What are the odds of each fighter winning, given this information?

Let's say one wishes to place a bet on one of two fighters. You are given the following information:

(1) Fighters taller than their opponents by 3% will more likely win. (2) Fighters younger than their opponents by 3 or more years will more likely win.

You are also given this information:

Statement (1) is 58% accurate. Statement (2) is 60% accurate.

If Fighter1 is 4% taller than Fighter2, and Fighter2 is 4 years younger than Fighter1, who would you bet on, and why?

To be clear, I'm really interested in a more general case, where you have $n$ statements with $x_n$% accuracy. The above example is a bit easy. :)

Allow me to ask this question from another angle:

Let's say you have n people and each of them gives a prediction on who will win the fight between two fighters. Each person's prediction can be considered to be $x_n$% accurate based on their previous predictions. How would you consolidate all n predictions to get a more accurate prediction? Would you consolidate them at all, or just go with the best predictor? What if everyone disagreed with the best guy?

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how is the above example easy? Taller and younger fighters are more likely to win, but by how much? Does 4% taller stand a better chance than 3% or is it constant past that threshold? The problem doesn't seem well posed as it's currently written. – Robert Mastragostino Jul 12 '12 at 18:22
Assume that it's constant past that threshold. – Korgan Rivera Jul 12 '12 at 18:23
When you say Statement(1) is 58% accurate, do you mean that "fighters taller than their opponents by $\geq$3% win 58% of the time"? Because this is different than, say, "58% of taller fighters win 1% more often" – Robert Mastragostino Jul 12 '12 at 18:27
I mean that "fighters taller than their opponents by ≥3% win 58% of the time." In other words, out of 100 fights, 58 fights were won by opponents who were ≥3% taller. – Korgan Rivera Jul 12 '12 at 18:34
By that way of counting, the statement "after getting heads once, the next flip of a fair coin is more likely to be tails" would be considered 50% accurate. I would rather call it 0% accurate. – Henning Makholm Jul 12 '12 at 19:15

We don't have enough information. Let's say for simplicity we divide the universe of fighters into four groups: 1 (short and old), 2 (short and young), 3 (tall and old), 4 (tall and young), and $p_{ij}$ is the probability of a fighter from group $i$ winning against a fighter from group $j$. We also should know what fraction of all fights involve a fighter from group $i$ against a fighter from group $j$: say $w_{ij}$ (where $w_{ij} = w_{ji}$ and $\sum_{i \le j} w_{ij} = 1$). Then statement (1) says $p_{31} w_{31} + p_{32} w_{32} + p_{41} w_{41} + p_{42} w_{42} = 0.58 (w_{31} + w_{32} + w_{41} + w_{42})$, and statement (2) says $p_{21} w_{21} + p_{23} w_{23} + p_{41} w_{41} + p_{43} w_{43} = 0.6 (w_{21} + w_{23} + w_{41} +w_{43})$. But those are just two equations in six unknowns.