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A person, $A$, speaks the truth $4$ out of $5$ times. The person throws a die and reports that he obtained a $6$. What is the probability that he actually rolled a $6$?

I know there is a similar question like this but my doubts are different from it and also I want to identify and solve total probability theorem questions so I posted a side doubt also. In my attempt, I defined the events

\begin{align*} E_1&: \text{The person tells the truth.} \\ E_2&: \text{The person lies.} \\ E_3&: \text{The person reports that the die landed on a 6.} \end{align*}

I noted that $P(E_1)=\frac{4}{5}$, $P(E_2)=\frac{1}{5}$, $P(E_3|E_1)=6^{-1}$ and $P(E_3|E_2)=0$ and obtained

\begin{align*} P(E_3) = \frac{4}{5} \cdot \frac{1}{6} + \frac{1}{5} \cdot 0 = \frac{2}{15}. \end{align*} However, the correct answer is, $\frac{4}{9}$. What did I do wrong?

Side doubt: Even though the first experiment (truth and lying) is different from the second experiment, can we still apply total probability theorem? In my book the dependent experiment lies inside the sample space associated with the mutually and exhaustive events.

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marked as duplicate by r.e.s., Henning Makholm, Graham Kemp probability Nov 20 '17 at 8:47

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ Can you please put all relevant information in the body of your post? Now you have an important part of that in the title. $\endgroup$ – Bram28 Nov 15 '17 at 3:17
  • $\begingroup$ In "$A$ speaks $4$ out of $5$ times", do you means "speaks the truth"? $\endgroup$ – madprob Nov 15 '17 at 3:23
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    $\begingroup$ Unless the book added some very specific language that is not in your question (something like, "We asked him if the roll was a 6 and he said yes"), your book is wrong. Your attempt also is wrong--it is quite possible the person rolls some number that is not $6,$ then the person lies and says it was a $6.$ So it is very plausible that $P(E_3\mid E_2) > 0,$ contradicting what you wrote. $\endgroup$ – David K Nov 15 '17 at 4:16
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    $\begingroup$ This question is essentially a duplicate. $\endgroup$ – r.e.s. Nov 15 '17 at 4:26
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    $\begingroup$ 2/15 are the odds of him actually rolling a 6 and then reporting it truthfully. But the question doesn't ask for that. It focuses only the subset "person reports a 6" and wants to know how probable the truth in this case is. And since the person is honest 4 out of 5 times, then these are exactly the odds of the 6 being correct. So, what G Tony Jacobs said. $\endgroup$ – Alex Nov 15 '17 at 15:02
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There is some missing information, namely: When he lies, does he lie at random, or with some pattern? In particular, if the die shows up as a $2$, for example, is he equally likely to report $1,3,4,5$ or $6$ as the outcome?

If we assume that all of these are equally likely, then the probability that he reports a $6$, given that a number other than $6$ was rolled, would be $\frac15\cdot\frac15=\frac1{25}$. (The first $\frac15$ is the probability of lying; the second $\frac15$ is the probability that "$6$" is the chosen lie.) On the other hand, the probability that he reports a $6$ given that a $6$ was rolled, is $\frac45$. Thus, when he reports a $6$, the probability that a six was actually rolled is:

$$\frac{\frac16\cdot\frac45}{\frac16\cdot\frac45 + \frac56\cdot\frac1{25}}=\frac45$$

If that's not the answer, then there is some other assumption at work here that we don't know about. Perhaps he consistently lies by reporting “$6$” when non-$6$ numbers are rolled? If that's the case, then our calculation changes:

$$\frac{\frac16\cdot\frac45}{\frac16\cdot\frac45 + \frac56\cdot\frac15}=\frac49$$

The problem should really be asked in a way that makes it clear how lying actually works in this situation. If we ask the question, "is the outcome a $6$?", then we're in the latter case, because then it's clear exactly what a lie would look like.

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  • $\begingroup$ no info like this in book $\endgroup$ – user501655 Nov 15 '17 at 4:32
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    $\begingroup$ Well, it wouldn’t be the first time an ill-posed problem is published in a book. $\endgroup$ – G Tony Jacobs Nov 15 '17 at 5:15
  • $\begingroup$ can u tell me about my side doubt $\endgroup$ – user501655 Nov 15 '17 at 6:46
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    $\begingroup$ There is no reason that the theorem would not apply here, but we are still missing information. We should be told, for example, whether lying or not is independent of the die roll. In all the answers I see posted here, we’re assuming it is. $\endgroup$ – G Tony Jacobs Nov 15 '17 at 6:56
  • $\begingroup$ But since the second event doesn't fall in to the sample space of first experiment ,will it be still valid(bc in my book ,during the derivation they showed the event which is associated with all other events of the sample space having a n intersection with all other events where the sample space enclosed all other events(partition of events).Does this event satisfy condition like that. $\endgroup$ – user501655 Nov 15 '17 at 7:08
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There are two serious problems with your work:

  • $P(E_3|E_2)$ is probably not zero. You don't provide the details about how $A$ lies, but it's certainly plausible, for example, that $A$ rolled a $4$ and lied saying he rolled a $6$.
  • You were asked to find the probability of the event "$A$ rolled a $6$", but that event doesn't even appear in your model! You tried to find $P(E_3)$ instead.

As an aside, your answer key seems to assume $A$ lies in odd way — that when he lies about the die roll he will always say "I obtained a $6$" if possible, rather than any of the other four ways to lie.

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  • $\begingroup$ Kay got that so basically when he lies he can say it is 6 for any other number that has also come, right $\endgroup$ – user501655 Nov 15 '17 at 12:17
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Let $D_{6}$ denote the event that the die lands on a $6$ and $R_{6}$ denote the event that the person reports that it landed on a $6$. You know that \begin{align*} P(R_{6}|D_{6}) &= 0.8 \\ P(D_{6}) &= 6^{-1} \end{align*} In order to obtain the correct answer, you also need to assume that $P(R_{6}|D_{6}^{c})=0.2$, that is, if the person lies and doesn't obtain a $6$, then he says he obtained a $6$.

Since you wish to determine $P(D_{6}|R_{6})$, which is an inverse probability of $P(R_{6}|D_{6})$, a strategy that usually works is applying Bayes' theorem. \begin{align*} P(D_{6}|R_{6}) &= \frac{P(D_{6})P(R_{6}|D_{6})} {P(D_{6})P(R_{6}|D_{6})+P(D_{6}^c)P(R_{6}|D_{6}^c)} \\ &= \frac{6^{-1} \cdot 0.8}{6^{-1} \cdot 0.8 + 5 \cdot 6^{-1} \cdot 0.2} = \frac{4}{9} \end{align*}

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    $\begingroup$ I have not reached Bayes theorem,still stuck practicing problems of probability theorem and just randomly tried to solve this problem bc it looked like total probability question.Also is there something wrong with my attempt $\endgroup$ – user501655 Nov 15 '17 at 3:43
  • $\begingroup$ also can u comment on my side doubt $\endgroup$ – user501655 Nov 15 '17 at 3:48
  • $\begingroup$ Ops... this solution is actually incorrect. $\endgroup$ – madprob Nov 15 '17 at 3:54
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    $\begingroup$ We don't know that $\mathsf P(R_6\mid D_6^\mathsf c)=0.2$. If the die shows 5, then any report of $1,2,3,4,$ or $6$ is a lie. $\endgroup$ – Graham Kemp Nov 15 '17 at 3:54
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    $\begingroup$ Yes! I'll just delete this. $\endgroup$ – madprob Nov 15 '17 at 3:54
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Several answers already show that the correct solution is 4/9, but I want to answer the question "What did I do wrong?"

Let's look at your calculation. Even if the person would tell the truth all the time, your probability would be 1/6. But if he is telling the truth, he MUST have thrown a 6.

You forgot to take into account that the person already told you that he has thrown a 6. Instead, you calculated the beforehand probability that the person will come up to you, telling you truthfully that he has thrown a 6. This is 2/15.

(You can check this easily: The same probability holds for any other number, so the probability that he will tell you any number and speak the truth is 12/15 or 4/5, which is the probability of him speaking the truth.)

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I see that the question has been edited a lot; I'm answering the current version (revision 7), which reads like this:

"A person, A, speaks the truth 4 out of 5 times. The person throws a die and reports that he obtained a 6. What is the probability that he actually rolled a 6?"

A rolls the die, looks at the result, and then states that the result is a 6. Since A knew what the roll was when stating "it is a 6", and it is given that A tells the truth 4/5 of the time, there are only two alternatives:

  1. A told the truth, with probability 4/5. The roll is a 6.
  2. A lied, with probability 1/5. The roll is not a 6.

Therefore the probability that A rolled a 6 is 4/5, and the 4/9 you report as "the correct answer" is wrong. There is no reason to start by considering what the roll is, because if A is lying, the roll cannot be a 6.

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  • $\begingroup$ what do u mean by-"the roll as a prior" $\endgroup$ – user501655 Nov 16 '17 at 0:43
  • $\begingroup$ Sorry for the jargon... I meant, there's no reason to consider cases of A's answer based on what the roll is. (Because the probability of lying is given, and independent of the roll.) $\endgroup$ – alexis Nov 16 '17 at 9:04
  • $\begingroup$ thnx,i get it now $\endgroup$ – user501655 Nov 16 '17 at 9:40
  • $\begingroup$ This is not the right way to think about it. To see that, change the numbers: A tells the truth 1/2 of the time, and they play a lottery with a 1/1 million chance of winning. If they say they won, by this logic, the probability that they really won is 1/2. Do you see how that's wrong? There are two ways that could report a win: winning and telling the truth, or losing and lying. These two events are not equally likely. $\endgroup$ – G Tony Jacobs Nov 16 '17 at 13:12
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There are 4 outcomes: speaks and 6, does not speak and 6, speaks and not 6, and does not speak and not 6. We know that he speaks so there are 2 outcomes: speaks and 6, and speaks and not 6. The probability of the first one is 1/6 and the other one is 5/6. So it is 1/6.

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    $\begingroup$ I think this answer is based on an earlier edit of the problem. $\endgroup$ – G Tony Jacobs Nov 15 '17 at 4:01
  • $\begingroup$ @GTonyJacobs Admittedly, the question is incoherent even with the edits so far. That is all the more reason not to answer at all. $\endgroup$ – David K Nov 15 '17 at 4:04
  • $\begingroup$ Yes, I tried to at least address with my answer what information is missing... $\endgroup$ – G Tony Jacobs Nov 15 '17 at 4:05
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Your error would appear to be in P(E3|E2)=0

You are missing out the fact that the user may report a 6 being throw when in fact one has not

In fact your use of E3 is not helpful as it covers two scenarios.

What you are interested in is

User throws a six AND tells the truth User lies AND reports that they threw a 6.

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  • $\begingroup$ thanx..this helped a lot $\endgroup$ – user501655 Nov 16 '17 at 1:47