What Is The Probability?
A man is known to speak truth 3 out of 4 times. He throws a 6-sided die and reports that it is a six. If he was lying, he would randomly say it's one of the other 5 values.
Find the probability that it is actually a six.
The answer is 0.75.
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3 solutions
Why is it 1/8?
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What is the probability that the man rolls a 6 and he truthfully says he rolled a 6?
Wasn't that what you calculated? 6 1 × 4 3 = 8 1 .
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So if you use Bayes theorem here , will it still be 1/8??
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@Nashita Rahman – This solution does not involve Bayes theorem at all.
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The table calculates P( man rolls n and man reports m) from the rule of product like I did above.
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From there, you can calculate P(man rolls 6 and man reports 6), and P(man reports 6).
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From there, you can calculate the conditional probability P(man rolls 6|man reported 6) as P( man rolls 6 and man reported 6) / P(man reported 6)
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@Calvin Lin – Yes but in this question you have to use Bayes theorem and using that the answer is coming out to be 3/8.
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@Nashita Rahman – We don't have to use Bayes theorem.
Your usage of Bayes theorem is flawed because you made an assumption about how the lying works. In particular, instead of calculating "P(E|B)=Probability that the man reports that six occurs given that six has actually not occurred", you were calculating "Probability that the man reports that six occurs given that six has actually not occurred AND that the man reports six". IE If you look at your working, you went one step further to claim "it is equal to the probability that (he didn't roll a 6) and he is lying". However, we've been saying that "There are many ways to lie, like by saying it is an elephant".
We've been trying to get you to calculate the former, and not the latter. (I agree that the probability of the latter is indeed 5/24).
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@Calvin Lin – I am totally confused now. I need to think hard and more and once I am convinced I'll come back to this. Till then let's see how the community responds to this !
@Calvin Lin – Btw we can also use Bayes theorem here !! It will work that's what I think.
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@Nashita Rahman – Yes, we can use Bayes theorem and it does work. The correct probability for P ( E ∣ B ) is 1/24, and not 5/24 like you are claiming. IE As such, the second term in the denominator is not 5 / 6 × 1 / 4 , but 5 / 6 × 1 / 4 × 1 / 5 .
You seem to be unconvinced of that, regardless of what I'm saying. That is why I am using another (non Bayes theorem) approach to show you that the answer is indeed 3/4.
According to the question , 2 cases are there. Firstly it's either he is lying or telling the truth. Why do u need to consider all six?
Let E be the event that the man reports that six occurs in the throwing of the die and let A be the event six actually occurs and B be the event six actually does not occur.
P(A) = Probability that six actually occurs = 1/6
P(B) = Probability that six actually does not occur = 5/6
Using conditional probability ,
P(E|A) = Probability that the man reports that six occurs given that six has actually occured = Probability that he is telling the truth = 3/4
P(E|B) = Probability that the man reports that six occurs given that six has actually not occured = Probability that he is lying = 1-3/4=1/4
Using Bayes' Theorem ,
P(A|E)= Probability that six has occured given that the man has reported six = P ( A ) P ( E ∣ A ) + P ( B ) P ( E ∣ B ) P ( A ) P ( E ∣ A ) = 1 / 6 × 3 / 4 + 5 / 6 × 1 / 4 1 / 6 × 3 / 4 = 3/8 = 0.375
As expressed by others, you assumed that "if the dice shows a 1 and the man lies, then the man has to say that it is a 6". This is not a fair assumption. What is reasonable is that he "would randomly report one of the other 5 values", which I've added to the problem.
As such, the second term in the denominator is not 5 / 6 × 1 / 4 , but 5 / 6 × 1 / 4 × 1 / 5 .
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The question is not saying the man is lying . The question is asking if the man say it's a six so what is the probability that it is actually a six. Using logic and Bayes theorem , the answer has to be 3/8.
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That is precisely our point. What is the probability that the man says it is a 6, given that it is not a 6?
- First, the man has to roll a not-6.
- Second, the man has to decide to lie.
- Third, the man has to choose to say it is a 6 (as opposed to any other valid value that makes it a lie)
You are ignoring the third step.
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@Calvin Lin – But in the question it is given the he says it's a six. The question is saying to find the probability that it is actually a six. So we have the probability that it's actually a six so he is telling the truth in this case. And then we proceed . Sir please read my solution again.
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@Nashita Rahman – What is the probability that the man says "It is a 6", given that "he rolled a 1"?
According to your solution, you are saying that the probability is 1/4.
If so, what is the probability that the man says "It is a 5", given that "he rolled a 1"?
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@Calvin Lin – It is 1/4 because he is lying and it is given that the probability that man lies is 1/4. I think the confusion here is that you have considered that he does not know the outcome which is not the case . The man knows the outcome of the fair dice and he can either lie or say the truth about the outcome. In the question , he is saying that it is a six and we have to find the probability that it is actually a six so we have to consider the probability that he is telling the truth.
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@Nashita Rahman – Given that the man rolled a 1, what is the probability that:
- He says "It is a 1".
- He says "It is a 2".
- He says "It is a 3".
- He says "It is a 4".
- He says "It is a 5".
- He says "It is a 6".
- He says "It is an elephant".
Note that these probability should sum to 1.
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@Calvin Lin – Nashita, please answer these 7 probability questions before we proceed.
@Calvin Lin – Considering whether the man is telling the truth or lying , the probability that it is 1 is 1/6 and it is not 1 is 5/6. To make the process not lengthy we take only 2 situations and not all six situations. Then we use Bayes theorem.
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@Nashita Rahman – But "it is not 1" does not imply that "It is 6". "It is 6" is a strict subset of "It is not 1". IE As expressed at the start, you assumed that "if the dice shows a 1 and the man lies, then the man has to say that it is a 6".
"It is not 1" has a probability 1/4 of happening.
"It is 6" does not have a probability 1/4 of happening. When restricted to "he would randomly say it's one of the other 5 values", it has a probability 1/20 of happening.
In short, if you're calculating the probability that "man did not roll 6 and man says it is 6", it is not equal to the probability that "man did not roll 6 and man lies".
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@Calvin Lin – There can be such a situation but the question is saying he reports it's a six sof no matter what the outcome is when the dice is rolled once , he reported it to be six. So 2 cases here either he is lying or telling the truth and not 6 cases here.
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@Nashita Rahman – "no matter what the outcome is when the dice is rolled once , he reported it to be six".
By saying the above statement, then his P(honest) = 0 instead of 3/4.
If he's going to disregard the result of the rolled dice and intentionally set his mind on reporting 6 from the get go, why are we caring and using that 3/4 in the calculations, then?
@Calvin Lin – And also the question is clearly saying that ithe man reports that it is six. So whatever the outcome actually is , he reports it to be six .
If the person always report that it is six then we will divide this in two ways
(1)The die will give 1 , 2 , 3 , 4 , 5 and he will report that it is six.
Let D i indicates that i appears on the die and p ( T ) indicates the probability of saying truth.We will find the probability that die will give other than six but he will say six appears which is false
p ( D i ∩ T c ) = p ( D i ) p ( T c ) = p ( D i ) ( 1 − p ( T ) ) = 6 1 × 4 1 = 2 4 1 .As 1 ≤ i ≤ 5 .So total probability is 5 × 2 4 1 = 2 4 5
Case(2) The die will give actually six and he will report six which is true.The probability is p ( D 6 ∩ T ) = p ( D 6 ) p ( T ) = 6 1 × 4 3 = 8 1
So, the probability that he reports six and actually six appeared is p = 8 1 + 2 4 5 8 1 = 8 3
Everytime he lies, he can lie with one of 5 other options. If the die shows 1 then he has 5 options to lie with. So probability of lying and telling 6 is (5/6) (1/4) (1/5) = (1/24)? So the probability is simply 3/4?
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Yes! That is exactly what I thought. Why should he always lie that the number is 6. He can tell its any one of the 5 remaining numbers.
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@Siva Bathula @saharsh rathi , the question is not saying that he lies it's always a six instead the question is asking that the man says it's a six and you have to find the probability that it is actually a six. If you Still cannot understand , just go through my solution , I have just uploaded one !!!
[This solution is incomplete.]
Let's consider the probability that the man rolls n and reports that it is m .
Fill in the probability of each of the other squares. Note that all of the probabilities should sum to 1, and that each column should sum to 1/6.
Hence, the probability that it is actually a six, given that he reports it is a 6, is ...