Sunday, May 20, 2012

Brainteaser with Montecarlo simulation

Here's a brain teaser.
A man speaks the 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?
Actually we know that the man has said he either got a 6 or he didn't. So the probability that we are interested in is $P(A|B)$, where the events A and B are A: "The man said 6", B: "He actually got a 6". This conditional probability is $P(A|B) = \frac{P(A,B)}{P(B)}$.

We work with the assumption of independence, so $P(A,B)=P(A)P(B)=\frac{3}{4}\frac{1}{6} = \frac{3}{24}=\frac{1}{8}$.

We decompose the event B in the intersection with A and the intersection with the complementary of A (being the last one defined as "he did not get a 6"), so that $P(B)=P(B,A) + P(B,\bar{A})$. We already have the first one so we compute $P(B,\bar{A})=\frac{1}{4}\frac{5}{6}=\frac{5}{24}$, making $P(B)=\frac{3}{24} + \frac{5}{24} = \frac{8}{24} = \frac{1}{3}$.

Therefore, $P(A|B) = \frac{P(A,B)}{P(B)} = \frac{\frac{1}{8}}{\frac{1}{3}}=\frac{3}{8} = 0.375$

Now we simulate the mentioned events in R. See that we generate the number of times the man tells the truth independently of the times he gets a six. We then count the times that a real 6 coincide with the man telling the truth in $num$. In the denominator $den$ we count the previous amount plus the number of times he says 6 but does not get a dice with a 6. This means that, among all the times the man reports a 6 (both true or false), we select only those times they are true.


lie=c(lie, rep(1,N*1/4))

den=sum((dice==6)*(lie==0)) + sum((dice!=6)*(lie==1))


> res
[1] 0.3739862
Note that the output is very close to the theoretical result of 0.375.

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