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As a question of pure probability, they are all equally likely; the list has just been narrowed from 100 numbers to 4, giving you a 25% chance of guessing right instead of 1%.
There was a question I have heard of the same structure that led to a paradox, and so actually is impossible. It goes like this:
Q. If you choose an answer to this question at random, what is the probability that you will guess correctly?
A. 25%
B. 50%
C. 25%
D. 0%
Hello Solvitur,
Which part do you think is unresolved? For your example, it is evident that Achilles can never break his word so long as it is possible for him to perform the task; therefore his promise is almost meaningless (it is equivalent to a statement that he will always be capable of performing the task once).
The advantage of understanding what you earlier called Conclusion 2 is that these kinds of consequences are associated with infinity, usually without confusion (:
2 + 9 + 7 - 8 + 3
How do you define knowledge? P:
2 + 9 - 7 + 1 + 8
Edit: 2 * 9 - 7 + 3 - 1
Hi zetafunc, no the triangle is acute, but the distance formula essentially uses Pythagoras by creating an artificial right triangle with the desired distance as the hypotenuse.
If I calculate the distance between (-5,-1) and (5,2) it is sqrt109 ~ 10.44, I will call that the base. I then note that the midpoint is (0,1/2), I calculate the distance between that and the third point (3,-5) and I get 1/2 * sqrt157 ~ 6.265, I call that the height.
1/2 * base * height gives 1/4 * sqrt17113 ~ 32.7
But a specific formula for these things gives exactly 32, and so does Heron's formula. Maybe I have not defined the height properly or something...
I don't know about your answer
Edit: Oh I am a twit sometimes. Of course I didn't define the height properly. It has nothing to do with the midpoint
What do you think of the Gettier problem, Agnishom? I have not thought about it for a while, but I remember finding it very difficult; and the more you think you have a resolution, the more it slips out of reach.
Hi;
Why is it that whenever I do this by finding the distance between points using Pythagoras, I always get an answer very close to but not quite consistent with the formula?
Well it looks like we all made some mistakes with this one, although it doesn't seem difficult enough to warrant it. My "solution" for six toggles, obviously, involved an illegal step. The problem is that with six toggles, you cannot get an even number of closed lockers; and six is even.
anna_gg - so how do we determine the minimum number of passes? Is it x, or is it 9+2x divided by our number of toggles?
For example, with 7 toggles, can it be done in three passes because 21/7 = 3, or can it only be done in six passes because x = 6
Edit: The answer is three passes.
First pass: Open 7 (2 closed)
Second pass: Open 1, close 6 (7 closed)
Third pass: Open 7
In fact, the answer is three passes for 3, 5 and 7 toggles, and trivially, 9 for 1 and 1 for 9, as the formula predicts.
-2 + 9 + 5 + 1 +/- 0
Lol, congratulations
171,270
Hi, by approximation rubbish I specifically meant rounding the fourth roots, which would work for any of at least the last 80 posts
2+9+2+5*0
82! * 123 = 58469813045257953505582999459696808437189035170363179419910794891439511352392595279399877212442937555681280000000000000000000
5.85 * 10^124
Hi! I find myself in complete agreement with your post about the paradox. At first, even besides ambiguity, it seems like statements 1 and 3 are not independent and it is hard to accept #1 without accepting #3... but as soon as you attempt to formalise that thought, the tension evaporates. It is simply that the pages never run out, and the days never run out, so there will always be a definite future.
Regarding Zeno's paradox, I would disagree with all of the statements as you mention - but I should mention that I have heard explanations that satisfy me even with infinitely divisible space. Without drawing on too much of my memory at this moment, the answer involves time and limits.
^ That number is in post #1676
1702*99 = 168,498
1704*99 = 168,696
That's a strange condition.
194 EveryoneTookTheGoodNames
Found a way for six lockers with only five passes though, and don't see the general principle just yet, so perhaps a lot is left to do!
Hi!
I found a way for five lockers with seven passes. Just tinkering at the moment though
What is all this approximation rubbish, and 0^0? Seems quite unnecessary, even lazy, lol
I think that has to do with my habit for excessive editing. lol
The square root of 289 would have been useful for a lot of these.
Oops double post. 28990
Thank you, I think I understand (: Sorry for my confusion!