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Re: A riddle

Posted: Fri Dec 30, 2016 8:19 pm
by Neil Edmond
It doesn't make any difference whether the initial choice was a goat door or the car door. The host will reveal a goat door after your initial choice, either way. Now all that's left is a 50/50 chance with either decision.

Re: A riddle

Posted: Fri Dec 30, 2016 9:34 pm
by slipstick
Neil Edmond wrote:It doesn't make any difference whether the initial choice was a goat door or the car door. The host will reveal a goat door after your initial choice, either way. Now all that's left is a 50/50 chance with either decision.
If your initial choice was a goat, and the host has to reveal the other goat, then changing your choice is a sure winner. If your initial choice was the car, and the host reveals a goat, then changing is a sure loser. When faced with the decision to change or not, you don't know where is the car and where the goat, but you do know that the odds are 2 out of 3 that your initial choice was a goat and so you can win by changing 2 out of 3 times.

It's like a game of dice - just because the sum on two dice can come up anywhere between 2 and 12, if you bet as though the probability of a 2 is the same as that of rolling a 7, pretty soon you will lose.

Re: A riddle

Posted: Fri Dec 30, 2016 9:56 pm
by all41
slipstick wrote:
Neil Edmond wrote:It doesn't make any difference whether the initial choice was a goat door or the car door. The host will reveal a goat door after your initial choice, either way. Now all that's left is a 50/50 chance with either decision.
If your initial choice was a goat, and the host has to reveal the other goat, then changing your choice is a sure winner. If your initial choice was the car, and the host reveals a goat, then changing is a sure loser. When faced with the decision to change or not, you don't know where is the car and where the goat, but you do know that the odds are 2 out of 3 that your initial choice was a goat and so you can win by changing 2 out of 3 times.

It's like a game of dice - just because the sum on two dice can come up anywhere between 2 and 12, if you bet as though the probability of a 2 is the same as that of rolling a 7, pretty soon you will lose.
Einstein's riddle
The situation

There are 5 houses in five different colors.
In each house lives a person with a different nationality.
These five owners drink a certain type of beverage, smoke a certain brand of cigar and keep a certain pet.
No owners have the same pet, smoke the same brand of cigar or drink the same beverage.

The question is: Who owns the fish?

for clues and solution see here:
https://udel.edu/~os/riddle.html

Re: A riddle

Posted: Fri Dec 30, 2016 11:56 pm
by Penn
What some of you are missing is, in the second choice there is no 3. 1/3 or 2/3 isn't relevant AT ALL.

Once one door with a goat is gone you have a NEW choice with new odds and the the first choice is COMPLETELY irrelevant. What are the possible outcomes of your current choice. One choice is "correct"* and the other is "wrong"* and there is only 2 choices, not 3. If you must bring the first choice into the equation then there is a 50% chance you chose the car the first time and a 50% chance you chose the OTHER goat.
chrisuk wrote:This is probably the original riddle/paradox (from the 1800s)... maybe makes it clearer.
Actually that one is completely different and more interesting because all 3 doors (actually chests in that situation) are still in play for the second choice. I have my view of it but I can see how a person could make a different argument.

My view is there is the first decision, keep it or move on. Keeping it is 50% chance it is both gold, 50% chance it is mixed and 0% chance of just silver. If you choose to keep it then there isn't a second decision with different odds. If you go with changing you have new odds. One of the 2 options has a definite 2 silver and the other either has 2 gold or mixed. So the second choice the odds are 50% silver and the other 50% is split between possibly just gold or mixed so 25% gold and 25% chance it is mixed.

* Correct and wrong are subjective. A person living in a third would place with no access to gas is going to get more use out of a goat.

Re: A riddle

Posted: Sat Dec 31, 2016 12:00 am
by Fred Barclay
all41 wrote: The question is: Who owns the fish?

for clues and solution see here:
https://udel.edu/~os/riddle.html
Just solved this... whew, not easy! :)
I'll PM you my answer as to not mess up the fun for the others.

Re: A riddle

Posted: Sat Dec 31, 2016 12:07 am
by all41
Fred Barclay wrote:
all41 wrote: The question is: Who owns the fish?

for clues and solution see here:
https://udel.edu/~os/riddle.html
Just solved this... whew, not easy! :)
I'll PM you my answer as to not mess up the fun for the others.
No Fred--it is not Professor Plum

Re: A riddle

Posted: Sat Dec 31, 2016 12:39 am
by sphyrth
Um... the German?

Re: A riddle

Posted: Sat Dec 31, 2016 12:49 am
by Schultz
Penn wrote:
What some of you are missing is, in the second choice there is no 3. 1/3 or 2/3 isn't relevant AT ALL.
But when you made that first choice, your odds of picking a goat was 2 out of 3, so it doesn't matter that a door is opened to reveal a goat, because whether you picked a car or a goat, there is always (at least) one goat left out. So the point is this: that you made your choice before a door was opened. And your odds were 2 out of 3 (higher than 50%) of picking a goat. Revealing a goat behind one of the unchosen doors doesn't change the odds you had when you made your pick . So by switching doors you are theoretically increasing your odds of picking the car. Take a look at Fred Barclay's diagram, noting the purple boxes on the bottom row. 2 out 3 are winners.

Re: A riddle

Posted: Sat Dec 31, 2016 4:46 pm
by mintage
Neil Edmund gets my vote.

Re: A riddle

Posted: Sat Dec 31, 2016 5:51 pm
by Schultz
I don't know how you can look at the purple boxes in the bottom row of Fred Barclay's diagram and still come to that conclusion.

[ I initially agreed with you guys, until I looked at that diagram. ]

Re: A riddle

Posted: Sat Dec 31, 2016 6:07 pm
by mintage
Whatever you pick - somebody is going to show you a goat. The first choice is already made, it is either a car or not. the fact that someone shows you a goat has no bearing on the fact that you either select a car or a goat. Changing your mind does not improve your odds, the second goat is just there so it can be taken away. Taking it away is irrelevant - you either picked a car or a goat. The second goat doesn't factor in. It's like a three sided coin and they say "now that the coin is at rest and you do not know which is up, we are going to take away the duplicate side, would you like to change your mind as to which of the two states the coin rests?". The third side is 100% irrelevant. What about the fish?

Re: A riddle

Posted: Sat Dec 31, 2016 6:17 pm
by Schultz
A door was chosen before any other door was opened. The odds were 1 out 3 that the car was chosen. Opening one of the goat doors doesn't magically change the odds that were there before that door was opened. The odds are still only 1 out 3 that the car was chosen (i.e., if you don't change doors). The green boxes in the bottom row of the diagram proves that.

Re: A riddle

Posted: Sat Dec 31, 2016 6:23 pm
by mintage
You are correct. It doesn't magically change the odds.

Re: A riddle

Posted: Sat Dec 31, 2016 6:24 pm
by Neil Edmond
mintage: Yep.

Shultz: Fred's chart is only applicable if the decision to stick or change is made while 3 doors are still in play. Once the first is revealed the chart is no good anymore.

Re: A riddle

Posted: Sat Dec 31, 2016 7:01 pm
by Schultz
Neil Edmond wrote:
Fred's chart is only applicable if the decision to stick or change is made while 3 doors are still in play. Once the first is revealed the chart is no good anymore.
That is a good point, but I'm not ready (yet) to go back to my original thoughts on this.

Re: A riddle

Posted: Sat Dec 31, 2016 8:20 pm
by sphyrth
The question is the difference between SWICTHING and STAYING.
You'll SWITCHING into the car 2/3 times.
You'll be STAYING with a goat 2/3 times.

People easily miss the how opening of the other goat helps in increasing the odds by a lot.

Re: A riddle

Posted: Sun Jan 01, 2017 12:48 am
by Portreve
Um...

Image
There is no spoon...

Re: A riddle

Posted: Sun Jan 01, 2017 2:09 pm
by Penn
Schultz wrote:I don't know how you can look at the purple boxes in the bottom row of Fred Barclay's diagram and still come to that conclusion.

[ I initially agreed with you guys, until I looked at that diagram. ]
Fine, you want pictures.

It starts with one alteration, I labeled the doors 1, 2 and 3.

Image

So if you choose door 1, you will be shown door 2 and that option is removed. You can't change your mind AWAY from door 3 since you didn't choose that. So your CURRENT possibilities are

Image

Not a thing erased affects your CURRENT choice and the resulting odds. That stuff is GONE.

This same logic is applicable if your first choice is 2 or 3.

Re: A riddle

Posted: Sun Jan 01, 2017 2:23 pm
by mintage
Good Luck Penn, you're a real soldier. :)

Re: A riddle

Posted: Sun Jan 01, 2017 3:42 pm
by Schultz
The simple solutions above show that a player with a strategy of switching wins the car with overall probability 2/3, i.e., without taking account of which door was opened by the host.
https://en.wikipedia.org/wiki/Monty_Hall_problem
"Marilyn vos Savant is an American who is known for having the highest recorded IQ according to the Guinness Book of Records, a competitive category the publication has since retired." https://en.wikipedia.org/wiki/Marilyn_vos_Savant

She says by switching your odds are now 2 out 3 of winning the car. Is your IQ higher than hers?