This is acquired from putting the property 'when differentiated, it remains the same' into a maclaurin series.
It's pretty clear, from this method, that there are no similar numbers.
Even though you're not him, yep.
Now,
combine these two facts!
Probability of getting wrong door = 2/3
Probability of winning if you change with the wrong door = 1
Probability of winning if you change with the right door = 0, probability of picking right door = 1/3
Chance of winning if you change = chance of choosing wrong door * chance of winning with wrong door + chance of choosing right door * chance of winning with right door
= 2/3 * 1 + 1/3 * 0
= 2/3
This has been demonstrated SO MANY WAYS. It's TRUE.
Yep, that's the basic idea. Draw out the probability tree of each possible choice of door, assuming a given one is the winning door (rather than doing the tree for all possibilities of winning door; it just overcomplicates things, as the doors are all identical for all intents and purposes, and thus interchangeable).
To be fair to him' date=' you two have got the magical I CAN'T READ attribute.
'Polish troops find Sarin gas shells' : July 1st 2004.
All of ATM's links: January, 2004.
Now, either the Beeb are prescient, or they're talking about [b']two different events![/b]
Woah!
This isn't to say that it was right to invade, just that your counterargument is factually inaccurate.
You have a 50% chance of picking the right door if you have no other information.
The main problem here is that you're interpreting it (which is the standard misinterpretation) as 'choosing a door'. You're not, you're choosing which probability subgroup you want to be a part of.
Yep. You appear to have missed that there's twice the likelihood of choosing the wrong door in the first place, though.
Lets try another tack. We've tried irrevocable proof, and you've ignored it.
How about this:
You have a 2/3 chance of choosing a 'losing' door.
If you change with a losing door, you ALWAYS win.
Therefore, if you change, you have a 2/3 chance of winning.
Or this:
If you decide to stay, the only way to win is by picking the winning door from the off. Therefore, the chance of winning by staying is 1/3. As the sum of the chances of the winning will be 1 (dur), the chance of winning by changing is 1 - 1/3, or 2/3.
Lets say that they remove the prize entirely! Then, change or not, you always lose! This is obviously a valid counterargument and I'm not just arbitrarily adding random irrelevencies to the argument at hand!
I'd say that controlling most of the country counts as 'winning'.
[edit]
That's the taliban and the warlords; they're not the same group, but I used the word 'taliban' as shorthand, as it didn't really matter.
Because it's true. Look at the post I quoted; there's no advanced concept there, it just lists every possible outcome, and the chance of you winning if you change is, indeed, twice that of you not.
As I said before, Atheism is the only really scientifically valid viewpoint.
Yes, science doesn't preclude the existance of god, but science doesn't preclude the existance of ANYTHING, so that's hardly a useful argument.
You can't patent pure mathematical algorithms, as far as I know.
And you haven't got this 'proof' thing down, have you. You have to prove it for all possible numbers, not just 1,000, 10,000 or whatever.
Because of the nature of infinity, with number crunching, you're not actually decreasing the possibility of it being wrong, you know.
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