Re: Probability of making a choice by a person, on the basis of earlier made choices?

On 27 May, 22:38, Paul Rubin <ru...@xxxxxxx> wrote:
Marcin wrote:
We are asking a man a few questions. After every response we can
check, whether he told the truth or deliberately lied. Lets say, at
the beginning, the probability, that the man will lie and the
probability the man will tell the truth are the same (0.5). If the man
answers the first question and lies, the probability that he will also
lie when asking a second question is (I guess) already greater than
0.5.

Why?

But, if the man will answer first five questions frankly, the
probability, that in the reply to sixth question he will also tell the
truth is already much higher than 0.5.

Why?

But how high is the propability
exactly, and how do you calculate that?
(sorry for english :) thx

You need more information.

If the man randomly and independently decides whether to answer each
question truthfully, his probability of a lie on question 6 is
unaffected by how many lies he told in the first five questions.  In
this case, the proportion of lies detected among questions whose answers
were checked converges to his probability of lying on any given question
as the number of questions checked increases (Weak Law of Large Numbers).
...

Let's stick with this version of the problem. The frequentist
response "ask a lot of questions" may not be practicable.

Alternatively you could take a Bayesian approach. With a symmetric
proper prior distribution, for example Beta(k,k) for some k>0, you
could say the probability that the first statement is true is 0.5.
Once there had been t true and f false statements this would modify
the posterior distribution, in the example to Beta(k+t,k+f) which
would make the posterior probability of the next statement being true
(k+t)/(2*k+t+f).

.

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