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A man speaks the truth 3 out of 4 times. A fair die is thrown and he reports it as six, find the probability that it is actually a six.

All that has been specified is to use Bayes' theorem.

I tried using Bayes' theorem provided 6 has already occurred, and then finding out the probability of the man speaking the truth.

The answer came out wrong. The answer says use provided condition as the man speaking the truth and then find probability of six occurring. I am confused by this. Please can someone explain?

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    $\begingroup$ We need additional assumptions: (i) the choice of whether to speak the truth or lie is independent of the result of the throw and (ii) if the person chooses to lie, the person chooses at random between the wrong answers, with all wrong answers equally likely. $\endgroup$ – André Nicolas Mar 16 '15 at 15:54
  • $\begingroup$ Well they told to use Bayes theorem.. So I thought we need to find probability of man speaking truth provided 6 has already occurred while the answer says to do it the other way round.. I am really confused $\endgroup$ – Maini who Mar 16 '15 at 15:58
  • $\begingroup$ Next time you can better put everything you know and explain why you are confused into your post, or your question can get downvoted for showing lack of effort. $\endgroup$ – Pedro Mar 16 '15 at 16:06
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    $\begingroup$ I will not type an answer, since the question may be closed while I am typing, which is irritating. But will try to give an outline in comments. We work under the additional assumptions I mentioned earlier. Let $B$ be the event the man reports $6$, and let $A$ be the event the number is actually $6$. We want $\Pr(A|B)$, which by the definition of conditional probability is $\frac{\Pr(A\cap B)}{\Pr(B)}$. I will use this definition rather than Bayes' Theorem, but it is easy to reword in Bayes' Theorem language. (to be continued). $\endgroup$ – André Nicolas Mar 16 '15 at 16:08
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    $\begingroup$ (Cont.) So we need to evaluate a couple of probabilities. Let us do $\Pr(B)$ since it is harder. The event $B$ can happen in 2 ways (i) there was a $6$ and the man told the truth or (ii) there was something else and the man falsely reported a $6$. The probability of (i) is $(1/6)(3/4)$. The probability of (ii) is $(5/6)(1/4)(1/5)$. Now put the pieces together. (There is an easier approach, that does not use Bayes machinery at all.) $\endgroup$ – André Nicolas Mar 16 '15 at 16:15
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Bayes' Theorem is all about destructing the problem and then constructing it in a logical manner.

Let

$E_{1}:$ Dice shows six.

$E_{2}:$ Dice does not shows six.

$ E:$Man reports it is six.

$P(E_{1})=1/6$

$P(E_{2})=5/6$

Truth:

$ P(E/E_{1})=3/4$

Lie:

$ P(E/E_{2})=1/4$

Now,

$P(E_{1}/E)= \frac{P(E/E_{1}).P(E_{1})}{P(E/E_{1}).P(E_{1})+P(E/E_{2}).P(E_{2})}$

$P(E_{1}/E)= \frac{({\frac{1}{6}}.{\frac{3}{4}})}{( {\frac{1}{6}}.{\frac{3}{4}})+( {\frac{5}{6}}.{\frac{1}{4}})} $

$P(E{1}/E)={\frac{3}{8}}$

Applying Bayes' Theorem

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  • $\begingroup$ Thanks,finally got it. Flowchart was of great help to visualize the things $\endgroup$ – Maini who Mar 16 '15 at 17:53
  • $\begingroup$ You're welcome. $\endgroup$ – Integrator Mar 16 '15 at 19:42
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Suppose there are $600$ dice throws.

  • An expected $100$ times this will show a six. On the assumption lying is random, an expected $75$ times he will will the truth report a six.

  • An expected $500$ times this will not show a six. On the assumption lying is random, an expected $125$ times he will lie, and of those an expected $25$ times he will report a six.

So he will report a six an expected $100$ times and be lying an expected $25$ times, making the probability $\dfrac14$.

This corresponds with intuition if the probability he lies is independent of the number thrown and is also independent of the number reported.

If you want to use Bayes' theorem, call the events $T_6$ throwing a six, $R_6$ reporting a six and $L$ lying. You want to find $$P(L|R_6)$$ $$=\dfrac{P(L \cap R_6)}{P(L \cap R_6) +P(L^c \cap R_6)}=\dfrac{P(R_6|L \cap T_6^c )P(L |T_6^c )P(T_6^c)}{P(R_6|L \cap T_6^c )P(L |T_6^c )P(T_6^c) +P(R_6|L^c \cap T_6 )P(L^c |T_6 )P(T_6)}$$ $$=\dfrac{\frac15 \times \frac14 \times \frac56}{\frac15 \times \frac14 \times \frac56 +\frac34 \times \frac16} =\dfrac14.$$

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