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Whatchya Gonna Do (2 goats and a car)


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196 replies to this topic

#31 Gambler

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Posted 06 February 2008 - 07:25 AM

... 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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#32 Martini

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Posted 06 February 2008 - 07:26 AM

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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#33 Gambler

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Posted 06 February 2008 - 07:37 AM

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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#34 Jkyle1980

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Posted 06 February 2008 - 07:55 AM

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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#35 Jkyle1980

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Posted 06 February 2008 - 08:17 AM

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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#36 Martini

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Posted 06 February 2008 - 08:23 AM

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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#37 Spyderzden

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Posted 06 February 2008 - 09:02 AM

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, 06 February 2008 - 09:04 AM.

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#38 vime

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Posted 06 February 2008 - 02:29 PM

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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#39 bonanova

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Posted 06 February 2008 - 02:31 PM

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

#40 Maximus

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Posted 06 February 2008 - 03:07 PM

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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