switch = lose.

Pick Goat A and switch = win

Pick Goat B and switch = win

Switching guarantees a 2/3 probability of winning.

Take some classes in probability and see what you come up with.

You need to cut it out with the presumptuous comments. You know nothing about my education, and your calling a moderator on a message board that specializes in brain teasers "Grasshopper", as if you're sure you're qualified to be his Master Po was out of line. I can tell you that bonanova's qualifications in the field of mathematics are rather impressive and judging by your misunderstanding of conditional probability, yours aren't.

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Posted · Report post

Dinghus, instead of jumping all over the place, how about we stick to one scenario and then maybe you can understand where you've gone wrong?

Now let's look at what you said here :

"This is wrong. Since the person doing the eliminating didn't do so randomly, but only eliminated those which are not winners, you have the same probability of having a winning box as you did before he started- 1/1,000,000. The probability that the other box is a winner is 999,999/1,000,000."

Not at all. Totally wrong. Why do you think all the odds pass to the other box? There is nothing that says this anywhere. In reality the boxes ALL keep their odds of being the right box. Thus you come down to 2 boxes both with 1/1,000,000 odds of being the car. I mean I could twist it like you are doing and say that the box you have chosen has a 999,999/1,000,000 chance of being right. It is the exact same flawed logic you are using.

Martini explained why you're wrong already, but lets look at the reasons why again and we can proceed to the original riddle from there.

Okay, there are a million boxes and one contains a winning prize. You apparently think that if you choose one at random, and someone comes along and non-randomly eliminates 999,998 losing boxes, both the box left and your box have an equal chance of being a winner. Wrongo!

Think about it, you almost certainly did not pick out the winning box. This means the person that did the eliminating, almost certainly possesses the winning box after eliminating 999,998 losing boxes.

If given the choice to keep your original box or switch, you wouldn't jump at the chance at switching? You really think both boxes have a 50-50 chance of being the winner?

Lets get to the bottom of this and then we'll move on to the car and goats.

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Posted · Report post

I just want to throw in a different way of explaining the problem:

Let's say there were ten doors

When you pick the first door, you have a 10% chance that you have the prize

Monty's has 90% chance that the prize is still in one of the doors.

He discards all the other doors that were empty, but the door he has left still has a 90% chance of having the prize.

When you are given the choice between two doors, you have a 50% chance of making the wrong choice but the fact of the matter is, your door has a 10% chance of having the price whereas Monty's door has a 90% chance.

This is the reason why you should switch the doors.

This is wrong. Since the person doing the eliminating didn't do so randomly, but only eliminated those which are not winners, you have the same probability of having a winning box as you did before he started- 1/1,000,000. The probability that the other box is a winner is 999,999/1,000,000.

You made the same mistake regarding the riddle. You said:

This is wrong for the same reason as above. Since Monty Barker did not eliminate doors randomly, he only eliminated one door that definitely had a goat behind it, the contestant's probability of having originally chosen the door with a car behind it remains 1/3, it was never 50/50 and never becomes 50/50. The probability of the remaining door having a car behind it is 2/3 and the rationale for this has been explained many times in this thread.

No, there are three possibilities for each of the two scenarios (switching or sticking) as unreality has shown in his last post.

If you choose to switch, these are the three scenarios. Let's say the car is behind A.

You pick A. You switch. You lose.

You pick B. You switch. You win.

You pick C. You switch. You win.

Switching gives you a 2/3 probability of winning.

If you choose to stick, these are the three scenarios. Let's say the car is behind A.

You pick A. You stick. You win.

You pick B. You stick. You lose.

You pick C. You stick. You lose.

Sticking gives you a 1/3 probability of winning and the fact that Monty decides to show you where a goat is after you pick a door doesn't change this.

Since you admitted to not reading the entire thread, maybe you missed this simulator where you can experiment with switching and tallying what percentage of the time you win a car.

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Posted · Report post

Why, Oh why, has this discussion extended to 16 pages?

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Why, Oh why, has this discussion extended to 16 pages?

Just set the posts/pg to 40 and it's only 4 pages. Problem solved!

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Posted · Report post

Just set the posts/pg to 40 and it's only 4 pages. Problem solved!

An amusing answer, ljb, but not what I meant regarding all the silliness. I had to correct an earlier post in which I inadvertently referred to you as "lbj" which has a different connotation.

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Just want to say Bravo to Martini for persevering on this and doing so in such a respectful and rational manner. I'm amazed you managed to do so! I read every entry in this thread. What a marathon. Glad to have come across conditional probability. People, the answer is she's better to switch, read the thread, read the authoritative sources on the web, and listen to Martini!

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Posted · Report post

Don't like to beat a dead goat, but I can't resist.

I will simplify this (I didn't read all the threads, not enough time, so maybe this is already in there.)

If you pick the car, then switch, you get the second goat and lose.

But if you pick a goat, the host shows you the other goat, so if you change your pick, you win the car.

Therefore, if you change your pick, you will win the car anytime you select a goat first. Since you select a goat 2/3 of the time, you will win the car 2/3 of the by changing your pick.

Q.E.D.

B.T.W., I love this puzzle. I think it is one of the all-time classics, and a great illustration of how statistics is not intuitive in most people.

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Posted · Report post

why is there so much talk of probability??!! Its making my head spin. Anyway, here is my reasoning and am not using probability here!

[Door 1] has goat

[Door 2] Picked by jennifer

[Door 3]

Jennifer should definitely switch since if the car is behind 2, she will get a dinner. But if the car is behind 3, she gets car wax. Now why would someone give you a car wax unless and until you have a car! the car is in 3.

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If Jennifer follows the strategy of always switching, she will win the Jag about 2/3 of the time.

Here is what will happen if she sticks with the "always switch" strategy:

About 1/3 of the time, unfortunately, Jennifer's first door will be the Jag door. She will lose, because

when Monty points out one of the goats, she will switch and select the door with the other goat.

About 2/3 of the time, fortunately, Jennifer's first door will be one of the goat doors. She will win, because

when Monty points out the other goat, she will switch and select the door with the Jag.

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It's 50/50. The problem in your logic is, after he shows a goat, your still figuring in the door that we now know has goat as a possibility, she can no longer choose that door which brings the possible selection down to two doors and since she can only choose one, that means she is choosing one door out of two possible doors, meaning a one in two chance... It can ONLY be a two out of three chance if all three doors are unknown and she can choose two of them. I'm not trying to be a troll here, but the level of stupidity required for adults not to understand 7th grade math is astounding.

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It's 50/50. The problem in your logic is, after he shows a goat, your still figuring in the door that we now know has goat as a possibility, she can no longer choose that door which brings the possible selection down to two doors and since she can only choose one, that means she is choosing one door out of two possible doors, meaning a one in two chance... It can ONLY be a two out of three chance if all three doors are unknown and she can choose two of them. I'm not trying to be a troll here, but the level of stupidity required for adults not to understand 7th grade math is astounding.

You have just proven that it is you that doesn't understand 7th grade math, and have thus been labeled a "troll."

To sum it up for you, you originally have a 1/3 chance of picking a winner, which means you have a two thirds chance of picking a loser. Odds are you picked a loser. If the host shows you the OTHER loser, then that means that it is a good bet that the third door is actually the winner, and you should switch. You would be a fool, or a troll, if you did not.

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Posted · Report post

I have spent a lot of time thing about this problem and I have read and fully understood the answer from many sources.

So let me put it to you like this, unless you are the smartest mathematician to ever ponder this question then there is no way that you are correct in saying that it is 50/50. Since mathematicians have actually assessed this problem and concluded that the answer is that you should switch and win 2/3 times. If you do not believe me then look it up or look at one of the many links already provided. Or read and understand what it is that the rest of these people are actually saying since you are obviously not.

Good day sir.

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Posted · Report post

I've been struggling with this question forever, but I've finally (and grudgingly) come to realize that the probability really does change if you switch your choice of doors. The number of doors doesn't matter, but the more there are, the more evident the logic is. So, based on a model of 100 doors:

1. You choose a door. you have a 1% chance of choosing the correct door.

2. All doors except the one you chose and one other are revealed. There are now two possibilities:

A - You picked the correct door the first time, at a 1% probability

B - You didn't pick the correct door the first time, at a 99% probability.

The only way that the other door can be the wrong door is if you picked the correct door the first time. The trick to this puzzle is that the options of the second choice are based on the decisions of the first choice - they are not independent events.

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you know...you can sit here and write out all the math you want, but in the end if there's 3 doors and you already know 1 to have the goat, then one door has a goat and 1 door has a car. That's a 50/50 chance. If there were 20 doors and I took away 18, you'd still be left with 2 and 1 would be the winning door. It's always going to be 50/50. It's pure logic. Mathematics may not agree with logic, but logic doesn't care. Logic is as logic does.

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Posted · Report post

you know...you can sit here and write out all the math you want, but in the end if there's 3 doors and you already know 1 to have the goat, then one door has a goat and 1 door has a car. That's a 50/50 chance. If there were 20 doors and I took away 18, you'd still be left with 2 and 1 would be the winning door. It's always going to be 50/50. It's pure logic. Mathematics may not agree with logic, but logic doesn't care. Logic is as logic does.

tonym144, you might want to read other posts in this thread and some more information on the web. There are links in this thread to some other resources related to this topic if you didn't find arguments in this thread convincing. Or you can just google "monty hall problem".

Your mistake is that you are thinking that if there are only 2 possible outcomes then they must have equal probability. That's not logical at all. If you buy a lottery ticket there are 2 outcomes - you win or you don't win. The probability is obviously not 50/50.

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you know...you can sit here and write out all the math you want, but in the end if there's 3 doors and you already know 1 to have the goat, then one door has a goat and 1 door has a car. That's a 50/50 chance. If there were 20 doors and I took away 18, you'd still be left with 2 and 1 would be the winning door. It's always going to be 50/50. It's pure logic. Mathematics may not agree with logic, but logic doesn't care. Logic is as logic does.

tonym144, it's a fun puzzle and one that's debated a lot because we like to follow our intuition instead of doing all that messy math.

Here's an intuitive path to the other answer.']

When you pick a door at the outset, that door gives you a 1/3 chance of winning.

Ask yourself how that probability could change by being shown a goat.

Remember Monty knows what's behind the other doors - he's not guessing about what he shows you.

And if your door has a 1/3 chance, the other 2/3 chance, instead of being split between the other two doors, now has to totally be with the third door, doesn't it?

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Posted · Report post

I will still stick to #2.

Earlier i had a blind guess of winning a car prob:1/3

Now i have prob:1/2

My hunch tell me however:: Had i selected a door with goat on first choice, the host would have happily opened it to let all know, the game would end there. They were always more interested in saving a jaguar for a goat.

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I will still stick to #2.

Earlier i had a blind guess of winning a car prob:1/3

Now i have prob:1/2

My hunch tell me however:: Had i selected a door with goat on first choice, the host would have happily opened it to let all know, the game would end there. They were always more interested in saving a jaguar for a goat.

Why did 1/3 become 1/2?

Remember Monty did not guess which door had a goat. He knew.

And he didn't open the door you chose, he showed a different goat.

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Sorry, I am not sure if the rules were changed in later posts, but i read it in every first post as

" .....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......"

so in my "second round", i only have to choose between door #2, #3 or stick with earlier selection (door#2).

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Sorry, I am not sure if the rules were changed in later posts, but i read it in every first post as

" .....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......"

so in my "second round", i only have to choose between door #2, #3 or stick with earlier selection (door#2).

You're fine on everything you've said. Three options have now become two options: Door 2 [stay] and Door 3 [switch].

Initially all the doors have 1/3 probability of having the car.

After we learn that Door 1 has 0 probability, where does that 1/3 go?

You assume it gets split equally between 2 and 3, making them both 1/2.

But why should Door 2 change, just because we are shown a goat behind one of the other two doors?

We know Doors 1 and 3 combined have 2/3. If we learn Door 1 has 0, doesn't that mean Door 3 now is 2/3?

Or,

If Door 2 has the car, switching to Door 3 loses.

If Door 2 has a goat, switching to Door 3 wins.

Are these cases equally likely?

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Posted · Report post

We know Doors 1 and 3 combined have 2/3. If we learn Door 1 has 0, doesn't that mean Door 3 now is 2/3?

No. Door#1 earlier had 1/3, now that it is open, it is out of equation. Only 2 doors are in race now and we have to re-assign prob. of 1/2, 1/2 for #2,#3. Both the cases you mentioned are equally likely (hence 1/2, 1/2) to happen.

Few may think that changing to #3 is better option but there is no place for finding prior prob. in this case (as #1 is removed out of equation instead of just having its prob changed).

You can read my older comment on why i still wish to stick with #2, even though it has only 1/2 of winning ..

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Posted (edited) · Report post

you know...you can sit here and write out all the math you want, but in the end if there's 3 doors and you already know 1 to have the goat, then one door has a goat and 1 door has a car. That's a 50/50 chance. If there were 20 doors and I took away 18, you'd still be left with 2 and 1 would be the winning door. It's always going to be 50/50. It's pure logic. Mathematics may not agree with logic, but logic doesn't care. Logic is as logic does.

Funny...all this time I thoughty math was logic...hmmm

Math may not agree with intuition, or what is believed to be common sense...but Math and Logic are attached at the hip

Edited by maurice
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