Notably, this effect could mean you differentially hire presties even if they’re slightly worse on average than the normies. I feel like practicing my math, so let’s look at the coin example.
If you flip once and get H, that’s an odds ratio of 2/31/3=2 for a heads-bias, which combine with initial odds of 1 to give odds for [a 2/3 heads-bias] of 2 which is a probability of 2/3, and your expectation of the overall bias is 23⋅23+13⋅13=59≈0.556.
Now suppose you flip twice and get HH. Odds for heads-bias is 4, probability 4/5, which gives expectation 45⋅23+15⋅13=3/5=0.6.
But suppose you flip twice and get HH, but instead of the coins being a mix of 2/3 and 1/3 heads-biased, they’re p and 1−p, for some known p. The odds ratio is now o:=(p1−p)2, probability of [a p heads-bias] is o1+o and the expected bias is op1+o+1−p1+o. I tried to expand this by hand but made a mistake somewhere, but plugging it into wolfram alpha we get that this equals 5/9 when
p=17±√1734≈{0.62,0.38}
Where the two solutions are because an equal mix of p and 1−p biases is the same as an equal mix of 1−p and p biases.
So you’d prefer an HH with a most-likely bias of 0.63 over an H with a most-likely bias of 0.66.
Notably, this effect could mean you differentially hire presties even if they’re slightly worse on average than the normies. I feel like practicing my math, so let’s look at the coin example.
If you flip once and get H, that’s an odds ratio of 2/31/3=2 for a heads-bias, which combine with initial odds of 1 to give odds for [a 2/3 heads-bias] of 2 which is a probability of 2/3, and your expectation of the overall bias is 23⋅23+13⋅13=59≈0.556.
Now suppose you flip twice and get HH. Odds for heads-bias is 4, probability 4/5, which gives expectation 45⋅23+15⋅13=3/5=0.6.
But suppose you flip twice and get HH, but instead of the coins being a mix of 2/3 and 1/3 heads-biased, they’re p and 1−p, for some known p. The odds ratio is now o:=(p1−p)2, probability of [a p heads-bias] is o1+o and the expected bias is op1+o+1−p1+o. I tried to expand this by hand but made a mistake somewhere, but plugging it into wolfram alpha we get that this equals 5/9 when
p=17±√1734≈{0.62,0.38}Where the two solutions are because an equal mix of p and 1−p biases is the same as an equal mix of 1−p and p biases.
So you’d prefer an HH with a most-likely bias of 0.63 over an H with a most-likely bias of 0.66.