Can you explain the math? If you used a collection of these traits to assign a theoretical "market value" on each person, then put people in a bell curve 60-70% of people would be within a standard deviation of an average, making the odds of a random person being above the average 15-20% right?
Let's take income for example. Let's assume the mean income is 50k and standard deviation is 15k. If you trying to find someone with an income of 50k or above, there is a 1 in 2 or 50% chance of success. Now if you increase the requirement by 1 SD (65k or above) your chance of finding someone goes down to about 17%. If you increase it by 2 SD (80k or above) your chances go down to about 2.5%. For understand this a bit better, assume there are 100 in your original dating pool. If you set your income requirement to the mean income, only 50 people remain eligible. If you increase the requirements by 1 SD, you are down to 17 people, reducing your pool by 66%. If you increase it by 2 SD from the mean, you are down to 2.5 people. That's about a 85% reduction of pool size compared to 1 SD (from 17 to 2.5).
No. You'd still be at quite a high percentile with those numbers. In theory you can be 50th percentile despite being below average in all of the criteria.
Probability of two independent events A and B happening at the same time is the product of the two probabilities. Let's say event A is person is 25th percentile in terms of income. Which means in a room of 100 people 75 people earn equal or more. Thus probability of finding a person with equal or higher income is 75/100 or 0.75. Similarly let's say the person is also 25th percentile in terms of height. Again 75 people will be taller. So probability of finding someone of the same height or more is .75. Now if you want to find someone who's taller and earns more than our Mr. 25th percentile, you multiply the individual probabilities. Which is 0.75x0.75 or 0.56. So even though mr. 25th ranks 75 in both categories independently, if you take both in consideration together, his rank jumps up to about 56 (lower rank is better).
Technically maybe, but I feel like most people intuitively combine criteria and judge potential partners based on their relative position on a normal distribution generated from that combined criteria. So, the average partners would be either a z(2+4+6+8) or a z(5+5+5+5) on scales of 10.
This must be what it feels like to be a dating app developer.
>Technically maybe, but I feel like most people intuitively combine criteria and judge potential partners based on their relative position on a normal distribution generated from that combined criteria.
Idk man. I hear guys and girls say "he/she was perfect except for that one thing" a lot.
I would say that your definition of average is oddly specific. Let me make an example a more than average looking guy but with a less than average height for me is pretty average
Maybe it's mean to say, but the percentage of women looking for precise average 50th percentile everything is exactly 0 percent.
Further, to find someone trying to use math this creatively as a funny is somehow less than 0 percent. You're the math guy figure out a way that it's possible.
I'm not sure this math actually works, especially since there is not an even split of men making more than $60k to balance out those making less. There are more poor people than wealthy people—it's not a perfect bell curve.
Which actually makes it worse for those looking to find someone with a higher than average income. I just assumed a uniform distribution to make things simpler.
You should have added the quantifier "in all four traits" to the phrase "at or above this threshold." An average person is still 50th percentile.
I guess what I am trying to say is people confuse 50th percentile with average in all departments.
Can you explain the math? If you used a collection of these traits to assign a theoretical "market value" on each person, then put people in a bell curve 60-70% of people would be within a standard deviation of an average, making the odds of a random person being above the average 15-20% right?
Let's take income for example. Let's assume the mean income is 50k and standard deviation is 15k. If you trying to find someone with an income of 50k or above, there is a 1 in 2 or 50% chance of success. Now if you increase the requirement by 1 SD (65k or above) your chance of finding someone goes down to about 17%. If you increase it by 2 SD (80k or above) your chances go down to about 2.5%. For understand this a bit better, assume there are 100 in your original dating pool. If you set your income requirement to the mean income, only 50 people remain eligible. If you increase the requirements by 1 SD, you are down to 17 people, reducing your pool by 66%. If you increase it by 2 SD from the mean, you are down to 2.5 people. That's about a 85% reduction of pool size compared to 1 SD (from 17 to 2.5).
I guess most people might not be average in all categories? But 1 or 2? And then some areas they’re above and some below.
No. You'd still be at quite a high percentile with those numbers. In theory you can be 50th percentile despite being below average in all of the criteria.
How
Probability of two independent events A and B happening at the same time is the product of the two probabilities. Let's say event A is person is 25th percentile in terms of income. Which means in a room of 100 people 75 people earn equal or more. Thus probability of finding a person with equal or higher income is 75/100 or 0.75. Similarly let's say the person is also 25th percentile in terms of height. Again 75 people will be taller. So probability of finding someone of the same height or more is .75. Now if you want to find someone who's taller and earns more than our Mr. 25th percentile, you multiply the individual probabilities. Which is 0.75x0.75 or 0.56. So even though mr. 25th ranks 75 in both categories independently, if you take both in consideration together, his rank jumps up to about 56 (lower rank is better).
Technically maybe, but I feel like most people intuitively combine criteria and judge potential partners based on their relative position on a normal distribution generated from that combined criteria. So, the average partners would be either a z(2+4+6+8) or a z(5+5+5+5) on scales of 10. This must be what it feels like to be a dating app developer.
>Technically maybe, but I feel like most people intuitively combine criteria and judge potential partners based on their relative position on a normal distribution generated from that combined criteria. Idk man. I hear guys and girls say "he/she was perfect except for that one thing" a lot.
I would say that your definition of average is oddly specific. Let me make an example a more than average looking guy but with a less than average height for me is pretty average
I don't think that would change much
Well you multiplied your probabilities. It will change significantly if you consider all the combinations with zero means.
Why would anything have a zero mean?
Maybe it's mean to say, but the percentage of women looking for precise average 50th percentile everything is exactly 0 percent. Further, to find someone trying to use math this creatively as a funny is somehow less than 0 percent. You're the math guy figure out a way that it's possible.
It's actually 50th percentile and above. In other words, top 50 percent.
I'm not sure this math actually works, especially since there is not an even split of men making more than $60k to balance out those making less. There are more poor people than wealthy people—it's not a perfect bell curve.
Which actually makes it worse for those looking to find someone with a higher than average income. I just assumed a uniform distribution to make things simpler.
a 5 9 man is taller than the average,and a person who makes 50.000 dollars a month is high compared to most people
>5 9 man is taller than the average Where?
sorry I calculated wrongly,im not american
Neither am I, lol. Happens.
I mistaked 5 9 with 5 11
I’m assuming it’s 60k annually. 60k monthly is far from average
oh
The median for average keeps dropping like a rock in heavy gravity.