Translation examples
Vartija tietää kuka vangeista on armahdettu, mutta hän ei saa kertoa sitä.
The warden knows which one is pardoned, but is not allowed to tell.
Kianto todettiin syylliseksi ja tuomittiin kuritushuoneeseen, mistä vapautui presidentti Kyösti Kallion armahdettua hänet.
This was discovered, and he was sentenced to death by hanging, but Gao Yang pardoned him.
Vanki A anelee vartijaa kertomaan kumpi vangeista B ja C on varmasti tuomittu kuolemaan (kumpaa siis ei ole armahdettu). ”Jos B on armahdettu, kerro minulle C:n nimi.
If A will be pardoned, the warden can tell A that either B or C is to be executed, and hence P ( b | A ) = 1 2 {\displaystyle P(b|A)={\tfrac {1}{2}}} ; whereas if C will be pardoned, the warden can only tell A that B is executed, so P ( b | C ) = 1 {\displaystyle P(b|C)=1} .
Nyt jos oletetaan, että vankia B ei armahdeta, on vangilla A siis edelleen 1/3 mahdollisuus tulla armahdetuksi kun taas vangin C mahdollisuus on 2/3, kaksinkertainen vankiin A nähden.
After he hears B will be executed, Prisoner A realizes that if he will not get the pardon himself it must only be going to C. That means there is a 2/3 chance for C to get a pardon.
Tapahtumat A = ”vanki A armahdetaan” B = ”vanki B armahdetaan” C = ”vanki C armahdetaan” b = ”vartija nimeää vangin B joutuvan teloitettavaksi (ei armahdettu)” Nyt Bayesin teoreemaa käyttäen saadaan, että vangin A mahdollisuus tulla armahdetuksi on: P ( A | b ) = P ( b | A ) P ( A ) P ( b | A ) P ( A ) + P ( b | B ) P ( B ) + P ( b | C ) P ( C ) = {\displaystyle P(A|b)={\frac {P(b|A)P(A)}{P(b|A)P(A)+P(b|B)P(B)+P(b|C)P(C)}}=} = 1 2 × 1 3 1 2 × 1 3 + 0 × 1 3 + 1 × 1 3 = 1 3 . {\displaystyle ={\frac {{\tfrac {1}{2}}\times {\tfrac {1}{3}}}{{\tfrac {1}{2}}\times {\tfrac {1}{3}}+0\times {\tfrac {1}{3}}+1\times {\tfrac {1}{3}}}}={\tfrac {1}{3}}.} Jokaisella vangilla on 1/3 mahdollisuus tulla armahdetuksi.
Call A {\displaystyle A} , B {\displaystyle B} and C {\displaystyle C} the events that the corresponding prisoner will be pardoned, and b {\displaystyle b} the event that the warden tells A that prisoner B is to be executed, then, using Bayes' theorem, the posterior probability of A being pardoned, is: P ( A | b ) = P ( b | A ) P ( A ) P ( b | A ) P ( A ) + P ( b | B ) P ( B ) + P ( b | C ) P ( C ) = {\displaystyle P(A|b)={\frac {P(b|A)P(A)}{P(b|A)P(A)+P(b|B)P(B)+P(b|C)P(C)}}=} = 1 2 × 1 3 1 2 × 1 3 + 0 × 1 3 + 1 × 1 3 = 1 3 . {\displaystyle ={\frac {{\tfrac {1}{2}}\times {\tfrac {1}{3}}}{{\tfrac {1}{2}}\times {\tfrac {1}{3}}+0\times {\tfrac {1}{3}}+1\times {\tfrac {1}{3}}}}={\tfrac {1}{3}}.} The probability of C being pardoned, on the other hand, is: P ( C | b ) = P ( b | C ) P ( C ) P ( b | A ) P ( A ) + P ( b | B ) P ( B ) + P ( b | C ) P ( C ) = {\displaystyle P(C|b)={\frac {P(b|C)P(C)}{P(b|A)P(A)+P(b|B)P(B)+P(b|C)P(C)}}=} = 1 × 1 3 1 2 × 1 3 + 0 × 1 3 + 1 × 1 3 = 2 3 . {\displaystyle ={\frac {1\times {\tfrac {1}{3}}}{{\tfrac {1}{2}}\times {\tfrac {1}{3}}+0\times {\tfrac {1}{3}}+1\times {\tfrac {1}{3}}}}={\tfrac {2}{3}}.} The crucial difference making A and C unequal is that P ( b | A ) = 1 2 {\displaystyle P(b|A)={\tfrac {1}{2}}} but P ( b | C ) = 1 {\displaystyle P(b|C)=1} .
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