I have a majority voting algorithm where there is a single voter with a probability $0.75$ of being correct and also 6 other voters that have a probably $0.6$ of being correct. Then I can determine the total probability by distinguishing two cases using the binomial distribution.
Either the single voter is correct and at least 3 of the other voters are correct:
$0.75 \cdot \sum_{x=3}^{6} \frac{6!}{6!(6-x)!}0.6^x \cdot 0.4^{6-x}$
Or the single voter is incorrect and at least 4 of the other voters are correct:
$(1 - 0.75) \cdot \sum_{x=4}^{6} \frac{6!}{6!(6-x)!}0.6^x \cdot 0.4^{6-x}$
The sum of those two is the total probability of the majority vote being correct.
Now this is all fine. But now what I would like to do is apply a weight to the single voter so it counts stronger. I now how I can do if there were concrete votes. Just make the vote counter more. But how can I achieve the same thing in terms of probabilities?
Any help is appreciated!
Assuming voters are independent, you start with the probability of the majority being correct of $$0.75 \cdot \sum\limits_{x=3}^{6} \frac{6!}{6!(6-x)!}0.6^x \cdot 0.4^{6-x} + 0.25 \cdot \sum\limits_{x=4}^{6} \frac{6!}{6!(6-x)!}0.6^x \cdot 0.4^{6-x} = 0.75168$$ which is not much better than making voter $1$ a dictator $(0.75)$
You have seven voters:
if you give voter $1$ less than a double weight then you will not affect majority decisions
if you give voter $1$ a double weight (or a quadruple or sextuple weight) then you might get a tie, which is presumably not good news
if you give voter $1$ between a double weight and a quadruple weight then voter $1$ and two others can outvote four other voters so your probabilities become $$0.75 \cdot \sum\limits_{x=2}^{6} \frac{6!}{6!(6-x)!}0.6^x \cdot 0.4^{6-x} + 0.25 \cdot \sum\limits_{x=5}^{6} \frac{6!}{6!(6-x)!}0.6^x \cdot 0.4^{6-x}=0.7776$$ which is slightly better
if you give voter $1$ between a quadruple weight and a sextuple weight then voter $1$ and one other can outvote five other voters so your probabilities become $$0.75 \cdot \sum\limits_{x=1}^{6} \frac{6!}{6!(6-x)!}0.6^x \cdot 0.4^{6-x} + 0.25 \cdot \sum\limits_{x=6}^{6} \frac{6!}{6!(6-x)!}0.6^x \cdot 0.4^{6-x}=0.758592$$ which is worse
if you give voter $1$ more than a sextuple weight then voter $1$ becomes a dictator and the probability is $0.75$ which is worse