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Another Solution for an Unsolvable Problem.

Notes on a problem of daily "household" maths

On the very interesting page Combining Probabilities a classical problem is described. This problem could easily brought up by a non mathematician in every days life and is unfortunately too incomplete to be solved by mathematicians:

Assume you have a event with a binary outcome (e.g. a tennis match or any other sport event where there is no draw). Further assume you have two experts. The first expert A is known to predict 75% correct and the second B still predicts 60% correct. Now there is a event where A predicts a loss, while B predicts a win. What is the probabilty for a win?

Well, the pure math answer is, there probability can be anything between zero and one, i.e. information is incomplete. To demonstrate that, look at the following table, discriminating 8 cases (2 different predictions by A, 2 by B and the real outcome):

Predictions		Real	case
by A	by B	result	probability
win	win	win	p₀
loss	win	win	p₁
win	loss	win	p₂
loss	loss	win	p₃
win	win	loss	p₄
loss	win	loss	p₅
win	loss	loss	p₆
loss	loss	loss	p₇

For those 8 probabilities of the 8 cases we know the usual probability things: p₀ +p₁ +p₂ +p₃ +p₄ +p₅ +p₆ +p₇ = 1, p_i >= 0. and additionally we have two equations, from the cases where the experts predict correct, these have to sum up to the corresponding expert total precision, as a formula: For A to be correct p₀ +p₂ +p₅ +p₇ = a and for B to be correct p₀ +p₁ +p₆ +p₇ = b with a = 0.75 and b = 0.60, just to be more general. This makes 3 equations for 8 unknowns which is far to less. Since p₁ is the probability of a win with the current prediction and p₅ of a loss, we actually want to obtain is the number p₁/(p₁+p₅) That is the conditional probabilty of a win outcome in the case the predictions are as described.

What we have

The focus on the Combining Probabilities page is to explore various assumptions and approaches to fill the gap and make the problem treatable by the mathematical apparatus. The mentioned approaches are:

Symmetry assumptions, which are not enough to make the problem solvable.
Statistical independence, which is enough in combination with symmetry, but not is practical unsound, the expert might well use partially the same information.
Bayesian approach, non correlated predictions, plus symmetry, as above.
A whole bunch of formulas which satisfy a minimal addition theorem and generate plausible results in certain situations.

This page is merely a footnote to the cited one, we describe one more approach which yields a solution.

Alternative approach

Consider the 8-dimensional probability range from above, truncated down to 5 dimensions by the 3 equations describing the probabilities properties to sum up to one and the a prioriy information. So, how about assuming that all cases in that 5 dimensional volumina are equally likely and take the expected value of the quotient

p₁/(p₁+p₅) as the solution of the problem?

Calculating the expected value with the equality assumptions breaks down to evaluating two integrals over 5 dimensional simplices, which is tedious, both to denote in HTML and to calculate. But the result

P(a,b) = Expected-value_{under-equidistribution-assumption} (p₁/(p₁+p₅)) has the properties:

P(a,b) + P(b,a) = 1, which says when both experts change their opinion simultaneously, the probabilty gets replaces by its inverse probability.
This especially means, that P(a,a) = 1/2 holds.
The expected monotony properties i.e. P(a₁,b) < P(a₂,b) for a₁ > a₂.

Unfortunately we haven't done the math for the 5 dimensional integrals yet (appart from the symmetry considerations), so we currently don't know a closed form of the solutions. But the caclulation can be done by a numerical approach. Actually we employed the simplest Monte Carlo Method known to obtain below values. The solution for the initially mentioned problem is:

P(0.75, 0.60) =0.273

Some more numerical evidences are (compare for the initially cited article for the detailed examples):

a	b	P(a,b)	comment	Heuristic Bayes
0.75	0.60	0.273	above example	0.333
0.60	0.75	0.727	changed opinions	0.667
0.50	0.75	0.778	a non expert	0.750
0.25	0.75	0.868	an anti expert	0.900
0.75	0.75	0.500	opposing experts	0.500
0.97	0.99	0.808	extreme numbers	0.753

All values obtaines numerically and approximations. Note that strangely P(0.50, b) does not equal b in the numerical simulations! It's fairly unclear, that the addition of a pure luck expert strengthens the expertise of the other expert!

Conclusion

So here we are and have one more solution. On the other hand we believe, that our approach is slightly more satisfying from science and practical point:

We make only one mathematical assumption.
The equidistribution assumption is kind of minimal and natural, assume that everything can happen with the same probabilty and average over all those cases.
The approach is tractable, i.e. yields numbers as results, though we didn't derived the closed solution here.
The solutions have plausible symmetry properties.
The demonstrated numbers look decent.

Further directions

The number of outcomes
The number of "experts", i.e. predictions available

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