If 2 expert pieces of advice are independent then both being wrong is lower than individual advice being wrong but isn't both being right also lower? I was watching this Coursera video on independence which ended up mentioning ensemble. I am trying to wrap my head around probability.
If we ask 2 experts whether or not a startup would fail, and we consider the case that they are independent then the probability of both being wrong is lower than either expert 1 or 2 being wrong individually.

Now on the flip side doesn't it mean that, if they are indeed independent isn't the probability of both being right lower than the probability that each is right individually? How is this desirable? It feels like a tradeoff. I feel hoodwinked.
 A: Intuition by using more experts
The case of 2 experts is a bit difficult (we explain later why).
It becomes more intuitive when you consider a choice based on a majority in multiple expert advice who are making a binary choice (that is: we count how many experts choose option A and how many choose option B and decide based on what the majority of experts call).
In this example, we can consider the probability of our decision being right (the decision based on the majority of experts) by the probability that the number of experts that are right is more than half (if the number of experts being right is more than half then our decision, which is based on the majority, is right).
Let's see how the number of experts that are right is distributed when the probability of an expert being right is independent and let the probability be $p=0.6$.

In the image, we highlighted the cases when more than half of the experts are right (half is emphasized by a dotted vertical line). You can see that this probability increases. So when we use the decision of the majority of the experts then the probability of being right increases and the probability of being wrong (which mirrors being right) is decreasing (provided p>0.5 for a single expert).
The case of 2 experts
The computations are correct.

*

*Indeed the probability of two false experts decreases. This is not wrong.


*indeed on the flip side the probability of two correct experts is also decreasing.
The problem with the case of 2 experts (and any other even number) is that there is also a probability of a tie. The case that one expert is saying something (false/right) that is different from the other expert (right/false).
So in this situation, it is unclear what to do with that tie situation (you could flip a coin) and you get indeed the situation where the probability of a majority of experts being right (which means both have to be right) is decreasing. So maybe the example with 2 experts is not so great. We can do better with 3 experts.
The case of 3 experts
So now the distribution for the number of experts being right is:
 number right      probability
 0                 1*(1-p)^3*p^0
 1                 3*(1-p)^2*p^1
 2                 3*(1-p)^1*p^2
 3                 1*(1-p)^0*p^3

And the probability that more than the majority of experts are right (2 or 3) is
$$P(\text{majority right})= \underbrace{(3p-2p^2)}_{\text{$>1$ if $p>0.5$}} p$$
And the probability that the minority is right (0 or 1) is the opposite.
So now you don't have the situation that you look at all right versus all wrong (probabilities which are indeed both decreasing). And maybe now the example makes more sense.
A: Yes, the probability that they are both right is also lower, as long as the individual probabilities $P(E_1^c)$ and $P(E_2^c)$ are less than one, i.e. $P(E_1)$ and $P(E_2)$ are more than zero.
To show why this is useful, we follow the video, and assume we are in the situation where the experts agree with each other. The probability that they are right is multiplied by a factor of $P(E_2^c)$, and the probability that they're wrong is multiplied by $P(E_2)$. If $P(E_2) < 1/2$, then the probability of their being right is now larger, relative to the probability of their being wrong, since $P(E_2^c) > P(E_2)$. The total probability of the experts agreeing is decreased, but the probability of their being right if they do agree is increased.
Again, this ignores the case where the experts disagree, which you need to handle if you want to show the ensemble performs better in general. For the given example, where the experts are giving binary judgements so they can't both be wrong, we could just go with the expert with the higher probability of being right, or flip a coin if they're equally likely. This is clearly at least as good as the case where we only consult one expert, since the highest probability of being right is at least as large as the first expert's probability of being right.
Handling this case also lets us ensure that $P(E_2) \leq 1/2$, since if $P(E_2) > 1/2$ we can swap that expert's opinion to be the opposite of what they said.
A: Yes, because
$$(1-P(E_1))(1-P(E_2))\leq 1-P(E_1)$$
It's natural because $N$ people agreeing upon something is lower than one people having the idea. In reality, of course, expert opinion's aren't independent given the case.
