3 ms·

Given it's all multiplied it doesn't matter if the luck is from 0-1000 or 0-1, it has the same effect on the outcome. a * (b * 1000) = (a * 1000) * b Sorry,

Given it's all multiplied it doesn't matter if the luck is from 0-1000 or 0-1, it has the same effect on the outcome. a * (b * 1000) = (a * 1000) * b Sorry, I hate bad maths, I realise it is just trying to imply that luck is more important, but in that case it should be * luck^n.

deleted 10y ago
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ketzu 10y ago
But luck in [0,1] would even make less sense as luck^n (as it goes to 0), while as luck in [0,1000] could scale a successful startup by the factor of up to 1000.
- nawitus 10y ago
  It doesn't matter if you're comparing "0.001" success to "1" success or "1" success to "1000" success.
- throwawayReply 10y ago
  What I meant by 'n' was a weighting. I wasn't clear. Let's say you have 4 factors a, b, c, d. If we are summing them then we can just sum them straight: score = a + b + c + d If we want to place more importance on some of them we weight them, with (constant) weights w, x, y, z by doing this: score = w * a + x * b + y * c + z * d. If it's a multiplicative relationship however then we multiply straight: score = a * b * c * d If we tried add weighting in the same way: score = w * a * x * b * y * c * z * d Then all we've done is scale the previous score by (w * x * y * z) which is constant, so hasn't actually affected any ranking. To weight a multiplicative relationship requires the weights to be exponents: score = a^w * b^x * c^y * d^z The scores can be in [0..1] or [1..N], the increased weights still increase relative power of the input parameter. If you take logs it's clear why. log(score) = w * log(a) + ... + z * log(d) which is our weighting when summing again, and log is monotonic (proof of which is left to the reader), log(S1) < log(S2) implies S1 < S2. Edit: Fixed formatting, it thinks * is italic.
BigMan555 10y ago
How does it have the same impact on outcome? You're talking about the associative property of multiplication but he's saying the luck factor can either reduce your success to 0 or multiply the outcome of your endeavor by 1000. It obviously does have an impact (note: the results of .9 * .9 * 1000 and .9 * .9 * 1 are different numbers).
- throwawayReply 10y ago
  They're different numbers, but it makes no difference in terms of how far along the possible scale you are. If I score money as 0-1 and luck as 0-1000, then I have a possible overall score out of 1000. If I score 0.3 on money and 600 on "luck" then I've scored 180 out of 1000, or 18%. Or if both are 0-1 then I'd have scored 0.3 and 0.6, and overall scored 0.18 out of 1, which is still 18%, it's just the absolute number that has changed. But the absolute number isn't important in a scale which doesn't have meaningful units.
  - BigMan555 10y ago
    I understood the absolute number to be of significance as the goal (eg. the absolute number is however many millions $ you'll make as outcome). But point taken, I don't think this was thought through :)

deleted 10y ago

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ketzu 10y ago

But luck in [0,1] would even make less sense as luck^n (as it goes to 0), while as luck in [0,1000] could scale a successful startup by the factor of up to 1000.

nawitus 10y ago
It doesn't matter if you're comparing "0.001" success to "1" success or "1" success to "1000" success.
throwawayReply 10y ago
What I meant by 'n' was a weighting. I wasn't clear. Let's say you have 4 factors a, b, c, d. If we are summing them then we can just sum them straight: score = a + b + c + d If we want to place more importance on some of them we weight them, with (constant) weights w, x, y, z by doing this: score = w * a + x * b + y * c + z * d. If it's a multiplicative relationship however then we multiply straight: score = a * b * c * d If we tried add weighting in the same way: score = w * a * x * b * y * c * z * d Then all we've done is scale the previous score by (w * x * y * z) which is constant, so hasn't actually affected any ranking. To weight a multiplicative relationship requires the weights to be exponents: score = a^w * b^x * c^y * d^z The scores can be in [0..1] or [1..N], the increased weights still increase relative power of the input parameter. If you take logs it's clear why. log(score) = w * log(a) + ... + z * log(d) which is our weighting when summing again, and log is monotonic (proof of which is left to the reader), log(S1) < log(S2) implies S1 < S2. Edit: Fixed formatting, it thinks * is italic.

BigMan555 10y ago

How does it have the same impact on outcome? You're talking about the associative property of multiplication but he's saying the luck factor can either reduce your success to 0 or multiply the outcome of your endeavor by 1000. It obviously does have an impact (note: the results of .9 * .9 * 1000 and .9 * .9 * 1 are different numbers).

throwawayReply 10y ago
They're different numbers, but it makes no difference in terms of how far along the possible scale you are. If I score money as 0-1 and luck as 0-1000, then I have a possible overall score out of 1000. If I score 0.3 on money and 600 on "luck" then I've scored 180 out of 1000, or 18%. Or if both are 0-1 then I'd have scored 0.3 and 0.6, and overall scored 0.18 out of 1, which is still 18%, it's just the absolute number that has changed. But the absolute number isn't important in a scale which doesn't have meaningful units.
- BigMan555 10y ago
  I understood the absolute number to be of significance as the goal (eg. the absolute number is however many millions $ you'll make as outcome). But point taken, I don't think this was thought through :)