https://www.explainxkcd.com/wiki/api.php?action=feedcontributions&feedformat=atom&user=Cc 2026-04-18T04:45:14Z User contributions MediaWiki 1.30.0 https://www.explainxkcd.com/wiki/index.php?title=Talk:1132:_Frequentists_vs._Bayesians&diff=16786 2012-11-09T18:20:21Z

<p>Cc: </p> <hr /> <div>Note: taking that bet would be a mistake. If the Bayesian is right, you're out $50. If he's wrong, everyone is about to die and you'll never get to spend the winnings. Of course, this meta-analysis is itself a type of Bayesian thinking, so [http://lmgtfy.com/?q=dunning-kruger+effect Dunning-Kruger Effect] would apply. - [[User:Frankie|Frankie]] ([[User talk:Frankie|talk]]) 13:50, 9 November 2012 (UTC)<br /> : You don't think you could spend fifty bucks in eight minutes? ;-) (PS: wikipedia is probably a better link than lmgtfy: {{w|Dunning-Kruger effect}}) -- [[User:IronyChef|IronyChef]] ([[User talk:IronyChef|talk]]) 15:35, 9 November 2012 (UTC)<br /> Randall has referenced the Labyrinth guards before: [http://xkcd.com/246/ xkcd 246:Labyrinth puzzle]. Plus he has satirized p&lt;0.05 in [http://www.explainxkcd.com/wiki/index.php?title=882:_Significant xkcd 882:Significant]--[[User:Prooffreader|Prooffreader]] ([[User talk:Prooffreader|talk]]) 15:59, 9 November 2012 (UTC)<br /> <br /> A bit of maths. Let event N be the sun going nova and event Y be the detector giving the answer &quot;Yes&quot;. The detector has already given a positive answer so we want to compute P(N|Y). Applying the Bayes' theorem:<br /> : P(N|Y) = P(Y|N) * P(N) / P(Y)<br /> : P(Y|N) = 1<br /> : P(N) = 0.0000....<br /> : P(Y|N) * P(N) = 0.0000...<br /> : P(Y) = p(Y|N)*P(N) + P(Y|-N)*P(-N)<br /> : P(Y|-N) = 1/36<br /> : P(-N) = 0.999999...<br /> : P(Y) = 0 + 1/36 = 1/36<br /> : P(N|Y) = 0 / (1/36) = 0<br /> Quite likely it's not entirely correct. [[User:Lmpk|Lmpk]] ([[User talk:Lmpk|talk]]) 16:22, 9 November 2012 (UTC)<br /> <br /> Here's what I get for the application of Bayes' Theorem:<br /> : P(N|Y) = P(Y|N) * P(N) / P(Y)<br /> : = P(Y|N) * P(N) / [P(Y|N) * P(N) + P(Y|~N) * P(~N)]<br /> : = 35/36 * P(N) / [35/36 * P(N) + 1/36 * (1 - P(N))]<br /> : = 35 * P(N) / [35 * P(N) - P(N) + 1]<br /> : &lt; 35 * P(N)<br /> : = 35 * (really small number)<br /> <br /> So, if you believe it's extremely unlikely for the sun to go nova, then you should also believe it's unlikely a Yes answer is true.</div>

Cc