Probability question, explanation does not make sense?!

Question

Q: A man speaks truth $3$ out of $4$ times. He throws a die and reports it to be a $6$. What is the probability of it being a $6$?

Provided Answer: There are two cases 1) he is telling truth that the die reports $6$, its probability $= 3/4\cdot 1/6 = 1/8$. 2) he is telling lie that the die reports $6$, its probability $= 1/4\cdot 5/6 = 5/24$ So required probability $= (1/8)/(1/8)+(5/24) = (1/8)/(1/3) = 3/8$

What I don't understand is if we know that the man speaks truth $3$ out of $4$ times, and if he says there is a $6$ on the die, doesn't that mean that there is a $3/4$ probability that there actually is a $6$.

Why does reporting what is on the die makes the man less trustworthy than he is supposed to be. What am I missing here?

Answer 1

Let X be the event where the man rolled a 6, !X be the event that the man did not roll a 6, and R be the event that the man said he scored a 6.

What you want is something like:

$$P(X|R) = \frac{P(R|X)}{P(R)} P(X)$$

I think you know $P(R|X)=3/4$, $P(X)=1/6$, $P(R) = P(R|X)P(X)+P(R|!X)P(!X)$. I am a bit bothered that $P(R|!X)=1/4$ seems to suggest that if the man doesn't roll a 6 he will lie and claim a 6 rather than claiming one of the other four possible values. This would lead to the answer that the text implies... $$P(X|R) = \frac{3/4}{8/24}1/6 = 3/8$$

This seems wrong because the man will have 5 options to lie with. It seems like it would give up the game if the man said more than one option. If instead we plug in $P(R|!X) = (1/4)(1/5)=1/20$ we have: $$P(X|R) = \frac{3/4}{4/24}1/6 = 3/4$$

Answer 2

This is the very point of Bayes's formula: you know how often the man speaks truth, but you have no way to know whether he said $6$ because it was true and it was one of those $3/4$ times or because we had a different thing on the die and he happened to lie.

What the Bayes's formula does is precisely this: it helps you reverse the probabilities and so you can know how likely it is that the man said truth depending on the number he said and the number he said alone (mostly because that was the one and only thing you could know).

Answer 3

I think the provided answer is wrong.

There are two cases 1) he is telling truth that the die reports $6$, its probability $= 3/4\cdot 1/6 = 1/8$.

That would be the probability of the man speaking the truth and the die being $6$, a priori. But here we are given the additional data "the man has told that the die was '6". Under that data (condition), the events are not independent.

Let $T$ be the event that the man spoke the truth, let $S$ be the event that the die was a $6$.

Then, assuming as a global implicit condition the event "the man has told that the die was '6" we have

$$ \begin{align} P(S) &= P(S , T) + P (S, \overline{T})\\ &= P(S | T) P(T) + P (S | \overline{T}) P ( \overline{T})\\ &= 1 \times \frac34 + 0 \times \frac14\\ &= \frac34 \end{align} $$

so your intuition is right.

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Edit: My answer relies in the following two assumptions:

1) the man lies with probability $1/4$, and he decides to lie or tell the truth independently of the die result (say, before the die is tossed).

2) when the man decides to lie, he says one of the $5$ incorrect die results with uniform probability

I think that, though the question is slightly ambiguous and other interpretations are possible, these are the most reasonable assumptions.

Answer 4

The answer by @KitterCatter is excellent. It pinpoints the issue. The question is ambiguous re: what he would do if he didn't roll a $6$ and decided to lie.

• (A) If he always lied it's $6$ then the answer is $3/8$.

• (B) if he always lied uniformly randomly then the answer is $3/4$.

• But frankly (C) if he always lied that it's the lowest number different from the roll, i.e. he never lied that it's $6$, then the answer is $1$! Because if he said it's $6$ you know he's not lying.

• And (D) there are other lying models where the answer is again different from all of the above.

Here are some examples. Each example comprises $24$ instances, he rolled each number $4$ times, and told the truth $3$ out of those $4$ times. But the answer is greatly affected by what he said when he didn't roll a $6$ and decided to lie.

    (A)               (B)               (C)               (D)
Roll | He said    Roll | He said    Roll | He said    Roll | He said
  1    1,1,1,6      1    1,1,1,2      1    1,1,1,2      1    1,1,1,6
  2    2,2,2,6      2    2,2,2,3      2    2,2,2,1      2    2,2,2,6
  3    3,3,3,6      3    3,3,3,4      3    3,3,3,1      3    3,3,3,2
  4    4,4,4,6      4    4,4,4,5      4    4,4,4,1      4    4,4,4,2
  5    5,5,5,6      5    5,5,5,6      5    5,5,5,1      5    5,5,5,1
  6    6,6,6,1      6    6,6,6,1      6    6,6,6,1      6    6,6,6,1
Ans = 3/8         Ans = 3/4         Ans = 3/3 = 1     Ans = 3/5

All these interpretations are allowed by the question statement, because it said he told you it's a $6$, but it left entirely unspecified how often / under what conditions he would do so if he didn't roll a $6$ and decided to lie.

Instead, suppose the question statement were as follows: You asked him, "Is it a $6$?" and he answered "Yes." This would correspond to (A) above, because he would answer "Yes" every time he rolled not $6$ and decided to lie. Thus all ambiguities would be removed.

Unfortunately, this kind of ambiguity is rather common in my experience. The person who wrote the question had in mind some Bayesian model, and didn't think through all the assumptions necessary to narrow down the possible value of Prob(said it was $6$ | rolled not $6$, and lying).

E.g. you can find similar issues infecting many carelessly worded discussion of the Monty Hall problem -- in the sense that the behavior of the host is often not fully specified in the question statement.