conditional probability calculation confusion Suppose I have 3 friends, each of them are independent and each of them have the probability of p to tell the truth about something (e.g. company, weather, stock, etc.).
Suppose something happened (e.g. company is holding a party other than not holding a party, weather is good other than bad, stock is rising other than drop, etc.), 
I want to calculate what is the probability 2 of them telling truth, and 1 of them telling wrong?
I think the answer is 3*p*p*(1-p), correct?
But what is the probability of something is true, if 2 of the friends telling me truth, but 1 of friends telling me it is not truth?
P (true | T, T, F) = P (T,T,F|true) * P(true) / P(T,T,F)
demominator P (T,T,F) = P(true) * P(T,T,F|true) + P(false) * P (T,T,F|false)
Suppose prior P(true) = 1/2 and P(false) = 1-1/2 = 1/2, then it become
P (T,T,F) = x*3*p*p*1/2 + 1/2*3*p*(1-p)*(1-p)
numerator = x*3*p*p*(1-p)1/2
final result = p

Suppose I consult more friends, but still they have the same ratio telling the true or not, so for 6 friends, 4 of them telling truth and 2 of them telling false
The result becomes,
P (true | T,T,T,T,F,F) = P (T,T,T,T,F,F|true) * P(true) / P(T,T,T,T,F,F)
numerator = C(2,6)*1/2*p^4*(1-p)^2
demominator = C(2,6)*1/2*p^4*(1-p)^2 + C(2,6)*1/2*p^2*(1-p)^4
C(2,6) means select 2 from 6, which is 15
final result is p^2 / (p^2 + (1-p)^2)

If I want to compare if asking 3 friends has higher probability of truth, or asking 6 friends, I just need to compare between p and p^2 / (p^2 + (1-p)^2),
I did calcualation, which gets as long as (2p-1)*(p-1) > 0, asking more friends, is better, which result in p<1/2, it seems counter intuitive?
My intuitive is, asking more friends is better if any individual friend has high confidence (p>1/2)? Anything wrong in my calculation?
 A: I'm not sure if I'm understanding what you're saying, because of your English and terminology, but it sounds like you're supposing you have $n$ independent binary signals of a single binary event, each of which has probability $p$ of being correct, and you want to compute the probability that the event actually happened. This is equivalent to supposing you have $n$ independent draws from a Bernoulli distribution whose parameter $θ$ is either $p$ or $1 - p$, and you want to estimate $θ$. This formulation makes it obvious what's missing from your scenario: a prior distribution for $θ$. Although we know $θ$ can only have one of two possible values, we need to decide on their relative prior probability in order to get a posterior estimate for $θ$.
So, choose a prior. Let $α ∈ [0, 1]$ such that a priori, $θ$ has probability $α$ to be $p$ and probability $1 - α$ to be $1 - p$. Given the $n$ Bernoulli trials, let $k$ be the observed number of successes. Then by Bayes's theorem,
$$\begin{align*}
&\phantom{=} P(θ = p | X = x) \\
&= \frac{P(X = x | θ = p) P(θ = p)}{P(X = x)} \\
&= \frac{p^k(1 - p)^{n - k} α}
   {(αp + (1 - α)(1 - p))^k (α(1 - p) + (1 - α)p)^{n - k}}.
\end{align*}$$
A: For the 4 of 6 case to be better than the 2 of 3 case,
$\frac{p^2}{p^2+(1-p)^2}>p$
$p^2>p^3+p\cdot(1-p)^2$
$p>p^2+(1-p)^2\space\space\space\space\space\space\space, p>0$
$0>p^2+(1-p)^2-p$
$0>p^2+1-2p+p^2-p$
$0>(2p-1)\cdot (p-1)\space\space\space$,I think you have the product as > 0
for p < 1, divide by the negative quantity (p-1), meaning the inequality must reverse
$0 < 2p-1$
$p>\frac{1}{2}$
