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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 3 4)+(\frac 5 6\cdot 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.

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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.

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  • $\begingroup$ no I want to ask which one is more appropriate? $\endgroup$ – dknight Oct 16 '13 at 17:26
  • $\begingroup$ 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. $\endgroup$ – David Yang Oct 16 '13 at 17:28
  • $\begingroup$ ok thanks, btw it is given that he will always say 6:) $\endgroup$ – dknight Oct 16 '13 at 17:31
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Your teacher's answer, ie 3/4 is the correct one. The wrong part in the other solution as another reader pointed out is this:

P(E|S2)= Prob. that the man reports six when six has not occurred = 1-3/4=1/4

Above is not correct since you've made a false assumption that:

Prob. that the man reports six when six has not occurred = Prob. that the man is lying

Above is NOT true since:

Prob. that the man reports six when six has not occurred ≠ Prob. that the man is lying

The prob. that the man is lying (ie not speaking the truth) = 1 - Prob. that the man is speaking the truth = 1 - 3/4 = 1/4

BUT

Finding out the prob. that the man reports six given that a siz has not actually occured is a bit tricky.

To make things clear, we'll first write the sample space of this experiment: { (1,1), (1,2), (1,3), (1,4), (1,5), 1,6), (2,1), (2,2), (2,3), (2,4), (2,5), 2,6), (3,1), (3,2), (3,3), (3,4), (3,5), 3,6), (4,1), (4,2), (4,3), (4,4), (4,5), 4,6), (5,1), (5,2), (5,3), (5,4), (5,5), 5,6), (6,1), (6,2), (6,3), (6,4), (6,5), 6,6) } In the above, the first number in each of the 36 ordered pairs is the number that acually occurs on the dice while the second number in the ordered pair is what the man reports it to be. So, we have 36 elements or 36 elementary events in the sample space. Let's see what would be the prob. of each of these 36 elements. Out of these 36 pairs, there are 6 pairs in which the man speaks the truth which are: (1,1), (2,2), (3,3), (4,4), (5,5), (6,6) Now, it is given that the prob. of the man speaking the truth is 3/4. And since we know that the set of above 6 pairs forms the event of the man speaking the truth, we MUST divide 3/4 EQUALLY into these 6 pairs, ie (3/4)/6 = 1/8. So, 1/8 is the prob. of a single pair in these 6 pairs.

The rest of the 30 pairs taken together form the event of the man NOT speaking the truth. And the prob. of the man NOT speaking the truth = 1 - prob. of the man speaking the truth = 1 - 3/4 = 1/4. So, dividing 1/4 equally amongst each of these 30 pairs, each pair gets a prob. of (1/4)/30 = 120.

Now, the prob. that the man reports six given that a siz has not actually occured = P(E|S2) = (sum of prob. of elements (or ordered pairs as told above) in the set (E ∩ S2)) / (Sum of prob. of elements in set S2)

Now, let's enumerate the all possible elements in the set E = {(1,6), (2,6), (3,6), (4,6), (5,6), (6,6)}

Set S2 (of 30 elements) is { (1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (5,1), (5,2), (5,3), (5,4), (5,5), (5,6), }

And finally, the set E ∩ S2 is {(1,6), (2,6), (3,6), (4,6), (5,6)}

Now, P(E ∩ S2) = (1/120) * 5 = 1/24 since the prob. of each if the elements in E ∩ S2 as told in the beginning is 1/120.

P(S2) = P(1,1) + P(2,2) + P(3,3) + P(4,4) + P(5,5) + ...(remaining 25 elements) = 1/8 + 1/8 + 1/8 + 1/8 + 1/8 + (1/120) * 25 since as told in the beginning the prob. of each of the pairs (1,1), (2,2), (3,3),(4,4), (5,5), (6,6) is 1/8 while the prob. of each pair in the remaining 30 pairs is 1/120

So, P(E/S2) = P (E ∩ S2) / P(S2) = (1/24) / (1/8 + 1/8 + 1/8 + 1/8 + 1/8 + (1/120) * 25) = (1/24) / (5/8 + 5/24) = (1/24) / ((15 + 5)/24) = (1/24) / (20/24) = (1/24) * (24/20) = 1/20 which is the prob. that the man reports 6 given that a six has not actually occured.

If you substitute 1/20 for P(E/S2) in Bayer's formula, you'll get the same answer as your teacher.

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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)

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  • 1
    $\begingroup$ 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. $\endgroup$ – hardmath Mar 22 '14 at 12:42

protected by Daniel Fischer Sep 4 '15 at 18:43

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