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 Lying gambler (Posted on 2010-10-21)
A gambler throws a die and tells the score. He tells truth 3/4 of the times that he speaks and randomly lies 1/4 of the times that he speaks. If he says it is a six, what is the probability that he actually got a six?

 No Solution Yet Submitted by Vishal Gupta No Rating

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 Probability doesn't lie either (spoiler) | Comment 4 of 6 |
OK, I'll jump in with conditional probability.

probability of having a 6 and telling the truth about it =
(probability of having a 6)
*(probability of telling truth)
= (1/6)*(3/4) = 3/24

Probability of not having a 6 but claiming that you do =
(probability of not having a 6)
*(probability of lieing)
*(probability that the number claimed, which must differ from actual roll, is 6) =
(5/6)*(1/4)*(1/5) = 1/24

(3/24)/(3/24 + 1/24) = 3/4

 Posted by Steve Herman on 2010-10-22 20:32:13

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