Gut: vg’f tbvat gb or fhssvpvragyl pybfr gb bar gung gur cebonovyvgl vf whfg tbvat gb ybbx yvxr n ohapu bs avarf gb zr.
Fermi estimate: VVEP bar qrpvory vf nobhg svsgl fvk creprag, naq crepragf punatr snfgre jura qrpvoryf ner fznyy, fb svsgl bar creprag vf yrff guna unys n qrpvory. Fb yrg’f fnl gung svsgl bar creprag vf nobhg mreb cbvag bar qo. Fb rnpu urnq cebivqrf abhtug cbvag bar qrpvoryf sbe pbva bar, naq rnpu gnvy cebivqrf artngvir gung sbe gur fnzr. Lbh unir nobhg gjragl gubhfnaq zber urnqf guna gnvyf, fb nobhg gjb gubhfnaq qo rivqrapr sbe pbva bar. Rirel gra qrpvoryf pbeerfcbaqf gb vapernfvat gur bqqf ol n snpgbe bs gra, fb cebonovyvgl vf gura fbzrguvat yvxr bar zvahf gra gb gur zvahf gjb uhaqerq.
Calculation: svsgl bar creprag vf abhtug cbvag bar frira qrpvoryf. Gurer ner gjb gvzrf rvtug gubhfnaq, fvk uhaqerq naq guvegl sbhe zber urnqf guna gnvyf. Gung znxrf nyzbfg rknpgyl guerr gubhfnaq qrpvoryf sbe urnqf. Fb gur bqqf ner gra gb gur guerr uhaqerq, tvivat cebonovyvgl nyzbfg rknpgyl bar zvahf gra gb gur zvahf guerr uhaqerq.
Compared to Manfred’s answer: Jr qvssre ol n snpgbe bs gra gb gur gjb uhaqerq naq guerr. V qba’g pheeragyl unir gur gvzr gb jbex bhg jul, naq jura V qb unir gvzr V znl abg pner rabhtu.
Edit: I think I did it wrong. No time to correct it currently, but the true answer should be higher than mine.
Edit 2: Maybe not? I thought I needed to use both P(H|C1)/P(H|C2) and P(H|C1)/P(T|C1), which are confusingly identical. But when I actually put it on paper, it looks correct.
I have us being different by a factor of 10^40, but yeah, that’s a bit surprising. Maybe we’re far enough out in the tails that the normal approximation is breaking down?
I don’t offhand have a model for why we expect your method to work, so I don’t know why it fails. But another approach using the normal approximation gets within a factor of 10, so that shouldn’t be it.
Um, I think you’re just counting standard deviations in the wrong direction? You’re counting standard deviations from 500,000 and doubling them, but the relevant distribution means are 510,000 and 490,000.
But no, those should be equivalent.
Oh! You’re squaring a sum, not summing a square. You’re counting the correct number of standard deviations in total, but you need the correct number for each distribution.
Gut: vg’f tbvat gb or fhssvpvragyl pybfr gb bar gung gur cebonovyvgl vf whfg tbvat gb ybbx yvxr n ohapu bs avarf gb zr.
Fermi estimate: VVEP bar qrpvory vf nobhg svsgl fvk creprag, naq crepragf punatr snfgre jura qrpvoryf ner fznyy, fb svsgl bar creprag vf yrff guna unys n qrpvory. Fb yrg’f fnl gung svsgl bar creprag vf nobhg mreb cbvag bar qo. Fb rnpu urnq cebivqrf abhtug cbvag bar qrpvoryf sbe pbva bar, naq rnpu gnvy cebivqrf artngvir gung sbe gur fnzr. Lbh unir nobhg gjragl gubhfnaq zber urnqf guna gnvyf, fb nobhg gjb gubhfnaq qo rivqrapr sbe pbva bar. Rirel gra qrpvoryf pbeerfcbaqf gb vapernfvat gur bqqf ol n snpgbe bs gra, fb cebonovyvgl vf gura fbzrguvat yvxr bar zvahf gra gb gur zvahf gjb uhaqerq.
Calculation: svsgl bar creprag vf abhtug cbvag bar frira qrpvoryf. Gurer ner gjb gvzrf rvtug gubhfnaq, fvk uhaqerq naq guvegl sbhe zber urnqf guna gnvyf. Gung znxrf nyzbfg rknpgyl guerr gubhfnaq qrpvoryf sbe urnqf. Fb gur bqqf ner gra gb gur guerr uhaqerq, tvivat cebonovyvgl nyzbfg rknpgyl bar zvahf gra gb gur zvahf guerr uhaqerq.
Compared to Manfred’s answer: Jr qvssre ol n snpgbe bs gra gb gur gjb uhaqerq naq guerr. V qba’g pheeragyl unir gur gvzr gb jbex bhg jul, naq jura V qb unir gvzr V znl abg pner rabhtu.
Edit: I think I did it wrong. No time to correct it currently, but the true answer should be higher than mine.
Edit 2: Maybe not? I thought I needed to use both P(H|C1)/P(H|C2) and P(H|C1)/P(T|C1), which are confusingly identical. But when I actually put it on paper, it looks correct.
I have us being different by a factor of 10^40, but yeah, that’s a bit surprising. Maybe we’re far enough out in the tails that the normal approximation is breaking down?
Oh, I misbracketed your formula. Yes, 10^40.
I don’t offhand have a model for why we expect your method to work, so I don’t know why it fails. But another approach using the normal approximation gets within a factor of 10, so that shouldn’t be it.
Um, I think you’re just counting standard deviations in the wrong direction? You’re counting standard deviations from 500,000 and doubling them, but the relevant distribution means are 510,000 and 490,000.
But no, those should be equivalent.
Oh! You’re squaring a sum, not summing a square. You’re counting the correct number of standard deviations in total, but you need the correct number for each distribution.
Dammit LW, stop nerd sniping me.
Oh yeah, whoops.
And also, muahaha, complete success.