3囚人問題
恩赦がランダムに決まるという仮定より
Pr [ A ] = Pr [ B ] = Pr [ C ] = 1 3 {\displaystyle \Pr[A]=\Pr[B]=\Pr[C]={1 \over 3}} ...2
であるので、ベイズの定理より、
Pr [ A ∣ b ] = Pr [ b ∣ A ] Pr [ A ] Pr [ b ] = Pr [ b ∣ A ] Pr [ A ] Pr [ b ∣ A ] Pr [ A ] + Pr [ b ∣ B ] Pr [ B ] + Pr [ b ∣ C ] Pr [ C ] = ( 2 ) Pr [ b ∣ A ] Pr [ b ∣ A ] + Pr [ b ∣ B ] + Pr [ b ∣ C ] = ( 1 ) Pr [ b ∣ A ] Pr [ b ∣ A ] + 1 {\displaystyle \Pr[A\mid b]={\Pr[b\mid A]\Pr[A] \over \Pr[b]}={\Pr[b\mid A]\Pr[A] \over \Pr[b\mid A]\Pr[A]+\Pr[b\mid B]\Pr[B]+\Pr[b\mid C]\Pr[C]}{\underset {(2)}{=}}{\Pr[b\mid A] \over \Pr[b\mid A]+\Pr[b\mid B]+\Pr[b\mid C]}{\underset {(1)}{=}}{\Pr[b\mid A] \over \Pr[b\mid A]+1}}