Mudbill
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RE: A puzzle
(09182015, 05:13 PM)FlawlessHappiness Wrote: A guy from reddit said this
Guy from reddit Wrote:Chooses Door 1, host opens Door 2, you switch. You lose.
Chooses Door 2, host opens Door 3, you switch. You win.
Chooses Door 3, host opens Door 2, you switch. You win.
Yes, compare this. One of the doors can be excluded from the equation as it's always removed, leaving us with 50/50.
(This post was last modified: 09182015, 08:04 PM by Mudbill.)


09182015, 08:03 PM 

Googolplex
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RE: A puzzle
(09172015, 11:10 PM)Mugbill Wrote: You have good handwriting, I'll give you that. However, I completely understand the situation so the explaination isn't necessary. I just don't agree with it. Supposedly it IS a paradox, but I just think people are overthinking and overcomplicating it. The end result is that there are two options, so the first pick can be disregarded from the conclusion.
1 / 3 = 0.33 = 33% = chance of getting right pick first round (but you have no idea whether you got it right)
2  1 = 1 = remove one of the wrong picks to remain with 1 right and 1 wrong.
1 / 2 = 0.5 = 50% = chance of getting right pick second round.
Thanks for my handwriting. But you think wrong. The first pick will not be affected by removing one door. Here is an other example: In a lottery you have a chance of 1/140.000.000 to win the jackpot. It's like you say one number between 1 and 140.000.000. For example you take the 10. Your chance is 1/140.000.000. Now I will remove all wrong numbers except the 11. Now there are only two possible numbers remaining. Do you really believe you win the jackpot with your first pick? Wouldn't it make sense to switch to the 11? Your chance isn't 50/50 now. It's still 1/140.000.000, but when you switch you will have a chance of 1/139.999.999. That's the logic behind this puzzle.
(09182015, 05:13 PM)FlawlessHappiness Wrote: But it makes no sense to me.
If the host always reveals a wrong door before the final choice, then that door shouldn't even exist.
If there are 3 doors, but my final choice is between 2 of them, why does the third exist, when it has nothing to do with my choice?
The percentages that is shown in the video are from when there are 3 doors to pick from. When there are 3 doors, there are 3 different results. That means the 66% of winning the car, are 66% of 3 different results.
But Mugbill and I don't see 3 doors. We see 2 doors. With 2 doors there are 2 results. That gives us 50% chance of 2 results.
Taking your picture into account
End result
You are only choosing between 2 doors. One with a prize and one without. The last wrong choice is nonexistant.
Id does no matter, if a door get removed or not. When you pick a door, you know that one of the others must be wrong. So what does the showmaster do? He just remove that wrong door. Your chance is still 1/3.
My sketch is correct. You couldn't summarize these two choices. These are not the same. Either you pick the first door, the second or the third. The sketch shows what happens then.
(09182015, 08:03 PM)Mugbill Wrote: Yes, compare this. One of the doors can be excluded from the equation as it's always removed, leaving us with 50/50. No, 50/50 is wrong. The correct answer is 1/3 and 2/3. You would have 50/50 when you would mix the two remaining doors and pick again.
(This post was last modified: 09192015, 07:28 PM by Googolplex.)


09192015, 07:23 PM 

FlawlessHappiness
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RE: A puzzle
I have no doubt in your drawing. The math behind it is 100% correct.
What I don't like about this problem is the wording of the solution.
(09192015, 07:23 PM)Googolplex Wrote: He just remove that wrong door. Your chance is still 1/3.
How can the chance still be 1/3 if there are only 2 doors?
(09192015, 07:23 PM)Googolplex Wrote: You would have 50/50 when you would mix the two remaining doors and pick again.
Which is the end result. No matter whether one door was removed or not, you are "mixing" the two doors and picking again.
"Mixing" here is by not mixing them. You still don't know what's behind the doors, so whether they're mixed or not won't matter.
Trying is the first step to success.


09192015, 07:31 PM 

Mudbill
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RE: A puzzle
I can see how in the lottery scenario you'd have to be incredibly unlucky to pick the right option first, seeing as it's a 1/140m. It would make sense to swap. But this entire thing is a probability paradox, and is hard to actually apply to real life. Regardless, I do see where people are trying to go with it and it does make sense, I just don't like it.
I think I will just submit defeat. I understand why it would be a better chance, but effectively you're left with 2 options and 2 results. Oh well. I grow tired of discussing this one. Guess that's what "paradoxes" do to you.


09192015, 08:14 PM 

Romulator
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RE: A puzzle
(06082014, 10:00 PM)Nice Wrote: hmm that's interesting but I have a different theory
It's called "dogfoodolians theory of infitism"
basically imagine that [those doors] represents "X"
okay so [chance] is "1"
Now lets look how it works out it math
5+5x 1 + 1 = 0
5 + 5x = 0
5x = 5 /:5
x = 1
[therefore there is no solution]
Discord: Romulator#0001
(This post was last modified: 09202015, 11:03 AM by Romulator.)


09202015, 11:02 AM 

Googolplex
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RE: A puzzle
(09192015, 08:14 PM)Mugbill Wrote: I can see how in the lottery scenario you'd have to be incredibly unlucky to pick the right option first, seeing as it's a 1/140m. It would make sense to swap. But this entire thing is a probability paradox, and is hard to actually apply to real life. Regardless, I do see where people are trying to go with it and it does make sense, I just don't like it.
I think I will just submit defeat. I understand why it would be a better chance, but effectively you're left with 2 options and 2 results. Oh well. I grow tired of discussing this one. Guess that's what "paradoxes" do to you.
You can make the test. Take 3 cups with one coin under one of the cups. Let an other person pick a cup, say him which cup is wrong, then look at the results when he stays or switch.


09202015, 02:04 PM 

