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I have this question as an example in my maths school book. The solution given there is:-

E = the man reports six

P(S1)= Probability that six actually occurs = $\frac{1}{6}$

P(S2)= Probability that six doesn't occur= $\frac{5}{6}$

P(E|S1)= Probability that the man reports six when six has actually occurred = $\frac{3}{4}$

P(E|S2)= Probability that the man reports six when six has not occurred = $1-\frac 3 4=\frac 1 4$

Therefore, by Bayes' Theorem,
$P(S1|E)=\frac{(\frac{1}{6}\cdot\frac{3}{4})}{(\frac{1}{6}\cdot \frac{3}{4})+(\frac{5}{6}\cdot \frac{1}{4})} =\frac{3}{8} $

I have its solution but my teacher said that the solution given is incorrect and told that the actual solution would be something else:-

$P(S1|E)=\frac{\frac 1 6\cdot\frac3 4}{(\frac 1 6\cdot \frac 3 4)+(\frac 5 6\cdot \frac1 4\cdot\frac1 5)} = \frac 3 4$

So, I want to ask which one is correct. Thank you.

Ramasamy Kandasamy
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dknight
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    There is a fair bit of debate on this over here :) : http://math.stackexchange.com/questions/1084785/a-man-who-lies-a-fourth-of-the-time-throws-a-die-and-says-it-is-a-six-what-is-t – Shreyas Dec 05 '16 at 15:52

3 Answers3

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The difference in solutions comes in the estimation of the probability that the man reports six when six has not occurred.

If the man randomly chooses a number to report when he lies (which seems like a reasonable statement), then the probability he chooses 6 is 1/5. If you multiply your calculation of P(E|S2) by this, you get your teacher's solution.

David Yang
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  • no I want to ask which one is more appropriate? – dknight Oct 16 '13 at 17:26
  • The problem does not have enough information to say; if the man chooses 6 whenever he lies and the number isn't 6, then your solution is right. If he chooses randomly your teacher is right. Your teacher's solution seems more appropriate for this case. – David Yang Oct 16 '13 at 17:28
  • ok thanks, btw it is given that he will always say 6:) – dknight Oct 16 '13 at 17:31
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I know I am late, but this might help someone.

The problem arises in the case where the guy didn't get a six, but decides to lie.

So the probability of not getting a six is $\dfrac{5}{6}$.

and the probability that he lies is $\dfrac{1}{4}$.

Fair enough but we still don't know what is the faulty number that he replied with, it might be any number but the actual one.

So the probability that he replies with $6$ is going to be $\dfrac{1}{5}$.

And on solving using Bayes' theorem we get $\dfrac{3}{4}$.

This should be the correct answer for the exact wording of the question as you have provided.

Bean
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AspiringEngineer
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  • So the answer is $\frac{3}{4}$ when we assume that he reports a six when it is not a six, right? – Kaushik Apr 10 '21 at 11:20
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Let us define that,

$E_1=A$ speaks truth; $E_2=A$ tells a lie; $E= A$ reports a six

Given,

$P (E_1) =\frac{3}{4},~ P (E_2) =\frac{1}{4},~ P (E|E_1) =\frac{1}{6},~ P (E|E_2) =\frac{5}{6}.$

The required probability that actually there were six (by Bayes’) is

$P (E_1|E) = [P (E_1) \cdot P (E|E_1)] / [P (E_1) \cdot P (E|E_1) + P (E_2) \cdot P (E|E_2)] = \frac{3}{8}$.

(Gourav's answer is right)

Marcus M
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SOVIK ROY
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    The mistake, if I may call it that, which you and gaurav make, is in assuming $P(E|E2) = 5/6$, essential that $P(E \cap E2) = 5/24$. If the roll is not a six, then there are five ways the man could (plausibly) lie (he could say the roll was a bagel, which would be untrue but so obviously so as not to qualify as much of a lie). So the teacher's calculation takes that into account and uses $P(E \cap E2) = 1/24$ to arrive at the different answer. – hardmath Mar 22 '14 at 12:42