Many wrongs make a right
Let’s assume you have some analogue clocks, all of which are set at a random time. Assuming they’re all wrong by the minimum amount (i.e. if the clock is set at eleven o’clock and the actual time is 4am, it’s five hours slow rather than being seven hours fast), the average of all the clocks’ times will tend to the correct time as the number of clocks tends to infinity.
Nice.
March 19, 2008 | Filed Under Numbers and stuff, Random thoughts
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8 Responses to “Many wrongs make a right”
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Dan,
Are you sure? Isn’t the average time going to tend to 6 o’clock and the average difference will tend to zero?
With love mate,
Jon
Don’t think so, Jon. Assume the real time is 4am. Clocks will display times anywhere between twelve o’clock and 11.59, with an even spread across the times. For every clock that is an hour early (showing three o’clock, interpreted as 3am), there will on average be one that’s an hour late (5am). And for every one that’s five hours early (11pm the previous day), there’l be one five hours late (9am) By knowing the current time, you determine the time direction in which the clock is wrong, forcing this to be the midpoint and therefore the time to which the average tends.
so are you saying the times are rectangularly spread on either side of the correct time?
Yep!
So the average time will tend towards the current time, assuming that you already know the current time…
I cannot think of any situation where this information is useful.
More pondering on the bus.
This works because you are working on a modulo basis,ie 12=0, 13=1 etc?
Ie any time on a dial has half the clock (or circle) on either side of it, each filled by half of the infinite samples. More cirle than rectangle as 12=0?
Have you heard of the Central Limit Theorem Daniel?
I’m sorry Dan, but this is just utter BS!
What makes the clocks tend to the ‘correct time’ (whatever that is?) rather than some other, completely random time?
Nothing!