## A riddle

slipstick
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### A riddle

This is an old riddle and many of you have probably heard it before, but it seems to cause a lot of confusion for those hearing it for the first time. I read an article a couple of days ago which reminded me of this and so I will pose the riddle here to see how many people come up with right or wrong answers.

You are on a game show. In front of you there is a wall with three closed doors. Behind one of the doors is a new automobile; behind each of the other two doors is a live goat. You may choose to open any one of the doors and will win whatever is behind that door as a prize. After you make your selection, and before opening the chosen door, the host of the show, who knows what is behind each door, will open one of the doors which you did not select to reveal a goat. The host then asks you if you wish to change your decision and choose the other closed door. The question is this: should you change your decision? That is, can you improve your odds of winning the new car by choosing the other door instead of the one you originally chose?

For the present, please just post your advice to change, or not change without explanation. Then after 12:01 AM (00:01) UCT on 29 Dec., 2016 we can open this up for discussion and argument.
In theory, theory and practice are the same. In practice, they ain't.

Schultz
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### Re: A riddle

No, you cannot improve the odds.

Tech Freak
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### Re: A riddle

slipstick wrote:..........The host then asks you if you wish to change your decision and choose the other closed door. The question is this: should you change your decision?........
3 doors to choose from means you have 33% chance of choosing the correct door, right? Eliminating 1 door improves your changes from 33% to 50% assuming that the jackpot is still behind 1 of the 2 doors, right?

Why will the host ask you if you have any desire to change your decision? Simple, creating confusion. The change that you choose the correct door has increased and by creating confusion you overthink your decision trying to let you make the wrong decision (choosing the door which doesn't have a price behind it).

That's what i think.

deepakdeshp
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### Re: A riddle

Changing the choice doesnt matter I feel.
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chrisuk
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### Re: A riddle

Switching doors increases your chances of finding the car. It's a modern day version of an old (1800s) riddle/paradox
Chris

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Tech Freak
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### Re: A riddle

deepakdeshp wrote:Changing the choice doesnt matter I feel.
Yes it does change the matter. It's basically mathematics. Calculating the probabilities.
It is called the Month Hall problem, see here: https://en.wikipedia.org/wiki/Monty_Hall_problem

sphyrth
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### Re: A riddle

Yes, it's the Monty Hall Problem. It DOES feel counter-intuitive, but switching doors would increase your chances of getting that car from 33% to 66%.

The opening of the 2nd goat is what makes your chances better. Since you had 2 out of 3 chances of getting the a goat on your first pick, then that means you get to have 2 out of 3 chances switching to the car (since the second goat will ALWAYS be revealed before you are asked to make your second decision).
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Fred Barclay
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### Re: A riddle

Yep, you get a higher chance of getting the car if you switch (as counter-intuitive as it does feel).

If you change doors, you'll (theoretically) get the car 2/3 of the time, while if you stay with the door you'll only get the car 1 out of every 3 times.

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Tech Freak
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### Re: A riddle

@ Fred Barclay:

What about the influence of the show host? That person is trying to confuse you at the moment you made your decision. The diagram you have shown is only based on a theoretical calculation and only shows the theoretical possibilities. Nowhere in this possibilities the influence of the show host is taken in account.

Let's say you have choosen the correct door with the car begind it, the host knows behind which door the car is and is now trying to change your decision by asking you if you don't want to switch, are you sure you have choosen the correct door, even (and they do) let the audience put in their voice (people in the audience, which door would you have choosen)..... What influence will this have in your choice and winning the car?

greerd
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### Re: A riddle

As soon as Monty asks if you want to switch, its a new game with only two doors and a 50/50 chance.

sphyrth
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### Re: A riddle

Ahh.. The psychological aspect of the game. Of course, since switching doors only FEELS to be irrelevant, the host will surely put the player in a state of mental block. But it doesn't really change the fact that switching doors still increases your chances of winning the car.
You have only 1/3 chance of doing that if you stay after all.
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Schultz
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### Re: A riddle

I can now see how the odds do increase by switching doors thanks to Fred Barclay's diagram (specifically the bottom row). On the other hand, there's still an equal amount of "goats" and "cars" on that bottom row, so in a sense it's still 50/50.

chrisuk
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### Re: A riddle

This is probably the original riddle/paradox (from the 1800s)... maybe makes it clearer.
Chris

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Fred Barclay
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### Re: A riddle

Schultz wrote:I can now see how the odds do increase by switching doors thanks to Fred Barclay's diagram (specifically the bottom row). On the other hand, there's still an equal amount of "goats" and "cars" on that bottom row, so in a sense it's still 50/50.

Well, there are two options so there is a 1/2 chance in that regard (you will get either a goat or a car), but since the initial conditions are variable it ends up being (1/3)/(1/2) = (1/3)*2 = 2/3

Or something like that...

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slipstick
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### Re: A riddle

Schultz wrote:I can now see how the odds do increase by switching doors thanks to Fred Barclay's diagram (specifically the bottom row). On the other hand, there's still an equal amount of "goats" and "cars" on that bottom row, so in a sense it's still 50/50.

Look at it this way - in 2 out of three cases, your initial choice forced the host to choose a particular door to open. If you chose a goat, he had to choose the door with the other goat. So by changing your choice you are using the information that you have about your original choice. On the other hand, if you regard it as a 50/50 proposition because there are only two closed doors, then you are throwing away information - you are "forgetting" which door you originally chose. This makes it a different problem - by throwing away information, your odds of winning decrease to 1 out of 2.

If you simply count the number of goat boxes and car boxes on the bottom row of Fred's diagram, you are implicitly assuming that you choose one of the boxes at random. But if you use the information you have at hand (the rules of the game and your initial choice), then you can make an informed choice and improve your odds.
In theory, theory and practice are the same. In practice, they ain't.

Neil Edmond
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### Re: A riddle

After one option has been revealed, if you stick with first choice, you have a 1 out of 2 chance of winning. If you change your choice, you have a 1 out of 2 chance of winning. The odds are the same. The information gained from the first reveal might affect your decision to stick or change, but it doesn't change the odds.

slipstick
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### Re: A riddle

Neil Edmond wrote:After one option has been revealed, if you stick with first choice, you have a 1 out of 2 chance of winning. If you change your choice, you have a 1 out of 2 chance of winning. The odds are the same. The information gained from the first reveal might affect your decision to stick or change, but it doesn't change the odds.
No - 1 in 3 chance to win if you stick, 2 in 3 if you change. If you change, you are using the information from your first choice combined with the rules of the game - in 2 of 3 cases (if you initially choose a goat) the show host is forced to reveal the other goat and the car is behind the closed door which has not been chosen, so you win if you change - in 1 of three cases (you initially choose the car) the host reveals one goat and you lose if you change to the other goat (unless it's a better goat ).
In theory, theory and practice are the same. In practice, they ain't.

Schultz
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### Re: A riddle

Neil Edmond wrote:
After one option has been revealed, if you stick with first choice, you have a 1 out of 2 chance of winning.
Yes and no (really no). When the choice was made, you had a 1 out of 3 chance of winning. Those odds really don't change when one goat is revealed. Whether you picked the car or one of the goats, there will always be a goat to be revealed since there's two of them. So revealing a goat really doesn't change the original odds.

EDIT: slipstick, you beat me to it.

Neil Edmond
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### Re: A riddle

Schultz wrote:...revealing a goat really doesn't change the original odds.
Well...even though you have a choice between three options at first, there never was a 1 in 3 chance of winning. That's not the way the game is played. The game is the decision to stick or change once the initial reveal has been made. Now the odds are the same either way you go. It's a 50/50 chance, at that point, either way.

slipstick
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### Re: A riddle

Neil Edmond wrote:
Schultz wrote:...revealing a goat really doesn't change the original odds.
Well...even though you have a choice between three options at first, there never was a 1 in 3 chance of winning. That's not the way the game is played. The game is the decision to stick or change once the initial reveal has been made. Now the odds are the same either way you go. It's a 50/50 chance, at that point, either way.
The rules of the game force the host to open one particular door if your initial choice is a goat. You use these rules and the memory of your initial choice to make your odds of winning 2 out of 3. If you consider the odds to be 50/50, then you are playing a different game - either one where you didn't make an initial choice and the host randomly picked which goat door to open (a game with different rules), or one with the same rules; that is, that your initial choice (if a goat) forced the host to open a particular door, but you then "forgot" which choice you made initially. Either one of these changes does result in 50/50 odds.
In theory, theory and practice are the same. In practice, they ain't.