A man is known to speak truth 3 out of 4 times. He throws a die and reports that it is a six. The probability that it is actually six is
1. 1/8
2. 2/8
3. 3/8
4. 1/2

I think that the answer is indeed 3/8 (of course in case every time he lies he says that he got a 6, if it's not the case then the answer is simply 3/4), but the above solution is not right.

P=Favorable outcomes/Total # of possible outcomes.

Favorable outcome is that it's actually six and he's telling the truth \(=\frac{3}{4}*\frac{1}{6}=\frac{1}{8}\)

Total # of possible outcomes is: either it's six and he's telling the truth OR it's not six and he's telling the lie \(=\frac{3}{4}*\frac{1}{6}+\frac{1}{4}*\frac{5}{6}=\frac{1}{3}\)

\(P=\frac{\frac{1}{8}}{\frac{1}{3}}=\frac{3}{8}\).
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thanks bunuel for correct approach

Bunuel...I think the answer is still incorrect. I had a lengthy fight with a friend on this (admittedly he won ) and though the mathematics on this question is right logically the answer still feels should be 3/4 (though it isn't an option here). Applying probability....
p= (Number of time when it is actually 6)/(Number of times when it is and isn't a 6). And according to his truth factor of 3/4 he should have said 6 exactly 3 times out of 4. Still can't solve it mathematically using your equation though

So Ian, even in your 'all of that said...' part the underlying assumption that he claims (lies) he rolled a 6 is incorrect as I understand. A plausible lie should have multiplied the total outcomes by 1/5 and an improbable assumption should have by 1/infinity

I think in this case, this does actually hold through but you should be very careful in making generalizations like this one above. They often depend on assumptions which may not be obvious.

For instance, if let's say you were to consider that the dice is loaded and that the probability of it showing a 6 is 1/2 and any other number is 1/10. Then you will realize the answer to this problem becomes 15/16 instead of 3/4. So it is clearly hinged on a uniform probability of each outcome.
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bunuel...this question is strikingly similar to ones solved by Bayes theorem...is the knowledge of bayes theorem required for gmat?
btw +1 for solving the cheeky question!

can you show your work for the loaded dice variant

Responding to a pm:


As discussed above, the question is not very clear. We are not sure how he lies. If we have to make an assumption, we might assume that every time he lies, he calls it a 6 or we may assume that he lies consistently i.e. he will use every number equally while lying. So say if he throws a 1 twenty times, he will lie 5 times. When he lies, he will call it a 2 once, a 3 once, a 4 once, a 5 once and a 6 once.

Assuming he lies consistently:
So consider a case when he throws the die 24 times. He gets a 1 four times, 2 four times, 3 four times and so on since all are equally likely.
Of the 4 times he throws a 1, he lies once. Say he calls it a 2.
Of the 4 times he throws a 2, he lies once. Say he calls it a 3.
Of the 4 times he throws a 3, he lies once. Say he calls it a 4.
Of the 4 times he throws a 4, he lies once. Say he calls it a 5.
Of the 4 times he throws a 5, he lies once. Say he calls it a 6.
Of the 4 times he throws a 6, he lies once. Say he calls it a 1. He says the truth in other 3 instances and calls it a 6.

We have to find the probability that when he says it is 6, it actually is a 6.
So 4 times he says it is a 6. Out of these 4 times, it is actually 6 in 3 instances.
So the probability that it actually is 6 (when he says it is 6) is 3/4.


Assuming he calls it a 6 every time he lies

Of the 4 times he throws a 1, he lies once. Say he calls it a 6.
Of the 4 times he throws a 2, he lies once. Say he calls it a 6.
Of the 4 times he throws a 3, he lies once. Say he calls it a 6.
Of the 4 times he throws a 4, he lies once. Say he calls it a 6.
Of the 4 times he throws a 5, he lies once. Say he calls it a 6.
Of the 4 times he throws a 6, he lies once. Say he calls it something else. He says the truth in other 3 instances and calls it a 6.
So 8 times he says that it is a 6. Out of these 8 times, it is actually 6 in 3 instances.
So the probability that it actually is 6 (when he says it is 6) is 3/8.
You can calculate this as given by Bunuel above.


Where did you go wrong:
You are finding the probability that he throws a 6 AND then calls it a 6 (P(Throws a 6) - 1/6; P(Calls it 6) - 3/4; Does both = 1/6 * 3/4 = 1/8).
But that it not the question. You have to find the probability that it is actually 6 (after he has said that it is a 6). So we need to find
P(No of time it is a 6 when he calls it a 6)/(Total number of time he calls it a 6)
It is already given that he calls it a 6. Now we have to figure the probability that it actually is a 6.


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