How to merge several predictions into a single decision Disclaimer: I am not actually trying to predict the football match outcomes and I am not betting. Football is only taken for illustration.
Given there are several different independent predictors that predict  outcome of the same binary event, each with its own confidence defined by the probability of their terminal probability variable taking the given value.
What are the common ways, or how would you merge those recommendations into a single decision outcome or a single variable probability that predicts the outcome more accurately than each of the predictors individually?
To illustrate my question, here is an example.

*

*Predictor A predicts the win of the team t in a football match m with probability 0.6.

*Predictor B predicts the win of the team t in a football match m with probability 0.55.

*Predictor C predicts the win of the team t in a football match m with probability 0.71.

How would you merge those inputs into a single probability of an event outcome (team t winning the match m) to base your decision for whether to place a bet or not in order to obtain better probability of overall prediction success?
Additional Context
To help community readers with intentional ambiguities in the question and in an attempt to qualify as a valid question for the Mathematics StackExchange, here are some of my "thoughts" even though I'd really like to ask you for your suggestions on this topic.
I call "predictor" a process that calculates a probability of a given probability variable taking a given value, e.g. on the picture:
PA(E=e|A,B=b,C,...)
PB(E=e|D=d,F,G=g,...)
PC(E=e|H,I=i,J,...)
I am looking at possible strategies of making a decision based on the given input.
P(E=e|EA,EB,EC)
What is the additional information that I need to know about predictors A, B, C in order to improve the probability of successful overall prediction?
I am sure that the merging of multiple independent probability predictions for the same event is a very common situation in practice in sophisticated systems. How are they dealing with this situation?
My current "straightforward" guess would be to measure the a-posteriori probability of each of the predictors e.g. basing on historical data to determine how often their prediction P(E) matches the real event outcomes and use that as the weight during "merging", e.g. as weighted average.
Will that improve the quality of the eventual prediction?
Is that how the tasks of this class are being solved in practice (e.g. gaming, health, other ML domains)?
Where can I read about this in greater detail?

 A: "Weighting" the probabilities of different pundits requires you to understand mutual information. I'll just give some simple illustrative examples of mutual information and a reference in this answer.
Suppose we have probability of winning $P(W)$, and some pundits who try to guess the outcome. I'll calculate the mutual information between their guesses and the outcomes.
Pundit A (Mr Average) predicts completely at random, by throwing a coin. Because the probability of a win and the probability of Mr Average predicting a win are independent, $p(w,a)=p(w)p(a)$, where $a$ is the guess of Mr Average, the mutual information $I(W,A)$ between Mr Average and the actual outcome is given by
$$
I(W,A)=\sum_{w,a} p(w,a)\log_2{p(w,a)\over p(w)p(a)}\\
=\sum_{w,a} p(w,a)\log_2(1)\\
=0.\\
$$
Pundit C (Mr Clever) will predict completely accurately the outcome of any event. The "mutual information" $I(W,C)$ between Mr Clever and the actual outcome is
$$
I(W,C)=\sum_{w,c} p(w,c)\log_2{p(w,c)\over p(w)p(c)}\\
=\sum_{w} p(w)\log_2(p(w))\\
=1
$$
since $p(w|c)=1$ and $p(\bar{w}|\bar{c})=1$, where $\bar{w}$ indicates a loss.
Pundit B (Mr Better-than-average) predicts with an outcome of 0.75 of being right, and his mutual information
$$
I(W,B)=\sum_{w,b} p(w,b)\log_2{p(w,b)\over p(w)p(b)}\\
=-\sum_{w}p(w)\log_2(p(w))+\sum p(w,b)\log_2{p(w|b)}\\
=1+2\times0.5\times(0.25\log_2(0.25)+0.75\log_2(0.75))\\
\approx 0.2.
$$
Incidentally, pundit D (Mr Doofus) predicts worse than average with a 0.25 chance of being right, and his mutual information is exactly the same,
$$
I(W,D)\approx 0.2,
$$
since if Mr Doofus guesses a win, we bet on a lose, and vice-versa.
There is a more extensive discussion of the relation between mutual information and betting on sports outcomes in Cover and Thomas's book "Elements of Information Theory", chapter 6.
A: Maybe a weighted average of the probabilities might work. Give A, B, C weights $w_1$, $w_2$, $w_3$, (positive numbers ) and then say the probability of a win is
$$\frac{1}{w_1 + w_2 + w_3} (w_1 \cdot .6 + w_2 \cdot .55 + w_3 \cdot .71)$$
Now, $w_1$, $w_2$, $w_3$ would be determined from some other data. Say the more we trust  $A$, the bigger would be $w_1$ compared to the other $w_i$. Just an idea...
