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Whatchya Gonna Do (2 goats and a car)


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196 replies to this topic

#151 Dinghus

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Posted 21 October 2008 - 10:45 PM

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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#152 bonanova

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Posted 21 October 2008 - 11:30 PM

Take your pills.
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The greatest challenge to any thinker is stating the problem in a way that will allow a solution.
- Bertrand Russell

#153 Martini

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Posted 22 October 2008 - 12:21 AM

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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#154 Kent

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Posted 22 October 2008 - 01:03 AM

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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#155 charlybaby

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Posted 02 October 2009 - 08:22 PM

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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#156 jerbil

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Posted 03 October 2009 - 12:50 AM

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

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Posted 03 October 2009 - 03:22 AM

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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#158 jerbil

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Posted 03 October 2009 - 09:33 PM

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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#159 jordan147

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Posted 18 October 2009 - 03:27 AM

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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#160 SPUD

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Posted 30 October 2009 - 05:07 PM

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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