3 ms·
You mean the statistics demonstration? I'm sure I've seen it in several places. I know of two tricks for detecting the students. The first is to look for six
by Dove 15y ago
You mean the statistics demonstration? I'm sure I've seen it in several places.
I know of two tricks for detecting the students. The first is to look for six or seven heads or tails in a row. Over a hundred tosses, a coin will probably do that, but humans "being random" won't. The other is to look at the page as a sequence of "HHH" and "TT" strings and estimate how many there are. A coin, of course, changes from heads to tails 50% of the time, but a human does it more like 70% of the time.
I'm sure there are other characteristics, too, but those two are sufficient to throw out most human attempts at a glance. It's actually kind of obvious, when you see the two side by side.
Me "being random" with the numpad: 10110101001010010101011010100101011010100101010110101
Computer-generated random: 1101110100000000011111110111011000010001010110011111
See? Here's a few more. Try it.
1000100111011001010000001001010110111000011011101011
1101010001010010101110100101000101010111101001001000
1101111010110111100010110001000100001001111001001110
1100000010000010101001000001101001101011111100111001
- jcitme 15y agoAnswers please?
- Cushman 15y agoCool! Makes sense, too— it feels unrandom to sit there hitting one key a bunch. How do you know when to stop? So here's what I got: figure I have a bias to switch keys. That means that 01 and 10 are more common than 11 and 00. So what I need to do is group 01 with 00, and 10 with 11. What I do is generate twice as many bits as I need, treat the string as a sequence of two-bit pairs, and reduce each one to its first bit. That looks like this: 01011110000010111000010011111011111111000000111010110001111000111100111100001100011010011101 Gets you past the litmus test, but looks like it goes too far the other way? Hard for me to tell, actually. Another thing would be something like a sequential xor of each bit in triples (i.e. 010 -> (0 ^ 1) ^ 0 = 1), which segments triples across probability like so: - + + - 000 001 010 011 100 101 110 111 0 1 1 0 1 0 0 1 You can do that quickly by counting the 1s— 1 or 3 is 1, 0 or 2 is 0. That looks like this: 00011011011011000000111111001011000101010000100100110011011010 I don't know if it adds much more (apparent) entropy, though. Data: 011101101010100100010100100010111001010100110000101011101101110 0 1 0 1 1 1 1 0 0 0 0 0 1 0 1 1 1 0 0 0 0 1 0 0 1 1 1 1 1 0 1 0 0 0 1 1 0 1 1 0 1 1 0 1 1 0 0 0 0 0 0 1 111010111010100101000101011110100011011011010001111011110101001 1 1 1 1 1 1 1 0 0 0 0 0 0 1 1 1 0 1 0 1 1 0 0 0 1 1 1 1 0 0 0 1 1 1 1 1 0 0 1 0 1 1 0 0 0 1 0 1 0 1 0 0 101110100101101010110101010001101101010010110111010110101001110 1 1 1 1 0 0 1 1 1 1 0 0 0 0 0 1 1 0 0 0 1 1 0 1 0 0 1 1 1 0 1 0 0 1 0 0 1 0 0 1 1 0 0 1 1 0 1 1 0 0 1 0